为什么 sizeof(x++) 不增加 x? [英] Why does sizeof(x++) not increment x?
问题描述
这是在dev c++ windows中编译的代码:
Here is the code compiled in dev c++ windows:
#include <stdio.h>
int main() {
int x = 5;
printf("%d and ", sizeof(x++)); // note 1
printf("%d
", x); // note 2
return 0;
}
我希望在执行 note 1 后 x
为 6.但是,输出是:
I expect x
to be 6 after executing note 1. However, the output is:
4 and 5
谁能解释为什么 x
在 note 1 之后没有增加?
Can anyone explain why x
does not increment after note 1?
推荐答案
来自 C99 标准(重点是我的)
6.5.3.4/2
sizeof 运算符产生其操作数的大小(以字节为单位),它可以是表达式或括号内的类型名称.大小由操作数的类型决定.结果是一个整数.如果操作数的类型是变长数组类型,则对操作数求值;否则,不计算操作数,结果是一个整数常量.
The sizeof operator yields the size (in bytes) of its operand, which may be an expression or the parenthesized name of a type. The size is determined from the type of the operand. The result is an integer. If the type of the operand is a variable length array type, the operand is evaluated; otherwise, the operand is not evaluated and the result is an integer constant.
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