passing argument makes pointer from integer without a cast

Question

I've read through several similar questions on Stack Overflow, but I've not been able to find one that helps me understand this warning in this case. I'm in my first week of trying to learn C though, so apologies if I've missed an obvious answer elsewhere on Stack Overflow through lack of understanding.

I get the following warning and note:

 warning: passing argument 2 of ‘CheckIfIn’ makes pointer from integer without a cast [enabled by default]
 if(CheckIfIn(letter, *Vowels) ){
 ^

 note: expected ‘char *’ but argument is of type ‘char’
 int CheckIfIn(char ch, char *checkstring) {

When trying to compile this code:

#include <stdio.h>
#include <string.h>
#define CharSize 1 // in case running on other systems

int CheckIfIn(char ch, char *checkstring) {
    int string_len = sizeof(*checkstring) / CharSize;
    int a = 0;

    for(a = 0; a < string_len && checkstring[a] != '\0'; a++ ){

        if (ch == checkstring[a]) {
            return 1;
        }
    }
    return 0;
}


// test function    
int main(int argc, char *argv[]){
    char  letter = 'a';
    char *Vowels = "aeiou";     

    if(CheckIfIn(letter, *Vowels) ){
        printf("this is a vowel.\n");
    }

    return 0;
}

To fix the type: CheckIfIn(letter, *Vowels) should be CheckIfIn(letter, Vowels). Then sizeof(*checkstring) / CharSize is not what you expect. — Jarod42
– Jarod42, Commented Aug 30, 2014 at 16:50
#define CharSize 1 // in case running on other systems—char has always a size of 1, even on non-8 bit systems. Thus, int string_len = sizeof(*checkstring) / CharSize; always evaluates to 1, on all systems. You need to pass the length of an array to a function as a separate parameter, or use a function like strlen if your array has some sentinel value (like a 0-byte for a string): size_t string_len = strlen(checkstring);. — mafso
– mafso, Commented Aug 30, 2014 at 17:12
Also. Since you don't want to modify the string pointed to by Vowels, you should declare it as a const char* (A pointer to a const string) and change the parameter type of CheckIfIn from char* to const char*. — John
– John, Commented Aug 30, 2014 at 18:18
Thanks for your help everyone - really helpful. As you can see I'm making quite basic mistakes. I wouldn't mind whoever downvoted explaining the downvote though. I'm not saying it's not legit to downvote, but if you don't tell people what you think they're doing wrong then a downvote doesn't achieve very much - kind of bad forum behaviour I think. — DanBennett
– DanBennett, Commented Aug 30, 2014 at 19:06

John · Accepted Answer · 2014-08-30 16:53:29Z

8

Vowels is a char*, *Vowels is just a char, 'a'. chars get automatically promoted to integers, which your compiler is allowing to be implicitly converted to a pointer. However the pointer value will not be Vowels, it will be the address equal to the integer encoding of the character 'a', 0x61 almost universally.

Just pass Vowels to your function.

answered Aug 30, 2014 at 16:53

John

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Sachin Mokashi · Accepted Answer · 2020-12-18 02:47:34Z

0

In your case, the type conversion is from char to integer pointer. In some cases, the function takes void pointer as the second argument to accommodate for all the data-types. In such cases, you would need to typecast the second argument as (void *)

This would be the function declaration in most well written modular functions:

int CheckIfIn(char ch, void *checkstring);

You would need to pass the argument as a void pointer, provided the Vowels is not a char pointer

 if(CheckIfIn(letter, (void *)Vowels) ){
        printf("this is a vowel.\n");
    }

answered Dec 18, 2020 at 2:47

Sachin Mokashi

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