C char appears as int when compiling

Question

I'm learning C, and have been trying to make a program that takes user input, and removes any double spaces in it, then prints it out again. We have not done arrays yet so I need to do this char by char. This is my code:

#include <stdio.h>

main()
{
    char c;
    int count;
    count = 0;

    while ((c = getchar()) != '\n')
        if (c == ' ')
            count++;
        if (c != ' ')
            count = 0;
        if (count <= 0)
            printf("%s", c);
}

This code does not work, however. The compiler returns the error

:15: warning: format ‘%s’ expects type ‘char *’, but argument 2 has type ‘int’

Any help? I have no clue what I am doing wrong.

score 8 · Answer 1 · edited May 23 '17 at 12:04

8

Use the %c format specifier to print a single char

printf("%c", c);

The %s format specifier tells printf to expect a null-terminated char array (aka a string).

The error message refers to c having type int due to default promotion of arguments (beyond the format string) passed to printf. This previous answer has a nice description of default promotion; this previous thread explains some of the reasoning for why default promotion is necessary.

edited May 23 '17 at 12:04

Community

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answered Jan 31 '13 at 13:15

simonc

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Can you please explain why the error message refers to `int`? – Martin v. Löwis Jan 31 '13 at 13:18
@Martinv.Löwis Good point. I've added links to previous threads which contain better explanations than I'd manage – simonc Jan 31 '13 at 13:25
Note that `getchar()` returns an `int`, not a `char`. That way you can distinguish a `char` read from `EOF` (can't be a legal `char` value, obviously). Not the brightest interface, but compact (and customary by now). – vonbrand Feb 01 '13 at 01:12

score 0 · Answer 2 · answered Jan 31 '13 at 15:18

0

You are using %s which is used for a string and which expects a terminating NULL character(\0)..

Using %c will print you char by char..

answered Jan 31 '13 at 15:18

Raghu Srikanth Reddy

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phuclv · Answer 3 · 2019-02-16T08:36:22.833

Your code has so many problems

First, you print a char with %s (which expects a char pointer, i.e. a string)

In C, char literals are of type int, so whether promoted or not, they're always int. In C++, char literals would be of type char, but after promotion like other answers said, again they'll be int. A plain char variable will also be promoted to int in expressions, and passed as int in vararg functions like printf. That's why the compiler warns you that argument 2 has type ‘int’, because it's expecting a char* and you're passing it an int

→ You must use %c to print a char
Your while loop's body is only the first if block, because in C block scope is defined by {}, not by indentation. So the code will run like this, which is not like what you intended
```
while ((c = getchar()) != '\n')
{
    if (c == ' ')
        count++;
}
if (c != ' ')
    count = 0;
if (count <= 0)
    printf("%s", c);
```
→ You need to put the code block in a pair of brackets. And using else instead of 2 separate ifs to make it more readable and faster (for dumb compilers)
```
while ((c = getchar()) != '\n')
{
    if (c == ' ')
        count++;
    else
        count = 0;
    if (count <= 0)
        printf("%s", c);
}
```
main() is wrong. The correct versions in C would be
```
int main(void)
int main(int argc, char **argv)
```
See What should main() return in C and C++?
c must be declared as int because getchar returns int. See Why must the variable used to hold getchar's return value be declared as int?

A minor point is that instead of int count; count = 0;, just initialize the variable while declaring int count = 0. Or better yet use an unsigned int, because count can't be negative

C char appears as int when compiling

3 Answers3