推荐回答（1个）
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如果你的 G函数的参数就想是个函数指针的话, 应该写成下面这种形式, 而不是使用 void*
int G(int (*)(int));
然后你就可以 G(F) 调用了.
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如何把void*转换成函数指针:
int G(void * f)
{
int value = 10;
((int (*)(int))f)(value); // 我觉得这样可以. 给f传的参数为 10.
}
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