getting wrong output when using char and int

Question

Writing a simple code that's suppose to scan an integer and a character and then write them out. my input is 1a and the output should be 1a but i'm getting 0 on the integer spot. have a pretty basic understanding of c so may have missed something that's pretty obvious thanks in advance.

#include <stdio.h> 
int main()
 { 
   int a; 
   char b; 
   scanf("%d", &a); 
   scanf(" %s", &b); 
   printf("%d", a); 
   printf("%s", &b); 
 }

Edit the question to include your code. Don't post it as a comment. — dbush, Oct 23 '18 at 12:58
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score 0 · Answer 1 · answered Oct 23 '18 at 13:02

0

b is a char, %s is for string input so adds trailing 0 after b and you can get a crash. Use %c to input char.

answered Oct 23 '18 at 13:02

Alexander Ekzhanov

162
2

And same goes for `printf("%s", &b);`. – kiran Biradar Oct 23 '18 at 13:03
so what you mean is I should change it to scanf("%c",&b); printf("%c",&b); ? because that gave me wrong output as well. The output was 1k instead of 1a. – Noncode101 Oct 23 '18 at 13:07

score 0 · Answer 2 · answered Oct 23 '18 at 13:16

You basically want this:

#include <stdio.h> 

int main()
{
  int a;
  char b[100];     // array of 100 chars

  scanf("%d", &a);
  scanf("%s", b);

  printf("%d", a);
  printf("%s", b);
}

To fully understand this, you need to read the chapters dealing with scanf and the one dealing with strings in your C text book.

score 0 · Accepted Answer · answered Oct 23 '18 at 13:20

b is a character so replace %s with %c, moreover

scanf() takes & before the variable as it want to store the variable refering to that address.

2.printf() Just outputs the value to the console present in that varaible. Thereby no need to use & inside it

CORRECTED CODE:

    #include <stdio.h> 
    int main()
    { 
     int a; 
     char b; 
     scanf("%d", &a); 
     scanf(" %c", &b); 
     printf("%d", a); 
     printf("%c", b); 
     }

score 0 · Answer 4 · answered Oct 23 '18 at 13:27

0

you can try it:

#include <stdio.h>
int main()
 {
  int a;
  char b;
  scanf("%d", &a);
  scanf(" %c", &b);
  printf("%d", a);
  printf("%c", b);
 }

answered Oct 23 '18 at 13:27

S. Hesam

5,266
3
37
59

getting wrong output when using char and int

4 Answers4