Variable scope

I have the following piece of code:

Code:

---

int main()
{
  int key = 1;
  getKey();
  printKey();
  printf("%s", "Key in main is: ");
  printf("%d", key);
  return 0;
}

int getKey()
{
  int *key;
  printf("Please enter a key: ");
  scanf("%d", &key);
}
int printKey()
{
  int *key;
  printf("%s", "You entered: ");printf("%d\n", key);
}

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How exactly do i set "key" in main to whatever is being entered in getKey()? I can't use the assign & operator on a pointer so i'm having trouble referencing it.

You could use a pointer:

Code:

---

#include <stdio.h>
void getKey(int *);
int main(void)
{
    int key = 1;
    getKey(&key);
    printf("key = %d\n", key);
    return 0;
}

void getKey(int *k)
{
    printf("Please enter a key: ");
    fflush(stdout);
    scanf("%d", k);
}

---

Or you could return a value:

Code:

---

#include <stdio.h>
int getKey(void);
int main(void)
{
    int key = 1;
    key = getKey();
    printf("key = %d\n", key);
    return 0;
}

int getKey(void)
{
    int k;
    printf("Please enter a key: ");
    fflush(stdout);
    scanf("%d", &k);
    return k;
}

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Find a tutorial on function paramters and pointers

Code:

---

#include <stdio.h>

void getKey(int *key){
  printf("Please enter a key: ");
  scanf("%d", key);
}

void printKey(int *key){
  printf("You entered: ");
  printf("%d\n", *key);}

int main(void){
  int key = 1;
  getKey(&key);
  printKey(&key);
  printf("Key in main is: ");
  printf("%d", key);
  return 0;
}

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