关于sizeof，下面程序谁能解释一下为什么？

zhsp1029

#include <iostream>
#include <string>
using namespace std;
int main()
{
        int iValue = 5;
        unsigned char uchValue = 255;
        int iLen = sizeof((iValue = uchValue + iValue));
        cout << iValue << endl;
        return 0;
}

上面这段程序，输出的iValue的值你认为是什么呢？但是我实际输出的值是5，谁能解释一下为什么，谢谢了.

tyc611

ISO14882, 5.3.3 sizeof
The sizeof operator yields the number of bytes in the object representation of its operand. The operand
is either an expression, which is not evaluated, or a parenthesized type-id. The sizeof operator shall not
be applied to an expression that has function or incomplete type, or to an enumeration type before all its
enumerators have been declared, or to the parenthesized name of such types, or to an lvalue that designates
a bit-field.

albeta

原帖由 tyc611 于 2009-11-8 20:50 发表
ISO14882, 5.3.3 sizeof
The sizeof operator yields the number of bytes in the object representation of its operand. The operand
is either an expression, which is not evaluated, or a parenthesize ...

楼上强

vime000

2楼的回答让我想起低俗小说中那个黑人大声背诵圣经的片段~

虑而后能得

这个我觉得是作用域的问题，sizeof是库函数 不是操作符
调用sizeof((iValue = uchValue + iValue));和使用fun((iValue = uchValue + iValue));属于传值调用，对主函数中的ivalue没有影响

churchmice

2楼正解
这问题以前讨论过，有印象

liu1061

为什么要这样用了，有点想不明白！

dhb118

个人理解 sizeof() 为编译期的操作符 编译器会把sizeof()里的表达式转换成一个常量 所以他对像这种iValue = uchValue + iValue 表达式只是取左值作为参考 等效于 sizeof(iValue) .

本人菜鸟,弱有错误请大侠指正.

cskyrain

c 中sizeof() 为操作符