Accepting Integer Array from Command Line - C Programming

Question

I need to accept an array of integers from the command line, such as: ./test 2 9 -5 17

Currently I have:

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

And then loops to find the two numbers within this array that create the maximum value possible starting with argv[1]. Obviously this does not work. Do I need it to be:

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

and then parse this? I am not sure how to covert characters to integers in C.

Thanks,

I am not sure how to covert characters to integers in C. atoi() or strtol() will work to convert a string to an int — ryyker
– ryyker, Commented Apr 11, 2016 at 18:30
There is no range. Any integers, positive and negative, large/small, are to be accepted. — JDeffo
– JDeffo, Commented Apr 11, 2016 at 18:31
JDeffo: "no range. Any integers" --> So values like 123456789012345678901234567890 are acceptable? — chux
– chux, Commented Apr 11, 2016 at 18:40
@Mike Holt OP may want to limit code's operation to integers within the range of some C integer type like int of long long or maybe not. It would be illuminating to know OP's thoughts on the coding goals. — chux
– chux, Commented Apr 11, 2016 at 18:47

FredK · Accepted Answer · 2016-04-11 18:32:40Z

1

Use strtol() to convert each argv[i] (for i starting at 1) to a long int.

answered Apr 11, 2016 at 18:32

FredK

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lost_in_the_source · Accepted Answer · 2016-04-11 18:51:03Z

0

int main(int argc, char *argv[])
{
    size_t i;
    int n;

    for (i = 0; i < argc; ++i) {
        if (sscanf(argv[i], "%d", &n) == 0) {
            fprintf(stderr, "Position %d is not an int\n", i + 1);
            return 1;
        }
        printf("%d\n", n);
    }

    return 0;
}

answered Apr 11, 2016 at 18:51

lost_in_the_source

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