关于错误的困惑:要求左值为一元'&'操作数

问题描述

我一直在尝试通过编写简单的代码来理解指针的概念，我遇到了一个错误问题，似乎我无法解决或理解它.

I have been trying to understand pointer concepts by writing simple code,and I got an error problem, and it seems like I couldn't solve it or understand it.

#include <stdio.h>

int *foo(void);

int main(void) {
    printf("%d\n", *foo());

    return 0;
}

int *foo(void) {
    static int num = 1;
    ++num;
    return &(++num);

}

这是错误消息.

error: lvalue required as unary ‘&’ operand
  return &(++num);

函数'foo()'返回一个指向int的指针，并且main应该是使用*运算符打印返回的int.对于foo()中的静态num，我认为通过放置静态限定符，num不再是临时变量，因此&"可用于num.

Function 'foo()' returns a pointer to int, and main is supposed to beprint the returned int by using * operator. For static num within foo(), I thought that by putting static qualifier, num is not temporary variable anymore, so '&' can be used to num.

推荐答案

相对于前缀增量运算符++，C和C ++之间存在区别.

There is a difference between C and C++ relative to the prefix increment operator ++.

在C中，结果是递增后的操作数的新值.因此，在此表达式&(++num)中，有一种尝试来获取临时对象的地址(右值).

In C the result is the new value of the operand after incrementation. So in this expression &(++num) there is an atttempt to get the address of a temporary object (rvalue).

在C ++中，程序将是正确的，因为在C ++中，结果是更新的操作数.它是左值.

In C++ the program will be correct because in C++ the result is the updated operand; it is an lvalue.

在C中，结果是一个新值，而在C ++中，结果是更新的操作数.

That is in C the result is a new value while in C++ the result is the updated operand.

因此在C语言中，您可能不会例如写

So in C you may not for example write

++++++x;

在C ++中，此表达式

while in C++ this expression

++++++x;

是正确的，您可以将一元运算符&应用于类似这样的表达式

is correct and you may apply the unary operator & to the expression like

&++++++x;

要使函数在C语言中正确无误，您必须将应用的运算符分开，例如

To make the function correct in C you have to separate the applied operators like

int *foo(void) {
    static int num = 1;
    ++num;
    //return &(++num);
    ++num;
    return #

}

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