C: asterisk sequence using for-loops doesn't quite work

Minimunk7 0
Viewed 61

I'm trying to create an output of sequential asterisks using for-loops. The idea is to use inputs by the user to determine the number of rows and the increase of asterisks between each row. I can't make the program print asterisks beyond the first row and when I have been able to it's been the same amount of asterisks. What am I doing wrong? I know there's some other issues, I'd like to try them on on my own so my question is specifically about the output.

Only one row:

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

int main(int argc, char *argv[])
{
    int arg1, arg2, x, y, z;
    scanf("%d, %d", &arg1, &arg2);
    arg1 = atoi(argv[1]);
    z = arg2 = atoi(argv[2]);
    x = 0;
    if (argc < 2 || argc > 3)
    {
        printf("Wrong number of arguments, input 2 arguments.\n");
    }
    else
    {
        for (y = 1; y <= arg1; y ++)
        {
            for (; arg2 > 0; arg2 --)    
            {
                printf("*");
            }
        arg2 = arg2 + arg2;
        printf("\n");
        }
    printf("Total: %d\n", x);
    return 0;
    }
}

ubuntu@lab1:~$ gcc p4.c -o p4
ubuntu@lab1:~$ ./p4 4 5
*****



Total: 0
ubuntu@lab1:~$

Intended example output:

$ gcc p4.c
$ ./a.out 3 2
**
****
******
Totalt: 12
$ ./a.out 0 25
Totalt: 0
$ ./a.out 4 4
****
********
************
****************
Totalt: 40
$ ./a.out
Usage: ./a.out rows growth
$
printf nested-loops command-line-arguments for-loop c
2 Answers

For starters this if statement

if (argc < 2 || argc > 3)
{
    printf("Wrong number of arguments, input 2 arguments.\n");
}

must precede these statements

arg1 = atoi(argv[1]);
z = arg2 = atoi(argv[2]);

that must be moved into the else part.

Otherwise the program can invoke undefined behavior.

And moreover the condition is incorrect. It must look at least like

if ( argc != 3 )
{
    printf("Wrong number of arguments, input 2 arguments.\n");
}

The variable x is not changed within the program. So its output does not make a sense

x = 0;
//...
printf("Total: %d\n", x);

After the inner for loop

for (; arg2 > 0; arg2 --)    
{
    printf("*");
}

the variable arg2 is equal to 0.

So this statement

arg2 = arg2 + arg2;

keeps the value of arg2 equal to 0.

The program can look for example the following way

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

int main( int argc, char * argv[] )
{
    if (argc != 3)
    {
        puts( "Wrong number of arguments, input 2 arguments." );
    }
    else
    {
        int n = atoi( argv[1] );
        int m = atoi( argv[2] );

        if (!( 0 < n ) || !( 0 < m ))
        {
            puts( "The arguments must be positive integeres. Try again" );
        }
        else
        {
            const char c = '*';
            long long int total = n * ( n + 1 ) / 2 * m;

            for (int i = 0; i < n; i++ )
            {
                for (long long int j = 0; j < ( i + 1ll ) * m; j++)
                {
                    putchar( c );
                }
                putchar( '\n' );
            }

            printf( "Total: % lld\n", total );
        }
    }
}

If to pass to the program the values 4 and 5 then the program output will look like

*****
**********
***************
********************
Total:  50

Within this for loop

for (long long int j = 0; j < ( i + 1ll ) * m; j++)

the variable j declared as having the type long long int instead of just int because the prodxct of n * m can be a big number that can not fit in an object of the type int.

Vlad from Moscow0

Have a look at this:

int main() { /* OP: "...so my question is specifically about the output." */

    // To generate output, use some values
    int rows = 5; // = atoi( argv[ 1 ] );
    int grow = 4; // = atoi( argv[ 2 ] );
    int star = grow;
    int total = 0;

    // in lieu of a command line...
    printf( "rows: %d, growth: %d\n", rows, grow );

    while( rows-- ) {

        // output the stars for this row
        for( int n = 0; n < star; n++ )
            putchar( '*' );
        putchar( '\n' );

        total += star; // accumulate a total

        star += grow; // and increase for the next loop
    }

    printf( "Total: %d\n", total );

    return 0;
}

Output

rows: 5, growth: 4
****
********
************
****************
********************
Total: 60
Fe2O30
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