How to read command-line arguments as integers instead of strings?

Question

Sorry for my question, I know there are a lot similars but I didn't found any that is simple enaugh to help me.

I've started coding in C and try to solve a simple exercise: Read an integers array from command line, sum the elements using the function array_sum and print result. (input example array of 3 elements: 3 0 1 2)

int array_sum(int *array,  size_t size);
int main(int argc, char **argv){
    int sum=array_sum(argv, argc);
    printf("array_sum: %i\n", sum);
    return 0;
}

my problem is that argv is a char array and the function want an integer array. Should I convert elements one by one in a new int array? There are better ways?

`Should I convert elements one by one in a new int array?` Yes. — Code-Apprentice, Feb 12 '17 at 10:48
@VladfromMoscow the first element was intended the lenght of the array — adev, Feb 12 '17 at 12:34

StoryTeller - Unslander Monica · Accepted Answer · 2017-02-12T10:58:36.647

5

argv is an array of pointers to C strings. You need to convert the strings into integers first. You can do something like this:

int array_sum(int *array,  size_t size);
int main(int argc, char **argv){
    int *num_arr = malloc((argc - 1) * sizeof *num_arr);

    for (int i = 0; i < argc - 1; ++i)
      num_arr[i] = atoi(argv[i+1]);

    int sum = array_sum(num_arr, argc - 1);
    printf("array_sum: %i\n", sum);

    free(num_arr);
    return 0;
}

The only way to make the code in main shorter is by moving the conversion loop into a separate function that returns the malloced pointer.

edited Feb 12 '17 at 10:58

answered Feb 12 '17 at 10:45

StoryTeller - Unslander Monica

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Thanks, there no need to use malloc because I know the array's lenght – adev Feb 12 '17 at 10:49
1

@adev - If you refer to `argc` then the array length is variable, and you most certainly need `malloc`. Although you can get away with a variable length array if `argc` isn't "too big". – StoryTeller - Unslander Monica Feb 12 '17 at 10:50
The call of `array_sum(argv, argc)` would not compile. Presumably you intended `array_sum(num_arr, (size_t)argc-1)`. – Peter Feb 12 '17 at 10:57
@Peter - I did, thank you. Got distracted before finishing the example proper. – StoryTeller - Unslander Monica Feb 12 '17 at 10:58

score 2 · Answer 2 · edited May 23 '17 at 12:10

In your code, char *argv[] is an array of char* pointers supplied from the command line. In order to convert the numbers supplied, you can use the following:

atoi(), which converts string arguement to an integer type.
Or strtol(), which converts the initial part of a string to a long int, given a base.
Other special functions from C99, alot of which are described in this post.

Since atoi() has no error checking, it is best to use strtol(), which allows extensive error checking.

You should store these converted numbers in a dynamically allocated int* pointer, which will need to be allocated on the heap using malloc(), which was suggested by @StoryTeller in his answer. You could also just declare an array on the stack, such as int arr[n]. The problem arises when you want to return this array in a function, which is not possible. Using a pointer in this case would allow more flexibility for abstraction.

malloc()allocates block of memory on the heap, and returns a void* pointer to it.

Note: malloc() should always be checked, as it can return NULL. You need to also free() this pointer at the end.

Here is some example code:

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

#define BASE 10

/* Guessed that your function would look like this */
int array_sum(int *array, size_t size) {
    int sum = 0;

    for (size_t i = 0; i < size; i++) {
        sum += array[i];
    }

    return sum;
}

int main(int argc, char *argv[]) {
    int *arr = NULL;
    char *endptr = NULL;
    int check, sum;
    size_t ndigits = (size_t)argc-1;

    /* allocate pointer */
    arr = malloc(ndigits * sizeof *arr);
    if (arr == NULL) {
        fprintf(stderr, "Cannot %zu spaces for integers\n", ndigits);
        exit(EXIT_FAILURE);
    }

    for (size_t i = 0; i < ndigits; i++) {

        /* sufficient checking for strtol(), more can possibly be added here */
        check = strtol(argv[i+1], &endptr, BASE);
        if (endptr != argv[i+1] && *endptr == '\0') {
            arr[i] = check;
        }
    }

    sum = array_sum(arr, ndigits);

    printf("array_sum: %d\n", sum);

    /* pointer is free'd */
    free(arr);
    arr = NULL;

    return 0;
}

Example input:

$ gcc -Wall -Wextra -std=c99 -o sumcommands sumcommmands.c
$ ./sumcommands 3 2 1

Output:

array_sum: 6

Note: You can use more error checking for strtol() on the Man page.

Jarvis · Answer 3 · 2017-02-12T10:54:12.103

1

Why do you need to pass an int array as argument to the function ? No need to create an extra int array when you can simply do this :

int array_sum(char **argv, int argc){
  int sum = 0;
  for(int i = 0;i < argc - 1;i++){
    sum += atoi(argv[i])
  }
  return sum;
}

edited Feb 12 '17 at 10:54

answered Feb 12 '17 at 10:50

Jarvis

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True, but that's only viable if the requirement of using `array_sum` can be relaxed – StoryTeller - Unslander Monica Feb 12 '17 at 10:52
yes @StoryTeller, I must use provided function. The purpose of exercise I think is the interfacing – adev Feb 12 '17 at 10:54

score 0 · Answer 4 · answered Feb 12 '17 at 10:53

0

You can use atoi() function to convert char ** array to **int . what i see here is each integer you type is converting into string rather than char.

answered Feb 12 '17 at 10:53

Prithvi Raj

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How to read command-line arguments as integers instead of strings?

4 Answers4