C program read input from command line

Question

I'd like this program to read instructions from the command line instead. It's a number converter for octal -> binary and binary -> octal. Can't I just tell it before running what to convert to? For example ./conv octal binary 11 would yield octal 3. I've run linux programs in the past where I used ./script input ouput.

#include <stdio.h>
#include <math.h>
int binary_octal(int n);
int octal_binary(int n);
int main()
{
    int n;
    char c;
    printf("Instructions:\n");
    printf("Enter alphabet 'o' to convert binary to octal.\n");
    printf("2. Enter alphabet 'b' to convert octal to binary.\n");
    scanf("%c",&c);
    if ( c=='o' || c=='O')
    {
        printf("Enter a binary number: ");
        scanf("%d",&n);
        printf("%d in binary = %d in octal", n, binary_octal(n));
    }
    if ( c=='b' || c=='B')
    {
        printf("Enter a octal number: ");
        scanf("%d",&n);
        printf("%d in octal = %d in binary",n, octal_binary(n));
    }
    return 0;
}
int binary_octal(int n)  /* Function to convert binary to octal. */
{
    int octal=0, decimal=0, i=0;
    while(n!=0)
    {
        decimal+=(n%10)*pow(2,i);
        ++i;
        n/=10;
    }

/*At this point, the decimal variable contains corresponding decimal value of binary number. */

    i=1;
    while (decimal!=0)
    {
        octal+=(decimal%8)*i;
        decimal/=8;
        i*=10;
    }
    return octal;
}
int octal_binary(int n)  /* Function to convert octal to binary.*/
{
    int decimal=0, binary=0, i=0;
    while (n!=0)
    {
        decimal+=(n%10)*pow(8,i);
        ++i;
        n/=10;
    }
/* At this point, the decimal variable contains corresponding decimal value of that octal number. */
    i=1;
    while(decimal!=0)
    {
        binary+=(decimal%2)*i;
        decimal/=2;
        i*=10;
    }
    return binary;
}

abelenky · Accepted Answer · 2014-09-29 21:01:59Z

3

Command line options are passed to main via argc and argv.

int main(int argc, char** argv)
{
    for(int i=0; i<argc; ++i)
    {
        printf("Option %d is %s\n", i, argv[i]);
    }
}

Try running just that mini-program with a variety of command-line params, you'll figure it out quickly!

answered Sep 29, 2014 at 21:01

abelenky

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1 Comment

pts Over a year ago

argv[0] is usually the name of the program, as typed on the command line.

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