参数不确定的函数指针 [英] Function pointer with undetermined parameter

问题描述

我想要一个可以使用各种类型参数的函数指针.我该怎么办?

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

下面的示例(第1行)中，我希望void (*fp)(int)也能够采用void (*fp)(char*).以下代码无法正确编译，因为我在期望将int传递给char *的地方，因此编译以下代码会给您警告(并且无法正常工作).

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;
}

注意:我知道尝试将整数5作为char *参数传递是愚蠢的，但这不是此问题的关注点.如果需要，可以将char *替换为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

我会说使用void (*fp)(void*)，然后将指针传递给变量并将其强制转换为void *.这很hacky，但是应该可以.

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);
}

这篇关于参数不确定的函数指针的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！