Use pointer notation to print a character array

Question

I know that for a given integer array, a pointer to that integer array, I can access the integer array using something like this:

int main(){

  int x[4] = {0,1,2,3};
  int *ptr;

  ptr = x;

  for(int i = 0; i < 4; i++){
    printf("%d", *(ptr+i));
  }
  return 0;
}

Now say I have a character array instead, doing the same thing doesnt work.

int main(){

  char x[4] = "Haha"; 
  char *ptr;

  ptr = x;

  for(int i = 0; i < 4; i++){
    printf("%s", *(ptr+i));
  }
  return 0;
}

Apparently *(ptr+i) in the first code increments the pointer to the integer 4 bytes each time. It doesnt work for the second code. How do I use the same notation for character array? I think the idea is to increment one char each time.

%s is for an array of chars (null terminated string), %c is for a single character. printf("%c", *(ptr+i)); — robthebloke
– robthebloke, Commented Feb 4, 2020 at 2:17

Hoang Nam · Accepted Answer · 2020-02-04 03:17:47Z

2

You print each letter after a loop so you should correct your source code: printf("%s", *(ptr+i)); into printf("%c", *(ptr+i)); I think you should change char x[4] = "Haha"; into char x[5] = "Haha"; because, maybe in this case it doesn't cause any errors but sometimes, you will get overflown, be careful!

answered Feb 4, 2020 at 3:17

Hoang Nam

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