Char isn't converting to int

c argv
limasxgoesto0 picture limasxgoesto0 · Mar 19, 2012 · Viewed 26k times · Source

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?

Answer

Scott Hunter picture Scott Hunter · Mar 19, 2012

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

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