C++基础FAQ [1]

最新推荐文章于 2025-12-12 16:37:00 发布

原创最新推荐文章于 2025-12-12 16:37:00 发布 · 355 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#c++ #pointers

类的函数为什么在sizeof中反映不出来。

#include <iostream>


class Base
{
public:
	void f();
	int g();
private:
	int i;
};


int main()
{
	Base b;
	int i = sizeof(b);

	return 0;
}

如上例，sizeof的值为4。

但是这并不代表不占空间，因为类的函数代码，放在代码段(Code Segment)。

摘自：http://topic.youkuaiyun.com/u/20120501/23/80fa77e8-dd63-4903-a405-a142271989b9.html?seed=789173484&r=78429170

The sizeof operator cannot be used with the following operands:

Functions. (However, sizeof can be applied to pointers to functions.)
Bit fields.
Undefined classes.
The type void.
Dynamically allocated arrays.
External arrays.
Incomplete types.
Parenthesized names of incomplete types.