关于函数声明函数指针 [英] About Pointers To Functions in function declarations
问题描述
#include<stdio.h>
#include<stdlib.h>
int fun1()
{
printf("I am fun1.");
return 0;
}
int fun2(int fun())
{
fun();
return 0;
}
int main()
{
fun2(fun1);
return 0;
}
以上程序可以运行。就我而言,我能理解 INT FUN2(INT(*有趣)())
,但我不知道该怎么 INT FUN2(诠释乐趣())
的作品。谢谢你。
The above program can run. As far as I am concerned, I can understand int fun2(int (*fun)())
, but I do not know how int fun2(int fun())
works. Thank you.
推荐答案
当你写 INT FUN2(INT乐趣())
,参数 INT乐趣()
转换成 INT(*有趣)()
,就变成完全等价的:
When you write int fun2(int fun())
, the parameter int fun()
converts into int (*fun)()
, it becomes exactly equivalent to this:
int fun2(int (*fun)());
一个更famiiar转换发生在阵的情况下,当你将它声明为函数参数。例如,如果你这样的:
A more famiiar conversion happens in case of array when you declare it as function parameter. For example, if you've this:
int f(int a[100]);
即使在这里的参数类型转换成为int *
,它变成这样:
int f(int *a);
为什么函数类型和数组类型转换为函数的指针的类型,指针类型,分别是因为标准不允许的功能和数组被传递给函数,也不能究其原因你从一个函数返回的功能和数组。在这两种情况下,它们衰变到其指针的版本。
The reason why function type and array type converts into function pointer type, and pointer type, respectively, is because the Standard doesn't allow function and array to be passed to a function, neither can you return function and array from a function. In both cases, they decay into their pointer version.
在C ++标准03说,在§13.1/ 3(和它在C ++ 11也是一样),
The C++03 Standard says in §13.1/3 (and it is same in C++11 also),
这区别,仅仅在于一个参数声明是一个函数类型,另一个是指向同一个函数类型为相当于。 也就是说,函数类型调整为成为功能类型(8.3.5)的指针
Parameter declarations that differ only in that one is a function type and the other is a pointer to the same function type are equivalent. That is, the function type is adjusted to become a pointer to function type (8.3.5).
和更有趣的讨论是在这里:
And a more interesting discussion is here:
- Reference to Function syntax - with and without &
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