未确定paramater函数指针 [英] Function pointer with undetermined paramater

问题描述

我想有一个函数指针，可以采取多种类型的参数。我该怎么做呢？

I want to have a function pointer that can take various types of parameters. How do I do that?

下面的例子（1号线），我想无效（* FP）（INT）来能够采取无效（* FP） （字符*）为好。下面code不正确编译，因为我传递的char *其中int有望使编译如下code会给你警告（并且将无法正常工作）。

The following example (line 1), I want void (*fp)(int) to be able to take void (*fp)(char*) as well. The following code does not properly compile because I'm passing char* where int is expected so compiling the following code will give you warnings (and won't work properly).

void callIt(void (*fp)(int))
{
    (*fp)(5);
}

void intPrint(int x)
{
    printf("%d\n", x);
}

void stringPrint(char *s)
{
    printf("%s\n", s);
}

int main()
{
    void (*fp1)(int) = intPrint;
    callIt(fp1);
    void (*fp2)(char*) = stringPrint;
    callIt(fp2);
    return 0;
}

注：我知道，试图通过整数5为char *参数是愚蠢的，但是这不是对这个问题的关注。如果你愿意，你可以替换的char *使用float。

Note: I know that attempting to pass integer 5 as char* parameter is stupid, but that's not the concern for this question. If you want, you can replace char* with float.

推荐答案

我想说的使用无效（* FP）（无效*），然后在三分球传给你变量和扮演他们为void *。这是pretty哈克，但它应该工作。

I would say use void (*fp)(void*), then pass in pointers to your variables and cast them as void*. It's pretty hacky, but it should work.

例如：

void callIt(void (*fp)(void*))
{
    int x = 5;
    (*fp)((void*)&x);
}

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