the code doesn't work properly at output C

Question

I tried to read 2 variables from the keyboard and to write them on the screen and I have a problem, the program display me only one..

#include <stdio.h>
#include <stdlib.h>

int main()
{
    short int n,x;
    scanf("%d",&n);
    scanf("%d",&x);
    printf("%d %d",n,x);
    return 0;
}

I introduced 14 and 15 and the program return me 0 and 15 can somebody tell me why?

Make the variables of data type `int` - `int n,x;` or e use `%hd` instead of `%d` — Rabbid76, Oct 14 '18 at 09:07

score 2 · Accepted Answer · answered Oct 14 '18 at 09:13

Use %hd format specifier for short int Use %hu format specifier for unsigned int Use %d format specifier for int Use %ld format specifier for long int

#include <stdio.h>
#include <stdlib.h>

int main() {
  short int n, x;
  scanf("%hd", &n); // Notice %hd instead of %d for short int
  scanf("%hd", &x); // Notice %hd instead of %d for short int
  printf("%hd%hd", n, x);// Notice %hd instead of %d for short int
  return 0;
}

Scheff's Cat · Answer 2 · 2018-10-14T09:27:49.370

The formatters in scanf() and printf() doesn't match the type of your variables n and x.

%d uses the variables as they were int while int has probably the double number of bytes as short int. (Integer types)

Hence, with the wrong formatters, scanf() uses the provided addresses wrong. For printf() it's a bit more complicated: The short ints are converted to int internally. (Default argument promotions) Hence, printing short int with %d (as they were int) doesn't fail.

So, it's the scanf() what must be fixed.

Either use correct formatters:

#include <stdio.h>

int main()
{
    short int n,x;
    scanf("%hd",&n);
    scanf("%hd",&x);
    printf("%d %d",n,x);
    return 0;
}

Live Demo on ideone

or use correct variable type for the formatters:

#include <stdio.h>

int main()
{
    int n,x;
    scanf("%d",&n);
    scanf("%d",&x);
    printf("%d %d",n,x);
    return 0;
}

Live Demo on ideone

The formatting of scanf() and printf() families are very powerful and flexible but unfortunately very error-prone as well. Using them wrong introduces Undefined Behavior. The compiler is (usually) unable to recognize errors as the evaluation of formatters happens at run-time and inside the scanf()/printf() functions. So, they have to be used caaarefully.

score 1 · Answer 3 · answered Oct 14 '18 at 09:11

%d assumes the variable to be of data type int.

Use the data type int:

int main()
{
    int n,x;
    scanf("%d",&n);
    scanf("%d",&x);
    printf("%d %d",n,x);
    return 0;
}

or use %hd instead of %d

int main()
{
    short int n,x;
    scanf("%hd",&n);
    scanf("%hd",&x);
    printf("%hd %hd",n,x);
    return 0;
}

Note, scanf("%d",&x); read a value and stores it to the memory addressed by &x. Since &x is treated as a 4 byte memory, 4 bytes are written to the address specified by &x.

the code doesn't work properly at output C

3 Answers3