Cc++ — Expression must be a modifiable lvalue (CC++)


Содержание

Выражение должно быть модифицируемым значением lvalue

У меня есть следующий код:

Но он выдает сообщение об ошибке:

Ошибка: выражение должно быть изменяемым значением

на этой строке «if». Я не пытаюсь изменить значение «match» или «k» здесь, но почему эта ошибка? если я пишу только как:

это нормально. Может ли кто-нибудь объяснить это мне?

Оператор присваивания имеет более низкий приоритет, чем && , поэтому ваше условие эквивалентно:

Но в левой части это значение r, а именно логическое значение, полученное в результате оценки выражения sub & shy; match == 0 && k , поэтому вы не можете назначить ему.

Напротив, сравнение имеет более высокий приоритет, поэтому match == 0 && k == m эквивалентно:

В C вы также столкнетесь с той же ошибкой, если объявите:

и попробуйте назначить значение без указания позиции индекса:

Вы указываете ранее объявленный/модифицируемый адрес.

Помните, что один = всегда является назначением на C или С++.

Ваш тест должен быть if ( match == 0 && k == M ) , вы сделали опечатку в тесте k == M .

Если вы действительно имеете в виду k=M (т.е. связанное с назначением присваивание внутри теста), вы должны прочитать код причины if (match == 0 && (k=m) != 0) , но большинство правил кодирования советуют не писать это.

Кстати, ваша ошибка предполагает запросить все предупреждения (например, -Wall опция g++ ) и обновить до последних компиляторов. Следующий GCC 4.8 даст вам:

и Clang 3.1 также сообщает вам ederman.cc:9:30: error: expression is not assignable

Поэтому используйте последние версии бесплатных компиляторов и включите все предупреждения при их использовании.

C++ — Error: Expression must be a modifiable lvalue

I’m creating a C++ application and I have a class which contains a char array variable, like this:

And I try to set this variable in a constructor like this:

but then I get an error ( Error: expression must be a modifiable lvalue ) on the line where I set the MyClass::myVariable variable on. I’m new to C++, so I have no idea about what this means. Any help would be appreciated!

Best How To :

You can not assign arrays using assignation operator ( = )

Otherwise change the member and argument to pointer or stl type as:

Explicit instantiation of > c++,templates,constructor,explicit-instantiation

When the constructor is a template member function, they are not instantiated unless explicitly used. You would see the code for the constructor if you make it a non-template member function. template class test < public: /*** template test(T param) < parameter = param; >; ***/ test(T param).

Implicit use of initializer_list

Your program is not ill-formed because is guaranteed to include (the same is true for all standard library containers) §23.3.1 [sequences.general] Header synopsis #include . Searching the standard for #include reveals the header is included along with the following headers .

How can I tell clang-format to follow this convention?

Removing BreakBeforeBraces: Allman Seems to do what you want (for me). I’m using SVN clang though. Although you probably wanted it there for a reason. According to the clang-format docs, the AllowShortBlocksOnASingleLine should do exactly what you want (regardless of brace style). This might be a bug in clang-format.

Make a triangle shape in C++

This code works fine for a right angled triangle — * ** *** But I guess you want a triangle like this — * *** ***** Try this — #include using namespace std; int main() < int i, j, k, n; cout c++,file,hdf5,dataformat

Speaking as someone who’s had to do exactly what you’re talking about a number of time, rr got it basically right, but I would change the emphasis a little. For file versioning, text is basically the winner. Since you’re using an hdf5 library, I assume both serializing and parsing are.

How can I access the members of a sub > c++,inheritance,constructor,subclass,superclass

This map: typedef map obj_map; only stores Object objects. When you try to put an Image in, it is sliced down and you lose everything in the Image that was not actually part of Object. The behaviour that you seem to be looking for is called polymorphism. To activate.

Type function that returns a tuple of chosen types

You can do this without recursion by simply expanding the parameter pack directly into a std::tuple: template struct Tuple < using type = std::tuple ::type. >; >; To answer your question more directly, you can declare a variadic primary template, then write two specializations: for when there are at least.

Having two arrays in variable php

The explode function is being used correctly, so your problem is further up. Either $data[$i] = mysql_result($result,$i,»data»); isn’t returning the expected string «2015-06-04» from the database OR your function $data[$i] = data_eng_to_it_($data[$i]); isn’t returning the expected string «04 June 2015» So test further up by echo / var_dump after both.

C++11 Allocation Requirement on Strings

Section 21.4.1.5 of the 2011 standard states: The char-like objects in a basic_string object shall be stored contiguously. That is, for any basic_string object s, the >

No match for ‘operator*’ error

As @101010 hints at: pay is a string, while hours_day is a float, and while some languages allow you to multiply strings with integers, c++11 (or any other flavor of c) doesn’t, much less allow strings and floats to be multiplied together.

C++ Isn’t this a useless inline declaration?

The Compiler can Access everything. The restrictions are only valid for the programmer. This means there are no restrictions for the Compiler to Access any variables! At the end every variable is just translated to an address which can be accessed. So for the Compiler it is no Problem to.

Incorrect Polar — Cartesian Coordinate Conversions. What does -0 Mean?

You are converting to cartesian the points which are in cartesian already. What you want is: std::cout . Using the key_char string, the values of the chars in it will serve as the initial value of the ints. std::vector key_num(key_char.begin(), key_char.end()); Then, iterate over each character of key_num and convert it to the equivalent int value for.

Issue when use two type-cast operators in template > c++

What you’re trying to do makes little sense. We have subclass . It is convertible to int&, but also to a lot of other reference types. char&. bool&. double&. The ambiguity arises from the fact that all the various overloads for operator #include #include template std::string stringulate(ValueType v) < std::ostringstream oss; oss c++,templates,template-specialization


The specialization still needs to be a template template argument. You passed in a full type. You want: template class random_gen < . >; Just std::uniform_real_distribution, not std::uniform_distribution . .

Marshal struct in struct from c# to c++

Change this: [MarshalAs(UnmanagedType.ByValTStr, SizeConst = 36)] private string iu; to this: [MarshalAs(UnmanagedType.LPStr)] private string iu; Note that this code is good only to pass a string in the C#->C++ direction. For the opposite direction (C++->C#) it is more complex, because C# can’t easily deallocate C++ allocated memory. Other important thing.

Checking value of deleted object

It is very bad, accessing deleted objects as if they were not deleted will in the general case crash. There is no guarantee that the memory is still mapped inside the process and it could result in a virtual memory page fault. It is also likely that the memory will.

undefined reference to `vtable for implementation’ error

I think you just misspelled CFLAGS in CFLAGES=-c -Wall I’m guessing this is the case since g++ ../src/main.cpp -I ../include/ does not have the -c option.

Delphi — Use a string variable’s name in assignfile()

Is it possible to use a variable in the AssignFile command? Yes. The second parameter of AssignFile has type string. The expression cFileDir + ‘\’ + sFile has type string. FWIW, AssignFile is known as a function rather than a command. Getting on top of terminology like this will.

Where are the local variables of an Andro > android,variables,local

From https://source.android.com/devices/tech/dalvik/dalvik-bytecode.html: Because, in practice, it is uncommon for a method to need more than 16 registers, and because needing more than eight registers is reasonably common, many instructions are limited to only addressing the first 16 registers. When reasonably possible, instructions allow references to up to the first 256.

Same function with and without template

The main reason to do something like this is to specialize void integerA(int x) to do something else. That is, if the programmer provides as input argument an int to member function abc::integerA then because of the C++ rules instead of instantiating the template member function the compiler would pick.

Method returning std::vector >

Your error is actually coming from: array.push_back(day); This tries to put a copy of day in the vector, which is not permitted since it is unique. Instead you could write array.push_back( std::move(day) ); however the following would be better, replacing auto day. array.emplace_back(); .

SQL Repeated Condition in Two Tables

you can achieve it like this DECLARE @Account >

Значения Lvalue и Rvalue (C++) Lvalues and Rvalues (C++)

Каждое выражение C++, с типом, а также входит в категории значений. Every C++ expression has a type, and belongs to a value category. Значение категории являются основой для правил, компиляторы должны следовать при создании, копирование и перемещение временные объекты во время вычисления выражения. The value categories are the basis for rules that compilers must follow when creating, copying, and moving temporary objects during expression evaluation.

Стандарт C ++ 17 определяет категории значение выражения следующим образом: The C++17 standard defines expression value categories as follows:

  • Объект glvalue представляет собой выражение, вычисление которого определяет удостоверение объекта, битового поля или функции. A glvalue is an expression whose evaluation determines the identity of an object, bit-field, or function.
  • Объект prvalue представляет собой выражение, вычисление которого инициализирует объект или битовому полю или вычисляет значение операнда оператора, как указано в контексте, в котором он отображается. A prvalue is an expression whose evaluation initializes an object or a bit-field, or computes the value of the operand of an operator, as specified by the context in which it appears.
  • Xvalue является glvalue, который обозначает объект или битового поля, ресурсы которых может многократно использоваться (обычно, поскольку он находится в конце его времени существования). An xvalue is a glvalue that denotes an object or bit-field whose resources can be reused (usually because it is near the end of its lifetime). Пример Некоторые типы выражений, включающих ссылки rvalue (8.3.2) дают xvalues, такие как вызов функции, возвращаемый тип которых является ссылкой rvalue или преобразование в ссылочный тип rvalue. Example: Certain kinds of expressions involving rvalue references (8.3.2) yield xvalues, such as a call to a function whose return type is an rvalue reference or a cast to an rvalue reference type.
  • Lvalue является glvalue, который не является xvalue. An lvalue is a glvalue that is not an xvalue.
  • Rvalue prvalue или xvalue. An rvalue is a prvalue or an xvalue.
Цукерберг рекомендует:  Обучение - Как проходит обучение

Следующей схеме показаны связи между категориями: The following diagram illustrates the relationships between the categories:

Lvalue имеет адрес, который можно получить доступ к программе. An lvalue has an address that your program can access. Примеры выражений lvalue включают имена переменных, включая const переменных, элементов массива, функция вызовы, которые возвращают ссылки lvalue, битовые поля, объединения и членов класса. Examples of lvalue expressions include variable names, including const variables, array elements, function calls that return an lvalue reference, bit-fields, unions, and class members.

Выражение prvalue не имеет адреса, доступном в программе. A prvalue expression has no address that is accessible by your program. Примеры выражений prvalue включают литералы, вызовы функций, возвращающих тип без ссылок и временные объекты, созданные во время вычислений выражения, но доступен только компилятором. Examples of prvalue expressions include literals, function calls that return a non-reference type, and temporary objects that are created during expression evalution but accessible only by the compiler.

Выражение xvalue имеет адрес, станут недоступными в программе, но может использоваться для инициализации ссылкой rvalue, который предоставляет доступ к выражению. An xvalue expression has an address that no longer accessible by your program but can be used to initialize an rvalue reference, which provides access to the expression. Примеры вызовов функций, которые возвращают ссылку rvalue и индекс массива, член и указателя выражения элементов, где массива или объекта является ссылкой rvalue. Examples include function calls that return an rvalue reference, and the array subscript, member and pointer to member expressions where the array or object is an rvalue reference.

Пример Example

В следующем примере показано несколько правильных и неправильных способов использования значений lvalue и rvalues. The following example demonstrates several correct and incorrect usages of lvalues and rvalues:

В примерах этого раздела показано правильное и неправильное использование при неперегруженных операторах. The examples in this topic illustrate correct and incorrect usage when operators are not overloaded. Перегрузив операторы, можно преобразовать выражение, такое как j * 4 , в значение lvalue. By overloading operators, you can make an expression such as j * 4 an lvalue.

Условия lvalue и rvalue часто используются при ссылке на ссылки на объекты. The terms lvalue and rvalue are often used when you refer to object references. Дополнительные сведения о ссылках см. в разделе декларатор ссылки Lvalue: & и декларатор ссылки Rvalue: & &. For more information about references, see Lvalue Reference Declarator: & and Rvalue Reference Declarator: &&.

Expression must be a modifiable lvalue

Technical Note 82120

Introduction

You may encounter the following message if you move code to another version of the IAR C/C++ Compiler:

This message occurs because a cast does not produce an lvalue (that is, a value that can be used on the left side of an assignment).

Also note that a casted expression that is used with the * operator does produce an lvalue, which is why the following is OK:

How to get rid of the error message

There are a couple of ways to rework the code to avoid this error message.

The best way is probably to rework the types so that the cast do not appear in the first place, if possible.

An alternative is to use a temporary variable of the desired type.

The following example fails:

You can rewrite it to:

Technical details

The reason why it does not work is that a cast does not produce an lvalue (ISO/ANSI 9899-1990 6.3.4 and Annex B which describes the syntax of C).

From ISO/ANSI 9899-1990 6.3.4, cast operators, footnote 44: «A cast does not yield an lvalue. [. ]».

An extract from Annex B follows below:

The assignment requires a unary-expression on the left side. A cast-expression is not part of a unary-expression. As mentioned above, you can use a unary-operator before the cast-expression to get a unary-expression, such as ‘*’.

Many compilers permit cast expressions on the left side of an assignment, including several produced by IAR Systems. We have since then switched to another C parser that is more restrictive on this, and besides, it is not proper ISO/ANSI C code.

All product names are trademarks or registered trademarks of their respective owners.


C++ — Expression must be modifiable lvalue [closed]

I am new to C++. I am trying to set up a very basic TCG game (with only minion with no text).

I think my issue might be linked with pointer.

Below are the functions and attributes:

Board inherits from Card_collection

And I have as a compiler error:

Please, where do you think is the error?

closed as off-topic by Bill Greene, spektr, GoHokies, Christian Clason, Anton Menshov ♦ May 12 ’17 at 15:52

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • «Questions about general programming topics (e.g., programming syntax, «how do I do X in language Y», «debug my code») are off-topic here, but can be asked on Stack Overflow. See Custom off-topic close reason for programming question.» – Bill Greene, spektr, GoHokies, Christian Clason, Anton Menshov

If this question can be reworded to fit the rules in the help center, please edit the question.

1 Answer 1

To this isn’t the right place to ask this question, so for future reference ask these types of questions on Stack Overflow.

You don’t really give enough code to be totally sure what’s wrong, but one thing that comes to mind given the error is the following. First, note that your get(int i) method is returning a copy of a Card , not the actual ith Card from the Card Collection. This means when you do:

You are actually setting the m_has_been_played_this_turn variable for a temporary Card instance, which may be where the error is coming from since this temporary Card instance should be an rvalue , not a lvalue .

One way to change this might be to change your code to the following:

But this is still not going to accomplish what I believe you want, since card_ is just a copy of the ith Card in your Card Collection. If you really want to modify the ith Card in your Card Collection, change the get(int i) definition to the following:

That & means you will pass a Card out of get(int i) by reference, allowing you to modify the Card within the Card Collection directly. Then the original code you wrote to set m_has_been_played_this_turn should be fine.

Выражение должно быть модифицируемым значением lvalue

У меня есть следующий код:

Но он выдает сообщение об ошибке:

Ошибка: выражение должно быть изменяемым значением

на этой строке «if». Я не пытаюсь изменить значение «match» или «k» здесь, но почему эта ошибка? если я пишу только как:

это нормально. Может ли кто-нибудь объяснить это мне?

Оператор присваивания имеет более низкий приоритет, чем && , поэтому ваше условие эквивалентно:

Но в левой части это значение r, а именно логическое значение, полученное в результате оценки выражения sub & shy; match == 0 && k , поэтому вы не можете назначить ему.

Напротив, сравнение имеет более высокий приоритет, поэтому match == 0 && k == m эквивалентно:

В C вы также столкнетесь с той же ошибкой, если объявите:

и попробуйте назначить значение без указания позиции индекса:

Вы указываете ранее объявленный/модифицируемый адрес.

Помните, что один = всегда является назначением на C или С++.

Ваш тест должен быть if ( match == 0 && k == M ) , вы сделали опечатку в тесте k == M .

Если вы действительно имеете в виду k=M (т.е. связанное с назначением присваивание внутри теста), вы должны прочитать код причины if (match == 0 && (k=m) != 0) , но большинство правил кодирования советуют не писать это.

Кстати, ваша ошибка предполагает запросить все предупреждения (например, -Wall опция g++ ) и обновить до последних компиляторов. Следующий GCC 4.8 даст вам:

и Clang 3.1 также сообщает вам ederman.cc:9:30: error: expression is not assignable

Поэтому используйте последние версии бесплатных компиляторов и включите все предупреждения при их использовании.

C/c++ — Expression must be a modifiable lvalue (C/C++)

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Здравствуйте:
При компиляции возникает ошибка:
Error: C:\CDAVR\KRR.c(73): the expression must be a modifiable lvalue

Хотя буквально год назад компилировал вроде как этот же код, и все проходило гладко. А сейчас достал его, и не могу понять где ошибка.

Код
#include
#include
#include
#include
#include

#define ADC_VREF_TYPE 0x00
#define V 5

// Инициализация LCD
#asm
.equ __lcd_port=0x18 ;PORTB
#endasm

//Функция для работы с АЦП
unsigned int read_adc(unsigned char adc_input)
<
ADMUX=adc_input | (ADC_VREF_TYPE & 0xff);
delay_us(10);
ADCSRA|=0x40;
while ((ADCSRA & 0x10)==0);
ADCSRA|=0x10;
return ADCW;
>

void main(void)
<
char acpread[5];
char acptry[5];
int j;

PORTB=0x00; //скорее всего нужно настроить на вывод
DDRB=0x00;

TCCR0=0x00; //Регистр управления таймером
TCNT0=0x00; //Счетный регистр
TCCR1A=0x00;
TCCR1B=0x00;
TCNT1H=0x00;
TCNT1L=0x00;
ICR1H=0x00;
ICR1L=0x00;
OCR1AH=0x00; //Регистр сравнения
OCR1AL=0x00;
OCR1BH=0x00;
OCR1BL=0x00;
ASSR=0x00; //Регистр таймера в ассинхроном режиме
TCCR2=0x00;
TCNT2=0x00;
OCR2=0x00;
MCUCR=0x00;
TIMSK=0x00; //Разрешение и запрет прерываний по таймеру
UCSRA=0x00; //Регистры управления и состояни
UCSRB=0x18; //Регистры управления и состояни
UCSRC=0x86; //Регистры управления и состояни
UBRRH=0x00; //Регистры ввода-вывода
UBRRL=0x4D;
ACSR=0x80;
SFIOR=0x00;
ADMUX=ADC_VREF_TYPE & 0xff; //Регистр управления мультиплексером
ADCSRA=0x85; //Регистр управления АЦП

lcd_init(16); // Инициализация ЖК
while (1)
<
itoa(read_adc(1), acpread);
acptry = acpread*V/1024;
j = 0; //цикл для вывода значения с фотодатчика
lcd_gotoxy(0,0);
while(j


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Mazzi
Дата 16.10.2011, 19:52 (ссылка) | (нет голосов) Загрузка .
Код
char acpread[5];
char acptry[5];

а использование не соответствует декларации

Код
acptry = acpread*V/1024;
Google
Дата 13.11.2020, 23:35 (ссылка)
Правила форума «Микроконтроллеры (MCU) и микропроцессоры (MPU)»

You should use the random header. #include std::default_random_engine generator; std::uniform_int_distribution dist(0, 5); int StringIndex = dist(generator); std::string ChosenString = characters[StringIndex]; The above will generate a random index into your array. If you want to limit the range, change the constructor of dist, for example (dist(0,2) would only allow for.

Val > c++,user-input

The simplest thing you can do is to use a for/while loop. A loop will basically repeat the same instruction for a number of n steps or until a certain condition is matched. The solution provided is pretty dummy, if you want to read the first name and last name.

Compiler modifying a variable without adressing it

What you are currently doing: Dict2 = Dict1; Is reference copying, so both Dict1 and Dict2 are pointing to the same location. You can create a new copy like: Dict2 = Dict1.ToDictionary(d=> d.Key, d=> d.Value); Remember, if your Key and Value are custom objects (based on some class), in that.

3 X 3 magic square recursively

Basically, you are finding all permutations of the array using a recursive permutation algorithm. There are 4 things you need to change: First, start your loop from pos, not 0 Second, swap elements back after recursing (backtracking) Third, only test once you have generated each complete permutation (when pos =.

Issue when use two type-cast operators in template > c++

What you’re trying to do makes little sense. We have subclass . It is convertible to int&, but also to a lot of other reference types. char&. bool&. double&. The ambiguity arises from the fact that all the various overloads for operator python,c++

For your andro ,stdout=subprocess.PIPE,shell=True) (output,err)=process.communicate() exit_code = process.wait() print output print err print exit_code return exit_code.

Get an ordered list of files in a folder

The fanciest way I’ve seen to perform what you want is straight from the boost filesystem tutorial. In this particular example, the author appends the filename/directory to the vector and then utilizes a std::sort to ensure the data is in alphabetical order. Your code can easily be updated to use.

Strings vs binary for storing variables ins > c++,file,hdf5,dataformat

Speaking as someone who’s had to do exactly what you’re talking about a number of time, rr got it basically right, but I would change the emphasis a little. For file versioning, text is basically the winner. Since you’re using an hdf5 library, I assume both serializing and parsing are.

Method returning std::vector >

Your error is actually coming from: array.push_back(day); This tries to put a copy of day in the vector, which is not permitted since it is unique. Instead you could write array.push_back( std::move(day) ); however the following would be better, replacing auto day. array.emplace_back(); .

Translating a character array into a integer string in C++

If you want a sequence of int, then use a vector . Using the key_char string, the values of the chars in it will serve as the initial value of the ints. std::vector key_num(key_char.begin(), key_char.end()); Then, iterate over each character of key_num and convert it to the equivalent int value for.

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This depends on what you want the behaviour (protocol) of your class to be. Since you’re logging into the error stream there, I assume you consider this an error condition to call pop() on an empty stack. The standard C++ way of signalling errors is to throw an exception. Something.

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Change this: [MarshalAs(UnmanagedType.ByValTStr, SizeConst = 36)] private string iu; to this: [MarshalAs(UnmanagedType.LPStr)] private string iu; Note that this code is good only to pass a string in the C#->C++ direction. For the opposite direction (C++->C#) it is more complex, because C# can’t easily deallocate C++ allocated memory. Other important thing.

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you can achieve it like this DECLARE @Account >

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This code works fine for a right angled triangle — * ** *** But I guess you want a triangle like this — * *** ***** Try this — #include using namespace std; int main() < int i, j, k, n; cout c++,templates,constructor,explicit-instantiation

When the constructor is a template member function, they are not instantiated unless explicitly used. You would see the code for the constructor if you make it a non-template member function. template class test < public: /*** template test(T param) < parameter = param; >; ***/ test(T param).

pointer to pointer dynamic array in C++

Implicit use of initializer_list

Your program is not ill-formed because is guaranteed to include (the same is true for all standard library containers) §23.3.1 [sequences.general] Header synopsis #include . Searching the standard for #include reveals the header is included along with the following headers .

create vector of objects on the stack ? (c++)

Yes, those objects still exist and you must delete them. Alternatively you could use std::vector > instead, so that your objects are deleted automatically. Or you could just not use dynamic allocation as it is more expensive and error-prone. Also note that you are misusing reserve. You either want to use.

Checking value of deleted object

It is very bad, accessing deleted objects as if they were not deleted will in the general case crash. There is no guarantee that the memory is still mapped inside the process and it could result in a virtual memory page fault. It is also likely that the memory will.


Where are the local variables of an Andro > android,variables,local

From https://source.android.com/devices/tech/dalvik/dalvik-bytecode.html: Because, in practice, it is uncommon for a method to need more than 16 registers, and because needing more than eight registers is reasonably common, many instructions are limited to only addressing the first 16 registers. When reasonably possible, instructions allow references to up to the first 256.

dispatch response packet according to packet sequence > c++,boost,boost-asio

You could use std::promise and std::future (or their boost counterparts if your are not yet on C++11). The idea is to store a std::shared_ptr > with the current sequence id as a key in the map whenever a request is sent. In the blocking send function you wait for the corresponding.

C/c++ — Expression must be a modifiable lvalue (C/C++)

I have this following code:

But it gives out an error saying:

Error: expression must be a modifiable value

on that «if» line. I am not trying to modify «match» or «k» value here, but why this error? if I only write it like:

it is ok. Could someone explain it to me?

c++ if-statement
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this question edited Nov 20 ’15 at 20:39 Brian Tompsett — 汤莱恩 3,787 13 32 82 asked Oct 5 ’12 at 11:45 E_learner 1,036 11 35 73 7 Did you mean k == M ? – Kerrek SB Oct 5 ’12 at 11:47 1 K == M not K = M – Shashwat Oct 5 ’12 at 11:48 1 You’re lucky that it resulted in a compiler error due to the operator precedences of && and = , instead of silently producing an incorrect program! – Rufflewind Aug 24 ’14 at 1:47 |

4 Answers
4

The assignment operator has lower precedence than && , so your condition is equivalent to:

But the left-hand side of this is an rvalue, namely the boolean resulting from the evaluation of the sub­expression match == 0 && k , so you cannot assign to it.

By contrast, comparison has higher precedence, so match == 0 && k == m is equivalent to:

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this answer answered Oct 5 ’12 at 11:49 Kerrek SB 321k 53 599 838 |

In C, you will also experience the same error if you declare a:

and than try to assign a value without specifying an index position:

e* input) < ((ptrtype*)Globalptr) = input;>I get Error on line ((ptrtype*)Globalptr) = input; says » expression must be a modifiable lvalue» c pointers casting void-pointers
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this question asked Oct 20 ’14 at 11:53 Islam

You’re specifying the accessible/modifiable address previously declared.

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this answer answered Apr 17 ’15 at 14:20 Alan 438 2 8 21 |

Remember that a single = is always an assignment in C or C++.

Your test should be if ( match == 0 && k == M ) you made a typo on the k == M test.

If you really mean k=M (i.e. a side-effecting assignment inside a test) you should for readability reasons code if (match == 0 && (k=m) != 0) but most coding rules advise not writing that.

BTW, your mistake suggests to ask for all warnings (e.g. -Wall option to g++ ), and to upgrade to recent compilers. The next GCC 4.8 will give you:

and Clang 3.1 also tells you ederman.cc:9:30: error: expression is not assignable

So use recent versions of free compilers and enable all the warnings when using them.

You test k = M instead of k == M .
Maybe it is what you want to do, in this case, write if (match == 0 && (k = M))

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this answer answered Oct 5 ’12 at 11:48 tomahh 7,973 2 18 47 yeah, I am really a numb, I forgot that one. Thank you all. – E_learner Oct 5 ’12 at 11:50 |

float centerX; float centerY; int flagMountain;>;eachElement cn[6];struct characters < eachElement each[6];>;characters chars[1500];float strtPt = 235.0;float initializer = strtPt;float endPt = 120.0;float h

C++: Expression must be a modifiable lvalue

I am making a text based simple operation/trig calculator. I am getting the error: «Expression must be a modifiable lvalue» when I attempt to sqrt the squared variables. Here is my current code: (the error highlights the sqrt)

Apologies for any formatting errors, first question asked here any help is appreciated. Thank you!

4 answers

  • answered 2020-08-10 20:42 Cameron637

You’ve switched around your assignment. You’re trying to assign the sqrt function the value of result; you want to assign the result to the value of the sqrt output:

answered 2020-08-10 20:43 Arnav Borborah

Modify your cin statement to cin >> a >> b; , this reads multiple values at once, since the comma operator doesn’t exist for cin . Also, switch the order of the result variable on the right side of the assignment to left side. The two operands to the = on the assignment are l-values, standing for left-values, supposed to be on the left side, and r-values, standing for right-values, supposed to be on the right side. Switch them, and you get an error.

Your new Modified Code

answered 2020-08-10 20:43 ForceBru

First, cin >> a, b does not do what you think it’s doing. You’d better write cin >> a >> b .

Then, you assign data to variables, not variables to data. So do result = sqrt(. ) .

answered 2020-08-10 20:43 Shravan40

The main problem is the line number 4 and 5 of your piece of code.


cin >> a,b; is wrong. You can take multiple input in a line like this cin >> a >> b;

And in the line number 5 you are doing assignment in wrong order. You should just reverse it result this way result = sqrt(pow(a, 2) + pow(b, 2)) ;

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Question

What is the computational cost of the remainder function, is there a specific instruction to calculate it in a cheap way in a specific case?

Description

I need to transform a mathematical variable x into the range of I=[-0.5; 0.5) from R=[-2; 2). While x is not element of I, then x is shifted towards I by repeatedly adding or subtracting 1 to the value of x. x is represented with double x in my code. I need the fastest way of this transformation for this I and R values but wider R ranges can be also interesting.

Ideas and speed comparison

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