Question

我正在学习C ++的参考资料，并尝试使用Thinking in C ++中的以下代码：

然而，我发现如果我没有将引用转换为'long'类型， f 和 g 的引用是相同的，我认为没有意义，它们的值都是 1 而不是十六进制显示的数字，有人可以解释一下吗？

感谢。

include <iostream>
using namespace std;
int dog, cat, bird, fish;

void f(int pet) {
    cout << "pet id number:" << pet << endl;
}
void g(int pet) {
    cout << "pet id number:" << pet << endl;
}
int main() {
    int i,j, k;

    cout << "f() normal: " << &f << endl;
    cout << "f() long: " << (long)&f << endl;
    cout << "g() normal: " << &g << endl;
    cout << "g() long: " << (long)&g << endl;  
    cout << "j normal: " << &j << endl;  
    cout << "j long: " << (long)&j << endl;
    cout << "k: " << (long)&k << endl;

    k=2;
    cout << "k: " << (long)&k << endl;  
} //

结果

f() normal: 1
f() long: 4375104512
g() normal: 1
g() long: 4375104608
j normal: 0x7fff6486b9c0
j long: 140734879939008
k: 140734879939004
k: 140734879939004

Answer 1

由于ostream operator<<的重载次数为void*而且任何数据指针都可以转换为void*，int的地址1}}像j一样打印出来。但是，函数指针无法转换为void*，因此这种特殊的重载已经过时了。

当另一个operator<<重载发挥作用时，在这种情况下，它将是bool的重载。函数指针可以转换为bool（true ==指针为非NULL）。指向f的指针为非NULL，因此在此转换中产生true，打印为1。

Answer 2

这与引用无关。该程序不使用任何引用。您正在使用地址运算符&。见https://stackoverflow.com/a/9637342/365496

f() normal: 1              the address of f is converted to bool 'true' and printed 
f() long: 4375104512       the address of f is converted to an integer
g() normal: 1              the address of g is converted to bool 'true' and printed
g() long: 4375104608       the address of g is converted to an integer
j normal: 0x7fff6486b9c0   the address of j is printed directly (there's an operator<< for this but not one for printing function pointers like f and g)
j long: 140734879939008    the address of j is converted to an integer
k: 140734879939004         the address of k is converted to an integer
k: 140734879939004         the address of k is converted to an integer

为什么函数的“地址”的值在C ++中是1？

2 个答案: