函数指针未返回值

Question

We just learnt in school about function pointers in C and i wanted to try them out in a test program. The idea is obviously pretty simple. However, when i try to use result = (*funcPtr)(); i get an STATUS_ACCESS_VIOLATION exception but i can't tell what i did wrong. Any ideas on what i'm missing?

#include <stdio.h>

int fun1();
int fun2();

int (*funcPtr)(void);

int fun1() {
  return 1;
}

int fun2() {
  return 2;
}

int main(void) {
  int input,result = 0;
  scanf("%d",input);
  if(input == 1) {
    funcPtr = &fun1;
  } else if(input == 2) {
    funcPtr = &fun2;
  }
  result = (*funcPtr)();
  printf("%d\n",result);
}

Answer 1

You forgot a & before input in the scanf . Therefore scanf writes not to the address of the input variable, but to its value (and the variable is uninitialized a this point).

Change that line to:

scanf("%d", &input);

to pass a pointer to input to scanf and not the value of input .

And as already pointed out by other users, do not forget to handle inputs other than 1 and 2 . Otherwise you will call your uninitialized funcPtr variable, which most likly results in a Segmentation fault.