Char isn't converting to int

Question

For some reason my C program is refusing to convert elements of argv into ints, and I can't figure out why.

int main(int argc, char *argv[])
{

    fprintf(stdout, "%s\n", argv[1]);

    //Make conversions to int
    int bufferquesize = (int)argv[1] - '0';

    fprintf(stdout, "%d\n", bufferquesize);
}

And this is the output when running ./test 50:

50

-1076276207

I have tried removing the (int), throwing both a * and an & between (int) and argv[1] - the former gave me a 5 but not 50, but the latter gave me an output similar to the one above. Removing the - '0' operation doesn't help much. I also tried making a char first = argv[1] and using first for the conversion instead, and this weirdly enough gave me a 17 regardless of input.

I'm extremely confused. What is going on?

Dover

I think you might mean argv[1][0] (type is char) <-- first character of first argument (after exec name). Of course, there is still no checking with that and there are better ways. Don't try to "cast away an error" (as argv[1] is typed as char*) because it often just doesn't work :-) – Regulus

Answer 1

argv[1] is a char * not a char you can't convert a char * to an int. If you want to change the first character in argv[1] to an int you can do.

int i = (int)(argv[1][0] - '0');

I just wrote this

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

int main(int argc, char **argv) {
    printf("%s\n", argv[1]);

    int i = (int)(argv[1][0] - '0');

    printf("%d\n", i);
    return 0;
}

and ran it like this

./testargv 1243

and got

1243
1

Ide

You dropped your reference to atoi... why? – Hewe

@Hewe I was never using atoi. you could use it if you want but my answer never had it. The answer from Scott Hunter uses it though. – Ide

Why is it that you had to do int i = (int)(argv[1][0] - '0'); as opposed to int i = (int)(argv[1][0]);? Why was the - '0' necessary? – Ownership

Because if your first character is '1', 1 in ascii is 49, so casting would return 49. You want to subtract by the first digit ('0' in ascii is 48), and then cast to get the correct answer. – Maryrose

Answer 2

Try using atoi(argv[1]) ("ascii to int").

Rosauraroscius

This should be the selected answer. – Rodarte

This right here. atoi() function was made for this. – Galle

Answer 3

You are just trying to convert a char* to int, which of course doesn't make much sense. You probably need to do it like:

int bufferquesize = 0;
for (int i = 0; argv[1][i] != '\0'; ++i) {
   bufferquesize *= 10; bufferquesize += argv[1][i] - '0';
}

This assumes, however, that your char* ends with '\0', which it should, but probably doesn't have to do.

Dineen

Actually it assumes the array of chars that your char * points to ends with '\0'. Whether the pointer "ends in '\0' (what would that mean anyway? that its representation ends in a 0 byte?) is not an issue. – Nonsmoker

Of, course. The sentence "a pointer to char ends with a certain symbol" just doesn't make sense. Thank you for the correction. :) – Dineen

Answer 4

(type) exists to cast types - to change the way a program looks a piece of memory. Specifically, it reads the byte encoding of the character '5' and transfers it to memory. A char* is an array of chars, and chars are one byte unsigned integers. argv[1] points to the first character. Check here for a quick explanation of pointers in C. So your "string" is represented in memory as:

['5']['0']

when you cast

int i = (int) *argv[1]

you're only casting the first element to an int, thus why you

The function you're looking for is either atoi() as mentioned by Scott Hunter, or strtol(), which I prefer because of its error detecting behaviour.

Hewe