Char isn't converting to int

问题

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:

-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?

回答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

回答2:

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

回答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.

回答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.

来源：https://stackoverflow.com/questions/9777117/char-isnt-converting-to-int

标签

argv