Converting a char * to int

Question

Hi i have the below code and for some reason am getting the error and can't seem to work out why.

warning: return makes integer from pointer without a cast

The code i have is:

long convertToInt(char *convert) {
        char *p = convert;
        while(*p){
            if(isdigit(*p)) {
                long val = strtol(p, &p, 10);
                return val;
            } else {
                p++;
            }
        }
    return NULL;
}

There's a code smell, its a `convertToInt`, meaning an int, not a long, which is misleading. — t0mm13b, Apr 29 '16 at 12:17
Could you show how you intend to use the `convertToInt` function? The answer may depend on this. — Jabberwocky, Apr 29 '16 at 12:21

score 2 · Accepted Answer · answered Apr 29 '16 at 12:18

2

NULL is a pointer, not an integer. It may be implemented as a #define that expands to 0 (the integer constant) or ((void *)0) the null pointer constant. If you want to return a value that means "an error occurred", you'll probably want to return an integer constant. Values 0 and -1 are traditional.

Even better is to returns a boolean status value for success/failure, and return the value via a pointer argument: bool_t convertToInt(const char *s, long *value). See the standard library function strtol for an example.

answered Apr 29 '16 at 12:18

1

`bool_t`? I guess you mean `bool`. – unwind Apr 29 '16 at 12:42
1

Yes, bool, not bool_t. – Apr 29 '16 at 13:00

score 1 · Answer 2 · answered Apr 29 '16 at 12:18

1

Your funtion returns long. So change the return statement to:

return 0L;

One possible definition of NULL is (void *)0 which is being converted to long when you return.

answered Apr 29 '16 at 12:18

Mohit Jain

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score 1 · Answer 3 · answered Apr 29 '16 at 12:22

1

The proper code would be, assuming convert is a string with the digits:

long convertToLong(char *convert) {
    long val = 0L;
    char *p = convert;
    while (isdigit(*p)){
        val = val * 10 + (*p - '0');
        p++;
    }
    return val;
}

answered Apr 29 '16 at 12:22

Paul Ogilvie

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score 0 · Answer 4 · answered Apr 29 '16 at 12:20

0

The problem is here:

return NULL;

where you return NULL, but the return type of the function is long.

Moreover your function is named convertToInt(), but it returns a long, which would imply a convertToLong() function name.

answered Apr 29 '16 at 12:20

gsamaras

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Converting a char * to int

4 Answers4