Solution for Exercise 08_06-allargsp.c

08_06-allargsp.c

#include <stdio.h>

int main( int argc, char **argv )
{
	int a,b;

	for( a=0; a<argc; a++ )
	{
		b = 0;
		/* the ugly expression is equivalent to
		   argv[a][b]
		 */
		while( *(*(argv+a)+b) )
		{
			putchar( *(*(argv+a)+b) );
			b++;
		}
		putchar('\n');
	}

	return 0;
}

Output

./a.out
this
is
a
test

Notes

* For the sample output (above), the command line typed is: ./a.out this is a test

* This solution involves using a nested loop. Pointer variable a counts the arguments and represents the address of each string typed at the command prompt. Pointer variable b represents the characters in the string, each one.

* Both pointers, a and b, are addresses. The character in each string is fetched by using this contraption: *(*(argv+a)+b) It represents individual character in each string in a pointer-pointer (char) array.

* The while loop continues until the value fetched from (*(argv+a)+b) is a null character. Remember that (*(argv+a)+b) is an address. Only when it's dereferenced — *(*(argv+a)+b) — is a character fetched.

Exercise 08_08-daysweek.c →