什么是各种风格的定义函数返回一个函数指针 [英] What are the various styles of defining a function returning a function pointer

问题描述

我有一个功能：

int compare(char * c1, char * c2){
...
...
}

什么是我可以写一个函数的各种风格的 INT ret_compare（void *的项目）返回一个指针进行对比？

What are the various styles in which I can write a function int ret_compare(void * item) that returns a pointer to compare?

推荐答案

有两种主要方式，一是使用的typedef 和一个没有（有两种变体的的typedef ）。您的比较应该不断的指针，如下：

There are two main styles, one using a typedef and one not (with two variants of the typedef). Your comparator should take constant pointers, as below:

int compare(const char *c1, const char *c2) { ... }

// Raw definition of a function returning a pointer to a function that returns an int
// and takes two constant char pointers as arguments
int (*ret_compare1(void *item))(const char *, const char *)
{
    // Unused argument - item
    return compare;
}

// More usual typedef; a Comparator2 is a pointer to a function that returns an int
// and takes two constant char pointers as arguments
typedef int (*Comparator2)(const char *, const char *);

// And ret_compare2 is a function returning a Comparator2
Comparator2 ret_compare2(void *item)
{
    // Unused argument - item
    return compare;
}

// Less usual typedef; a Comparator3 is a function that returns an int
// and takes two constant char pointers as arguments
typedef int Comparator3(const char *, const char *);

// And ret_compare3 is a function returning a pointer to a Comparator3
Comparator3 *ret_compare3(void *item)
{
    // Unused argument - item
    return compare;
}

---

请注意，这些比较器不能用使用bsearch（）和的qsort（）（除非你使用相当可怕蒙上），因为这些比较预期采取的常量无效* 参数。

---

Note that these comparators cannot be used with bsearch() and qsort() (unless you use fairly gruesome casts) because those comparators are expected to take const void * arguments.

请注意，那就是，对于比较字符串，而不是单个字符，使用函数的qsort（）或 bsearch（）应该是类似于：

Note, too, that for comparing strings, as opposed to single characters, the function used by qsort() or bsearch() should be similar to:

int string_comparator(const void *v1, const void *v2)
{
    const char *s1 = *(char **)v1;
    const char *s2 = *(char **)v2;
    return(strcmp(s1, s2));
}

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