函数指针的传递问题

#include <stdio.h>

typedef int (*func)(int);

int add (int a)
{
        return ++a;
}

int getfunc(func myfunc)
{
    myfunc = &add;
    return 0;
}

int main()
{
        int i;
        func myfunc;

        i = 10;
        getfunc(myfunc);

        printf(" a is %d/n", (*myfunc)(i));

        return 0;
}

I can't get what i want.the result is " a is 0".why is that??

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I think you're actually lucky that you get a is 0 instead of a crash. The problem is that getfunc takes the function pointer by value, so the myfunc = &add inside getfunc has no effect on the caller at all. Try

int getfunc(func *myfunc)
{
    *myfunc = &add;
    return 0;
}

and in main:

getfunc(&myfunc);