Why the first code work and second give wrong output?

Question

This is the first code

#include <stdio.h>
#include <stdlib.h>
int main()  
{   
   short int a;
   unsigned short int l;
   scanf("%d%u",&a,&l);
   printf("%d %u",a,l);
   return 0;
}

If I give the input(may be any other input)

5
9

The output is

0 9

The second code is

#include <stdio.h>
#include <stdlib.h>
int main()
{
   short int a;
   unsigned short int l;
   scanf("%u%d",&l,&a);
   printf("%d %u",a,l);
   return 0;
}

If I give the output (may be any other input)

9
5

The output is

5 9

Why scanf("%d%u",&a,&l); do not work and scanf("%u%d",&l,&a); works?

Because `%u` and `%d` are for normal `int`, not `short int`. — Oliver Charlesworth, Jan 05 '14 at 13:06
@OliCharlesworth Then how this worked in the first case? Or he got lucky? — Maroun, Jan 05 '14 at 13:06
Neither works since they both invoke undefined behavior for the reason that Oli provides. — Paul Hankin, Jan 05 '14 at 13:07
@ᴍarounᴍaroun it is undefined behavior, he is using memory outside the allocated for the variables. He got lucky. — SJuan76, Jan 05 '14 at 13:07
@ᴍarounᴍaroun: As this is UB, I'd imagine it's just chance. — Oliver Charlesworth, Jan 05 '14 at 13:07

score 3 · Answer 1 · edited Jun 20 '20 at 09:12

3

Using a wrong specifier for a data type invokes undefined behavior. Result may be expected or unexpected.

C11:7.21.6 Formatted input/output functions:

(p9) If a conversion specification is invalid, the behavior is undefined.282) If any argument is not the correct type for the corresponding conversion specification, the behavior is undefined.

and about h: p(7)

h Specifies that a following d, i, o, u, x, or X conversion specifier applies to a short int or unsigned short int argument (the argument will have been promoted according to the integer promotions, but its value shall be converted to short int or unsigned short int before printing); or that a following n conversion specifier applies to a pointer to a short int argument.

Use h before %d and %u for short type data.

scanf("%hd %hu",&a,&l);
printf("%hd %hu",a,l);

edited Jun 20 '20 at 09:12

Community

1
1

answered Jan 05 '14 at 13:09

haccks

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It's unnecessary for `printf`, though, due to default argument promotions. (although I'd argue that it's good for consistency...) – Oliver Charlesworth Jan 05 '14 at 13:13
@OliCharlesworth; But no harm in doing so :) – haccks Jan 05 '14 at 13:14
@user3162531; Read the very first line of my answer. – haccks Jan 05 '14 at 13:15
but still why this problem is only with %d not with %u – Jan 05 '14 at 13:18
@user3162531; I already explained that this is because the behavior of your program is not defined. – haccks Jan 05 '14 at 13:24

Why the first code work and second give wrong output?

1 Answers1

C11:7.21.6 Formatted input/output functions: