C converting character command line arguments to an integer array

Question

I'm new to C and trying to figure out arrays and command line arguments. I have:

int main(int argc, int **argv) {
    int vals[8];
    for(int i = 0;i < 8;i = i + 1) {
        vals[i] = atoi(argv[i]);
        printf("%d", vals[i]);
    }
}

I call it with ./file 1 2 3 4 5 6 7 8 and I would expect it to spit out 12345678, but instead, it spits out 01234567 which to me says that it's just printing the array positions. How do I get to actually print/access the value of vals[i], and/or make sure that the command line value is actually being properly assigned?

Thanks in advance.

Use any other example to test your program. Try ./file 8 7 6 5 4 3 2 1. — David Schwartz
– David Schwartz, Commented Apr 13, 2017 at 18:35
atoi issues 0 for your program name because it failed to convert to integer. — Jean-François Fabre
– Jean-François Fabre ♦, Commented Apr 13, 2017 at 18:36
Which is why, generally speaking, atoi() is a bad idea and you should use one of the strto___() functions. — John Hascall
– John Hascall, Commented Apr 13, 2017 at 18:38
@JohnHascall /strto[a-z]+/ will also return 0 but set errno in certain implementations — Ryan
– Ryan, Commented Apr 13, 2017 at 18:53
argv[0] is the name of the program (./file). atoi() will return 0 when given such a string. — Peter
– Peter, Commented Apr 13, 2017 at 23:34

Ryan · Accepted Answer · 2017-04-13 18:37:43Z

2

Start with argv[1] In order to exclude the first element of argv which is the program name. A simple way to do this is to increment argv at the top of the program.

int main(int argc, char **argv) {
    argv++; /* argv[0] is the program name */
    int vals[8];
    for(int i = 0;i < 8;i = i + 1) {
        vals[i] = atoi(argv[i]);
        printf("%d", vals[i]);
    }
}

On a side note, you should check the value of argc prior to accessing elements at index i in argv

edited Apr 13, 2017 at 18:37

answered Apr 13, 2017 at 18:35

Ryan

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3 Comments

Jean-François Fabre Over a year ago

why completely ignoring argc ?

Ryan Over a year ago

See my edit, please. Furthermore, I think for his particular question, telling him about the possiblity of undefined behavior is out of scope.

Peter Over a year ago

It might have worked but, rather than incrementing argv, it would have been better to start the loop with i = 1. And have the loop condition related to argc. Reading from argv[i] if no such argument has been given yields undefined behaviour.

user3344003 · Accepted Answer · 2017-04-13 18:35:46Z

0

argv [0] is the name of the program.

The arguments start at 1. You should also get in the habit of using argc in loops.

 int main(int argc, int *argv[]) 
 {
     for(int i = 1 ; i < argc ; ++ i ) 
     {
       int val = atoi(argv[i]);
       printf("%d", val);
     }
 }

answered Apr 13, 2017 at 18:35

user3344003

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