How do I store argv command line inputs in an array (C)

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Trying to input the argv command line inputs into an array.
All it's printing out is the memory location and not the argv inputs

What I have tried:

C

#define MAX 12

int main(int argc, char* argv[])
{
  char* argv_inputs[MAX];
  int array_of_ints[MAX];
  int i, size = argc;

  for(i = 2; i <= argc; i++)
  {
    argv_inputs[i] = argv[i];
    printf("%d ", argv_inputs[i]);
  }

Posted 11-Sep-17 4:11am

Member 13404013

Updated 16-Sep-21 22:58pm

CPallini

v2

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2 solutions

Solution 2

You don't need a copy of the argv input array but have to convert the textual arguments to numbers using atoi - C++ Reference[^] or strtol - C++ Reference[^] :

int main(int argc, char* argv[])
{
    int array_of_ints[MAX];
    int i;
    
    for (i = 1; i < argc && i < MAX; i++)
    {
        array_of_ints[i] = atoi(argv[i]);
        printf("%d ", array_of_ints[i]);
    }
    return 0;
}

Note that the above does not check if the input is a valid number and that I have corrected the loop. With C/C++, arrays start at the index zero and the last item is therefore accessed by size - 1.
[EDIT]
I have also added a check to ensure that no out-of-bound accesses occur.
[/EDIT]

Posted 11-Sep-17 4:29am

Jochen Arndt

Updated 11-Sep-17 4:31am

v2

Comments

CPallini 11-Sep-17 11:01am

5.Skipping the array item 0 in order to mimic VB? :-)

Jochen Arndt 11-Sep-17 11:27am

Thank you.

Skipping argv[0] as program path. I was thinking about storing starting at index zero but he has used the same index for both and so I decided to do it too.

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OriginalGriff · Accepted Answer · 2017-09-11T04:16:00

The input is an array of pointer-to-char values, and you are assigning them to an array of integers. That won't work: while the pointers are integer values (or long integer values in a 64 bit application, the numbers themselves are meaningless - they are the addresses of the first character in each string, so when you print them you get numbers which make no sense at all.
Try this:

int main(int argc, char* argv[])
    {
    int i;
    for(i = 2; i < argc; i++)
        {
        printf("%s\n", argv[i]);
        }
    }

That will print the inputs.