How to parse integer command line arguments in C?

Question

I would like a user to pass either two parameters or leave it blank. For instance:

./program 50 50

or

./program

When I tried to use int main(int argc, char *argv[]), first thing I have done was to change char *argv[] to int *argv[] but it did not work. What I want is from the user is just to enter two integers between 0 and 100. So if it is not two integers then it should give an error.

I was sort of thinking to give out an error with types (as I used to program on C#) but whatever I enter, argv[1] would be 'char' type all the time.

So what I have done is

for (int i = 0; i <= 100; i++) {
    //printf("%d", i);
    if (argv[1] == i) {
        argcheck++;
        printf("1st one %d\n", i);
    }
    else if (argv[2] == i) {
        argcheck++;
        printf("2nd one %d\n", i);
    }

This does not work as well. Also it gives warning when compiling, but if I change argv with atoi(argv[1]) for instance, then it gives a Segmentation fault (core dumped) error.

I need a simple way to solve this problem.

EDIT:

So I fixed with atoi(), the reason why it was giving segmentation fault was because I was trying it with null value when I have no parameter. So I fixed it up by adding an extra cond. But now the problem is if the value is let's say

./program asd asd

Then the output of atoi(argv[1]) would be 0. Is there a way to change this value?

Answer 1

You can not change the arguments of main() function. So you should just use atoi() function to convert arguments from string to integer.

atoi() has its drawbacks. Same is true for strtol() or strtoul() functions. These functions will return a 0 value if the input to the function is non-numeric i.e user enters asdf or anything other than a valid number. To overcome this you will need to parse the string before passing it to atoi() and call isdigit() on each character to make sure that the user has input a valid number. After that you can use atoi() to convert to integer.

Answer 2

Don't use atoi() and don't use strtol(). atoi() has no error checking (as you found out!) and strtol() has to be error-checked using the global errno variable, which means you have to set errno to 0, then call strtol(), then check errno again for errors. A better way is to use sscanf(), which also lets you parse any primitive type from a string, not just an integer, and it lets you read fancy formats (like hex).

For example, to parse integer "1435" from a string:

if (sscanf (argv[1], "%i", &intvar) != 1) {
    fprintf(stderr, "error - not an integer");
}

To parse a single character 'Z' from a string

if (sscanf (argv[1], "%c", &charvar)!=1) {
    fprintf(stderr, "error - not a char");
}

To parse a float "3.1459" from a string

if (sscanf (argv[1], "%f", &floatvar)!=1) {
    fprintf(stderr, "error - not a float");
}

To parse a large unsigned hexadecimal integer "0x332561" from a string

if (sscanf (argv[1], "%xu", &uintvar)!=1) {
    fprintf(stderr, "error - not a hex integer");
}

If you need more error-handling than that, use a regex library.

Answer 3

This will do:

int main(int argc, char*argv[])
{
   long a,b;
   if (argc > 2) 
   {
      a = strtol(argv[1], NULL, 0);
      b = strtol(argv[2], NULL, 0);
      printf("%ld %ld", a,b);
   }
   return 0;
}

Answer 4

The arguments are always passed as strings, you can't change the prototype of main(). The operating system and surrounding machinery always pass strings, and are not able to figure out that you've changed it.

You need to use e.g. strtol() to convert the strings to integers.

Answer 5

The things you have done so innocently are blunder in C. No matter what, you have to take strings or chars as your arguments and then later you can do whatever you like with them. Either change them to integer using strtol or let them be the strings and check each char of the string in the range of 48 to 57 as these are the decimal values of char 0 to 9. Personally, this is not a good idea and not so good coding practice. Better convert them and check for the integer range you want.

Answer 6

#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
main(int argc,char *args[])
{
    int i,sum=0;
    for(i=0;i<=argc;i++)
    {
        printf("\n The %d argument is: %s",i,args[i]);
    }
    printf("\nTHe sum of given argumnets are:");
    for(i=1;i<argc;i++)
    {
       int n;
       n=atoi(args[i]);
       printf("\nN=%d",n);
       sum += n;
    }
    printf("\nThe sum of given numbers are %d",sum);
    return(0);

  }

check this program i added another library stdlib.h and so i can convert the character array to string using function atoi.