Thread: while loop giving me a hard time

Hi I'm new to C and programming generally and have run into an obstacle with an easy while loop.

I want to add the same value to the variable spoj1min until I get the remainder of division 7.

spoj1min variable has an initial value of 17
spoj2min variable has an inital value of 12

could you tell me what is wrong with the below code that it doesn't increment the value by 17 for each cycle but keeps outputing 0 infinetly?

Thanks a lot for any help

Code:

while(x != 7){
    x = spoj1min % spoj2min;
    spoj1min = spoj1min + spoj1min;
    printf("%d\n", spoj1min,);

First, your code is not complete. I assume that it look like this:

Code:

#include <stdio.h>

int main (void) {

    int spoj1min = 17;
    int spoj2min = 12;
    int x = 0;

    while(x != 7) {
           x = spoj1min % spoj2min;
        spoj1min = spoj1min + spoj1min;
        printf("%d\n", spoj1min);
    }

    return 0;
}

The problem is in this line:

Code:

spoj1min = spoj1min + spoj1min;

This will double the value of spoj1min in every iteration and this will result in an overflow.
You can check this with an emergency break:

Code:

#include <stdio.h>

int main (void) {

    int spoj1min = 17;
    int spoj2min = 12;
    int x = 0;
    int em_brk = 0;

    while(x != 7) {
           x = spoj1min % spoj2min;
        spoj1min = spoj1min + spoj1min;
        printf("%d\n", spoj1min);
        if (++em_brk > 50) break;
    }

    return 0;
}

The program will stop and you can scroll up and see was happens.
I think you want increase spoj1min by 17 in every iteration.
The best way is to assing a new variable (call it 'add_on') and give it the value (in your case: 17).
In the loop add 'add_on' to 'spoj1min' in every iteration.
Then calculate the reminder, print booth out and finish.
If the reminder (x) is 7, the loop will stop.

Leave the emergency break in plaece. It could be that you give numbers that never fulfilled the break contion of the loop.

Originally Posted by RagingGrim

Unrelated question , which is faster
X += X;
Or X = 2*X;

Next time post in a new thread.

Anyway, after optimisations have been applied by the compiler, it is plausible that the resulting executable code might be exactly the same.

Code:

for(spoj1min = 17; (spoj1min % spooj2min) != 7; spoj1min += 17)

Do that instead.
It was hard for me to manage what you actually wanted, but I guess its that ^

PHP Code:

spoj1min: 17 | spoj1min % 12: 3
spoj1min: 34 | spoj1min % 12: 6
spoj1min: 51 | spoj1min % 12: 2
spoj1min: 68 | spoj1min % 12: 5
spoj1min: 85 | spoj1min % 12: 1
spoj1min: 102 | spoj1min % 12: 4
spoj1min: 119 | spoj1min % 12: 0
spoj1min: 136 | spoj1min % 12: 3
spoj1min: 153 | spoj1min % 12: 6
spoj1min: 170 | spoj1min % 12: 2