Question

This code doesn’t work but when I put print(7) this works fine:

#include<stdio.h>
#define print(i) printf("%"#i"c",ch);

int main(void) 
{
    char ch ='*';
    int N;
    scanf("%d",&N);

    if (N%2 == 0)
    {
            print(N); // print(7);
    }
    else
    {

    }
}

Here is the compilation output

 gcc draw.c -o draw
 draw.c: In function ‘main’:
 draw.c:11: warning: unknown conversion type character ‘N’ in format

How can I get the code to work for a user inserted input. Please help.

Answer 1

The problem is that the # operator is part of the preprocessor. It doesn’t understand run-time values.

The printf function recognizes a * syntax that tells it to accept an int argument that specifies a field width, rather than having the width as a constant in the format string.

For example, this function could replace your macro. Note that I’ve made ch a parameter; I think it’s much clearer that way.

int print(int n, char ch) {
    return printf("%*c", n, ch);
}

Personally, I probably wouldn’t bother with either a function or a macro for this; I’d just call printf directly:

printf("%*c", n, '*');