In this super easy program written in C won't let me read a variable of type "char"?

Question

This easy code in C won't work, and I don't get it why. If I read only "n", or only "ch" separately, it works, otherwise if I try to read them both, it won't let me read "ch". What happens and how could I make it work?

#include <stdio.h>

int main()
{
    int n;
    char ch;

    printf("n=");
    scanf("%d",&n);
    printf("ch="); //when i press Build and Run it won't let me read "ch"??? why?
    scanf("%c",&ch);
    return 0;
}

by 'wont let you' you mean it is reading a value you don't expect such as the newline ? — lostbard, Nov 12 '18 at 21:56

score 2 · Accepted Answer · answered Nov 12 '18 at 21:57

2

When you read in a number using %d, a newline is left in the input buffer. When you then read a character with %c, it reads that newline immediately so you don't get prompted for more input.

Unlike the %d format specifier, which discards any leading whitespace, the %c format specifier does not.

Add a leading space before %c to consume any leftover whitespace:

scanf(" %c",&ch);

answered Nov 12 '18 at 21:57

dbush

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it's working now, thanks a lot! – meiznub Nov 12 '18 at 21:59
@meiznub Glad I could help. Feel free to [accept this answer](https://stackoverflow.com/help/accepted-answer) if you found it useful. – dbush Nov 12 '18 at 22:00

In this super easy program written in C won't let me read a variable of type "char"?

1 Answers1