#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??
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);
No question here, but you need to pass by address, not by value. the problem seems to be getfunc(myfunc);
Fix getFunc to:
int getfunc(func *myfunc)
{
*myfunc = &add;
return 0;
}
and call it with getFunc(&myfunc);
It should be more like this (changes flagged with <<<):
#include <stdio.h>
typedef int (*func)(int);
int add(int a)
{
return ++a;
}
func getfunc(void) // <<<
{
return &add; // <<<
}
int main()
{
int i;
func myfunc;
i = 10;
myfunc = getfunc(); // <<<
printf(" a is %d\n", (*myfunc)(i));
return 0;
}
myfunc is a pointer. You created it but never assigned it a value. Then you call getfunc with a wild pointer!
Try this (your version, simplified):
int getfunc(func *myfunc)
{
*myfunc = add;
return 0;
}
int main()
{
func myfunc = NULL;
getfunc(&myfunc);
}
