Segmentation fault in C program using pointers as parameters

Question

Im trying to understand pointers as function parameters, and in one of the programs there is a segmentation error I can't fix. Firstly, why to use pointers in function arguments? and Why is this error showing?

#include <stdio.h>

void square_it(int* a)
{
  printf("The final value is: %d\n", *a * *a);
}

int main()
{
  int* input;

  puts("This program squares the input integer number");
  puts("Please put the number:");

  scanf("%d", &input);

  square_it(input);
  return 0;
}

Answer 1

This is the working code:

#include <stdio.h>

void square_it(int* a)
{
  printf("The final value is: %d\n", *a * *a);
}

int main()
{
  int i = 0;
  int* input = &i;

  puts("This program squares the input integer number");
  puts("Please put the number:");

  scanf("%d", input);

  square_it(input);
  return 0;
}

There are some errors in the original code:

According to the man-pages to scanf, it takes a format string and then the address of where to store the input.

You gave it the address of a pointer (eg. an int**), which is not what scanf expects.

Also you need to provide memory to store the input in. The scanf string tells that you want an integer as input. In the above code snippet that is i.

input points to i, so i can give the int*, that is input to scanf. scanf will then write into i. We can then go ahead and put the address of i into the sqare_it function.

Since we did not use the heap, we don't need to worry about memory management.