Return value of atoi (argv[1]) is always 0

Question

I'm running into a problem about return value of atoi().

I want to convert the char in command line argument argv[1] into int type and print it out.

Here is my code.

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

int main(int argc, char *argv[])
{
    int i;
    //print the char in *argv[]
    for(i = 0; i < argc; i++)
    {
            fprintf(stdout, "Arg %d: %s\n", i, argv[i]);
    }

    if (argc > 1)
    {
        i = atoi(argv[1]);      //convert char to int
        fprintf(stdout, "Int version of 1st arg: %d\n", i);
    }

    return 0;
}

I compile it by gcc and run it like ./a.out a b c

Other result is correct, but the atoi() result always displays as

Int version of 1st arg: 0

Could you give me some suggestion on this topic?

Answer 1

My suggestion is that if you want to convert a string to an int, provide as parameter a string that can actually be converted to an int. Your arguments are:

 ./a.out a b c

so no ints. What did you expect? What do you think a converted to an int is?

Answer 2

atoi() converts a char string to a number - but only if the string is a number.

If you want to print 'a' as an ASCII value just use %d and argv[1][0], i.e. the first character of string argv[1].

Answer 3

atoi has no way of signalling a conversion error, such as when trying to parse a. Use strtol instead, which gives you enough information to determine whether the conversion succeeded:

#include <stdlib.h>

char const * input = "abc";  // this could be argv[i]

char * e;

long int n = strtol(input, &e, 0);

if (*e != '\0') { /* conversion error! */ }
else            { /* n is valid        */ }