How do I?

[al_engler]

How can I reduce this to having only one function perform the job?

#include <stdio.h>

int i,g,q,p,c;
int liquid1 (int);
int liquid2 (int);
int liquid3 (int);
int liquid4 (int);
int main()
{
printf("Enter cups to be converted:");
scanf ("%d\n",&i);
printf ("gallons = %d",liquid1 (i));
printf ("quarts = %d",liquid2 (i));
printf ("pints = %d",liquid3 (i));
printf ("cups = %d",liquid4 (i));

system ("PAUSE");

return 0;
}

int liquid1 (int i)
{
int i,g;
g = i/16;
return (g);
}
int liquid2 (int i)
{
int i,q;
q=(i%16)/4;
return (q);
}
int liquid3 (int i)
{
int i,p;
p=((i%16)%4)/2;
return (p);
}
int liquid4 (int i)
{
int i,c;
c=((i%16)%4)%2;
return (c);
}

[ssharish2005]

#include <stdio.h>

int liquid1(int, int);  // have a choice as a second parameter

int main()
{
    int i,g,q,p,c;   // dont use global variables
    
    printf("Enter cups to be converted:");
    scanf ("%d\n",&i);

    printf ("gallons = %d",liquid1 (i, 1));
    printf ("quarts = %d",liquid1 (i,2));
    printf ("pints = %d",liquid1 (i,3));
    printf ("cups = %d",liquid1 (i,4));

    getchar(); // use getchar rather than system & PAUSE
    return 0;
}

int liquid1 (int i, int ch)
{
    int g;
    
    switch(ch)  // use switch to switch to the correspsonding case
    {
        case 1:
            g = i/16;
            break;
        case 2:
            g=(i%16)/4;
            break;
        case 3:
            g=((i%16)%4)/2;
            break;
        case 4:
            g=((i%16)%4)%2;
            break;
    }
    return g;    // return the result
}

u need proper indendation

ssharish2005

[al_engler]

How can I do it using pointers, the bane of my exsistance?

[ssharish2005]

well dint notice one thing in u'r code

scanf ("%d\n",&i);

u dont need that \n there. Which bascially means u got to enter the \n when every u enter a value for i.

How can I do it using pointers, the bane of my exsistance?

So what was your actual problem given to u. Which part need to be in pointer.

ssharish2005

[twomers]

Very simple example:

void pointer_function ( int *value )
{
	*value /= 10;
}

int main( void )
{
    int num = 4235232;

	printf ( "Before Function: %d\n", num );
	pointer_function ( &num );
	printf ( "After Function:  %d\n", num );
    
    return 0;
}

Just adapt that to your function and it should work.

[ssharish2005]

but note your actual i value will be altered. so for example if your i value had 10 and when u call i with the liquid1 function the i value becomes 0 ( (int) 10/16 == 0.652). And when u send the i value to liquid2 u are sending i which is now 0

ssharish2005

[CornedBee]

Your code will either fail to compile or produce undefined results, because your liquid* functions all contain a local variable i that either masks or conflicts with the parameter i.

That problem aside, all functions do different things. There is no sensible way to make them one function. (I do hope what ssharish posted was meant sarcastically.)

Of course, since all functions also are simply one calculation, you can substitute the calls with the expression. Oh, and (x % 2) is the same thing as ((x % 16) % 2) or stuff like that.

[al_engler]

Thanks for the help

[dwks]

Also, using system("PAUSE") is a bad idea.

• http://faq.cprogramming.com/cgi-bin/...&id=1043284385
• which leads you to http://faq.cprogramming.com/cgi-bin/...&id=1043284385

If you must use it, at least include <stdlib.h>.