C command line arguments to int

Question

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

int myatoi(char *str) {
    int i, res = 0;
    for (i = 0; str[i] != '\0'; ++i) {
        res = res * 10 + str[i] - '0';
        return res;
    }
}

int main(int argc, char *argv[]) {
    int j;
    char str[] = " i want to get argv[2] value here";
    int val = myatoi(str);
    printf("%d", val);
    if (argc == 3) {
        for (j = 0; j < val; j++) {
            printf("Hello, %s \n", argv[1]);
        }
    } else
    if (argc < 3) {
        printf("Not Enough Arguments\n");
    } else
    if (argc > 3) {
        printf("Too Many Arguments\n");
    }
    return 0;
}

How do I pass my value of argv[2] to char str[] so that it converts string to int and print the output.

argv[0] will be file name, argv[1] will be string, argv[2] should be int

example output: ./hello raju 2

Hello, raju!
Hello, raju!

as argv[2] is 2 it should print 2 times.

You overlooked the 0 output from the line printf("%d",val); — Weather Vane
– Weather Vane, Commented Feb 4, 2017 at 20:54

P.P · Accepted Answer · 2017-02-06 18:16:54Z

2

You can directly pass argv[2] to myatoi() or use a variable:

 char *str = argv[2];
 int val=myatoi(str);

You are also better off adding an input check before using argv like:

if (argc != 3) {
   printf("Expected 2 args\n");
   exit(1);
}

Your myatoi() suffers the similar problems like the atoi() standard function -- lacks the ability to detect and report bad inputs. Consider what happens if argv[2] is "xyz123" for example.

edited Feb 6, 2017 at 18:16

answered Feb 4, 2017 at 20:53

P.P

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1 Comment

raju Over a year ago

what should i do for check for bad input such as xyz123 @usr

chqrlie · Accepted Answer · 2017-02-04 21:54:52Z

1

Your function myatoi is broken: you should move the return statement out of the for loop.

Here is a corrected version:

int myatoi(const char *str) {
    int i, res = 0;
    for (i = 0; str[i] != '\0'; ++i) {
        res = res * 10 + str[i] - '0';
    }
    return res;
}

Be aware that your function only handles strings composed exclusively on digits. No sign, no spaces no other characters are allowed. You could use more flexible standard functions such as atoi or strtol declared in <stdlib.h>

Here is a simplified version of your code:

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

int main(int argc, char *argv[]) {
    if (argc == 3) {
        int val = atoi(argv[2]);
        for (int j = 0; j < val; j++) {
            printf("Hello, %s\n", argv[1]);
        }
        return 0;
    } else {
        printf("error: 2 arguments expected\n");
        return 1;
    }
}

answered Feb 4, 2017 at 21:54

chqrlie

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2 Comments

raju Over a year ago

thanks but i should not use built in functions for this program.is any other way of doing it,instead of my code @chqrlie

chqrlie Over a year ago

@raju: If you cannot use the library function atoi(), use the modified myatoi() I posted above.

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