今天看到一个小例子,发现了一个小trick。见代码:

#include <cstdio>
#include <cstdlib>

class base_1
{
public:
    int a;
};

class base_2
{
public:
    int b;
};

class base_3: public base_1, public base_2
{
public:
    int c;
};

int main(int argc, char* argv[]) {

    printf("&base_1::a = %p\n", &base_1::a);
    printf("&base_2::b = %p\n", &base_2::b);
    printf("&base_3::a = %p\n", &base_3::a);
    printf("&base_3::b = %p\n", &base_3::b);
    printf("&base_3::c = %p\n", &base_3::c);


    base_3 t;
    t.a = 1;
    t.b = 2;
    t.c = 3;

    typedef int (base_3::*tip);
    tip pm = NULL;

    printf("base_3::a = %d\n", t.base_3::a);
    printf("base_3::b = %d\n", t.base_3::b);
    printf("base_3::c = %d\n", t.base_3::c);

    pm = &base_3::a;
    printf("base_3::a(%p) = %d(ptr)\n", pm, t.*pm);
    pm = &base_3::b;
    printf("base_3::b(%p) = %d(ptr)\n", pm, t.*pm);
    pm = &base_3::c;
    printf("base_3::c(%p) = %d(ptr)\n", pm, t.*pm);

    return 0;
}

猜猜看这个代码输出什么?
答案是:

&base_1::a = 00000000
&base_2::b = 00000000
&base_3::a = 00000000
<u>&base_3::b = 00000000</u>
&base_3::c = 00000008
base_3::a = 1
base_3::b = 2
base_3::c = 3
base_3::a(00000000) = 1(ptr)
base_3::b(00000004) = 2(ptr)
base_3::c(00000008) = 3(ptr)

注意带下划线&base_3::b那一行,木有错,这个值不是4,而是0。我看了一下生成的汇编,很遗憾地他直接 push 0进栈了。
而下面赋值的pm = &base_3::b那里,汇编里直接 mov 了 4。
而且GCC和VC都是这样,所以不知道这么做的原因。

整理后记

原因:和编译优化有关。