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1

#include <stdio.h>
int main()
{
int a;
char b;
scanf("%d",&a);
scanf("%s",&b);
printf("%d",a);
printf("%c",b);
return 0;
}

when i give 1 and a as inputs why iam getting 0 and a ! as output need a reason

2

your program gives 0 and nothing as output because b is assigned ‘\n’ which is not visible in output .
try using this code , which gives correct output

#include< stdio.h >

int main()
{

int a;

char b[1];

scanf("%d",&a);

scanf("%s",b);

printf("%d\n",a);

printf("%s",b);

return 0;

}
3

Well, first of all u shouldn’t use %s placeholder to take a character as an input, use %c. Secondly, if u give input : 1 a into the program , it will store the ascii code of the " " character in b. Hence u can take input like this: scanf("%d %c",&a,&b);
This ignores the ’ ’ and stores the next value in the input into b.

4

Both the answer prevoiusly given are partially wrong.

You have two problems in the program.

  1. Always assign size of the character in use of char ( as it is assigned as arrays ), such as

                       char b[10];

  2. The char size must be greater than 1 here as one of the array is occupied by ‘\n’

    one more important thing, never use scanf("%c",&b), its wrong as b is a adress itself, you don’t have to reassign it. Thus it should be scanf("%c",b);

The correct program should be,

#include<stdio.h>

int main() {

int a;

char b[2];

scanf("%d",&a);

scanf("%s",b);

printf("\n%d\n",a);

printf("%s",b);

return 0;

}

5

@dheeru

#include <stdio.h>

int main()
{
    int a;
    char b;
    scanf("%d",&a);

    getchar();
    /* getchar() is used to consume '\n',
       left after integer input */

    /* If you will remove getchar() and test
       your code on input 1 a,
       Output will be 1 
       as '\n' is stored in b*/

    scanf("%c",&b);
    /* %c as format specifier to take a single
       character input from buffer */

    printf("%d ",a);
    printf("%c",b);
    return 0;
}
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