How to convert a string of numbers (characters) to integer values in c

atoi is a good function to use, however you can write your own atoi function in C.

int xatoi(char *s)
{
   int result=0;       

   while(*s)
   {

     result=result*10+(*s-48);
     s++; 
    }

  return result;

} the logic behind this is every character '1','2',......'s ascii value is stored for example '1' has ASCII value '49'. please compile above program and check for errors i haven't tested it but i am sure it will work.

please see the link below for your reference http://www.newebgroup.com/rod/newillusions/ascii.htm

argv is an array of string arrays and so argv[N] gives you an array. But argv[N][J] gives you element at position J of the array stored in argv[N]. However, it is strange that compiler did not complain about your code because atoi() function, in fact, expects a pointer (to the null-terminated string) and not a character value. The compiler should have said something like warning: passing argument 1 of ‘atoi’ makes pointer from integer without a cast. I recommend you not to ignore warnings. Personally I tend to compile the code with the highest warning level and not have a single warning, though our codebase at work is many millions of lines of code.

Also, prefer to use strtol over atoi. The problem with atoi is that it ignores invalid input (i.e. just returns you 0 if there is no problem, without affecting errno).

Anyway, here is an example of how to grab an array of base 10 integers from a command line:

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

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

    if (argc < 2) {
        fprintf(stderr, "Please specify some numbers\n");
        return EXIT_FAILURE;
    }

    n = argc - 1;
    v = alloca(sizeof(int) * n);

    for (i = 1; i < argc; ++i) {
        v[i-1] = strtol(argv[i][i], &e, 10);
        if (!v && errno) {
            fprintf(stderr, "Cannot convert '%s' into number: %s\n",
                argv[i], strerror(errno));
            return EXIT_FAILURE;
        } else if (*e != '\0') {
            fprintf(stderr, "%s is not a number\n", argv[i]);
            return EXIT_FAILURE;
        }
    }

    printf("You have entered the following numbers: %d", v[0]);

    for (i = 1; i < n; ++i)
        printf(", %d", v[i]);
    printf("\n");

    return EXIT_SUCCESS;
}

Hope it is useful.