#include int main() { char c; scanf("%c",&c) printf("%c",c); return 0; } why iam not able to give a character to 'c' at runtime..

Question

#include<stdio.h>

int main()
{
    char c;
scanf("%c",&c)
    printf("%c",c); 
    return 0;
}
why iam not able to give a character to 'c' at runtime..

harsimran_hs4 · Answer

Semicolon is missing at the end of the scanf statement , this would have been pointed out by the compiler 
Other than that everything is fine

anonymous · Answer

when you compile the code compile like this 
gcc -Wall -o name name.c
-Wall means that if your program has wrongs thing like syntax errors those are to be mention at compile time ,if it so you can identify where went wrong ...

rsmith6559 · Answer

With the addition of the semi-colon, your code works.  Would this addition help:

int main()
{
    char c;

    printf( "Enter a character: " );
    scanf("%c",&c);
    printf("%c
",c);
    return 0;
}