Using 'Printf' on a Variable in C [Closed]

Question

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int x = 1;

    printf("please make a selection with your keyboard\n");
    sleep(1);
    printf("1.\n");

    char input;
    scanf("%c", &input);
    switch (input) {
        case '1':
            x = x + 1;
            printf(x);
    }
    return(0);
}

I am trying a make a variable add to itself and then print that variable out, but I can't seem to get my code to work.

My output error is

newcode1.c: In function ‘main’:
newcode1.c:20:2: warning: passing argument 1 of ‘printf’ makes pointer from integer without a cast [enabled by default]
In file included from newcode1.c:1:0:
/usr/include/stdio.h:362:12: note: expected ‘const char * __restrict__’ but argument is of type ‘int’
newcode1.c:20:2: warning: format not a string literal and no format arguments [-Wformat-security]

Accepted Answer

Your printf needs a format string:

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

This reference page gives details on how to use printf and related functions.

Answer 2

As Shafik already wrote, you need to use the right format because scanf gets you a char.

Don't hesitate to look at printf - C++ Reference if you aren't sure about the usage.

Hint: It's faster/nicer to write x = x + 1; the shorter way is: x++;

Using 'Printf' on a Variable in C [Closed]

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Marcus Vance