指向函数的指针 - 一种不同的声明方法 [英] Pointer to a function - A different approach of declaring
问题描述
在阅读this(精神病患者给出的答案)时,我了解如何 typedef 和调用函数指针.但是在考虑了 typedef 之后,我对它们进行了一些试验,并且也能够以这种方式调用函数:
While reading this (the answer given by psychotic), I understood how to typedef and call a function pointer. But after thinking about typedefs I experimented with them a little and was able to call functions this way also:
typedef void func(unsigned char);
void test(unsigned char a);
int main()
{
unsigned char b=0U;
func *fp=&test;
while(1)
{
fp(b);
b++;
}
}
void test(unsigned char a)
{
printf("%d",a);
}
我不明白使用函数指针语法和这种方法有什么区别?两者似乎几乎产生相同的功能.
I don't get what is the the difference between using the function pointer syntax and this approach? Both seem to almost yield the same functionality.
推荐答案
风格
typedef void func_t (void);
...
funct_t* fp;
是声明函数指针最清晰的方法之一.清除,因为它与 C 的其余指针语法一致.
Is one of the clearest ways to declare function pointers. Clear because it is consistent with the rest of the pointer syntax of C.
相当于可读性稍差
typedef void (*func_t)(void);
func_t fp;
这又相当于可读性低得多
Which in turn is equivalent to the much less readable
void (*fp)(void);
当您将这些作为参数传递给函数时,第一种样式的优势变得明显:
The advantage of the first style becomes obvious when you pass these as parameters to a function:
1) void sort (func_t* callback); // very clear and readable!
2) void sort (func_t callback); // hmm what is this? passing by value?
3) void sort (void(*callback)(void)); // unreadable mess
通常,将指针语法隐藏在 typedef 后面是个坏主意.函数指针也不例外.
Generally, it is a bad idea to hide the pointer syntax behind typedefs. Function pointers are no exception.
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