CODEWITHSUNDEEP
cfunctionreturn-value
user4033167
user4033167

Reputation:

Function returns Int rather than char

I am working on some practice programs and have run into an issue with my function squeez below. It is returning an integer rather than a char array as intended. The code is meant that the user enters, first a character string, followed by a character he wishes to remove. The squeez function receives both and outputs a string without the given character.

Currently my function gives the warning:
return makes integer from pointer without a cast [enabled by default]

I know the function is returning type 'int' b/c it also displays this 2nd error:
format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘int’ [-Wformat=]

/* Squeeze Function */

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

char squeez(char s[], int c)
{
    int i, j;

    for (i = j = 0; s[i] != '\0'; i++)
    {
        if (s[i] != c)
        {
            s[j++] = s[i];
        }
    printf("%c %d", s[i], c);
    }
    printf("%s", s);
    s[j] = '\0';
    return s;
}

int main(void)
{
    int i, c;
    char str[10];

    for (i = 0; (c = getchar()) != '\n'; ++i)
    {
        str[i] = c;
    } 
    c = getchar();
    printf("%s", squeez(str, c));
    return EXIT_SUCCESS;
}

Any and all help would be appreciated thank you!

Upvotes: 0

Views: 3724

Answers (3)

200_success
200_success

Reputation: 7582

You should change the return type to char *:

char *squeez(char s[], int c) {
    …
    return s;
}

… because s is a pointer (the address of the first character of the string).

A second problem is that squeez() looks for a '\0' terminator, but you never wrote one there. A third problem is that you might overflow the 10-byte buffer. The loop in main() should look more like

#include <assert.h>

int main(void)
{
    int i;
    char str[10];

    for (i = 0; i < sizeof(str) - 1; i++)
    {
        if ((str[i] = getchar()) == '\n')
        {
            break;
        }
    }
    assert( (i == sizeof(str) - 1) || (str[i] == '\n') );
    str[i] = '\0';
    printf("%s", squeez(str, getchar()));
}

Upvotes: 3

Karthikeyan.R.S
Karthikeyan.R.S

Reputation: 4041

You have to return the character pointer. Change the returning char to char *.

If you are using the char then that will taken as a single character. So when you are using the char you are able to return the character only. like this. 

return s[0]; or else a character variable.

Upvotes: 0

sumithdev
sumithdev

Reputation: 331

You are actually returning a char pointer. So the return type of the function should be char *.

Upvotes: 0

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