Thread: Array problem

Do you know how to access any members of an array? If so, it shouldn't be that hard to figure out how to get the first two, should it?

Code:

foo = array[ elementnumber ];

Remember that arrays begin their indexing at 0, and not 1. Thus, if your array size is 10, you have accessable elements 0 - 9.

You could convert the number to a string, and then look at the last four digits. Or, you could use some math operators, such as the modulus operator to in effect mask off the last four digits. Or you could use division. For example, and I'm doing this because it's funner than simply using modulus, you could do this:

Code:

int foo = 12345679;
int bar = foo / 10000;
int baz = bar - foo * 10000;

printf( "baz is %d.\n", baz );

See, I could just show you the easy way, but this is funner. Anyway, that looks about right. If it's off, blame mixed drinks.


Quzah.

Originally Posted by bithub

Damn mixed drinks...

Thanks. I'll take out my wrath on this one here.

"Damn you, mixed drink!" *glug glug* "Damn you!"

Originally Posted by Quzah

See, I could just show you the easy way, but this is funner. Anyway, that looks about right. If it's off, blame mixed drinks.

Or maybe that was just part of the "funner" I was making it!


Quzah.

You're probably not doing it right then...

Code:

#include<stdio.h>
int main( void )
{
    int foo[] = { 1234567, 2345678, 3456789, 0x0 };
	int bar, baz, x;
	
	for( x = 0; foo[ x ]; x++ )
	{
	    bar = foo[ x ] / 10000;
            baz = foo[ x ] - bar * 10000;
            printf("baz = %d\n", baz );
	}
	
	for( x = 0; foo[ x ]; x++ )
	{
	    printf("foo[ x ] %% 10000 = %d\n", foo[ x ] % 10000 );
	}
	
	return 0;
}

Quzah.