函数指针与函数引用
在下面的代码中,函数指针和我认为的函数引用"似乎具有相同的语义:
In the code below, function-pointer and what i considered as "function-reference" seems to have identical semantics:
#include <iostream>
using std::cout;
void func(int a) {
cout << "Hello" << a << '
';
}
void func2(int a) {
cout << "Hi" << a << '
';
}
int main() {
void (& f_ref)(int) = func;
void (* f_ptr)(int) = func;
// what i expected to be, and is, correct:
f_ref(1);
(*f_ptr)(2);
// what i expected to be, and is not, wrong:
(*f_ref)(4); // i even added more stars here like (****f_ref)(4)
f_ptr(3); // everything just works!
// all 4 statements above works just fine
// the only difference i found, as one would expect:
// f_ref = func2; // ERROR: read-only reference
f_ptr = func2; // works fine!
f_ptr(5);
return 0;
}
我在 Fedora/Linux 中使用了 gcc 4.7.2 版
I used gcc version 4.7.2 in Fedora/Linux
更新
我的问题是:
为什么函数指针不需要解引用?为什么取消引用函数引用不会导致错误?- 是否在某些情况下我必须使用一种而不是另一种?
- 为什么
f_ptr = &func;有效?既然 func 应该衰减成一个指针?
虽然f_ptr = &&func;不起作用(从void *隐式转换)
Why function pointer does not require dereferencing?Why dereferencing a function reference doesn't result in an error?- Is(Are) there any situation(s) where I must use one over the other?
- Why
f_ptr = &func;works? Since func should be decayed into a pointer?
Whilef_ptr = &&func;doesn't work (implicit conversion fromvoid *)
推荐答案
函数和函数引用(即那些类型的 id-expressions)几乎立即衰减为函数指针,因此表达式 func 和 f_ref 在您的情况下实际上成为函数指针.如果你愿意,你也可以调用 (***func)(5) 和 (******f_ref)(6) .
Functions and function references (i.e. id-expressions of those types) decay into function pointers almost immediately, so the expressions func and f_ref actually become function pointers in your case. You can also call (***func)(5) and (******f_ref)(6) if you like.
在您希望 &-operator 像已应用于函数本身一样工作的情况下,最好使用函数引用,例如&func 与 &f_ref 相同,但 &f_ptr 是另一回事.
It may be preferable to use function references in cases where you want the &-operator to work as though it had been applied to the function itself, e.g. &func is the same as &f_ref, but &f_ptr is something else.
相关文章