Thread: Problem with while cycle

Here it is the code:

Code:

#include <stdio.h>

int main() {

	float cafe=0.27, troco, d_int=2.00;
	troco=d_int-cafe;

	printf("Moedas do troco:\n");

	while (troco>0) {

		if ((troco/2.00)>=1.00) {
			printf("1x 2 euros\n");
			troco-=2.00;
		}
		else if ((troco/1.00)>=1.00) {
				printf("1x 1 euro\n");
				troco-=1.00;
			}
			else if ((troco/0.50)>=1.00) {
					printf("1x 50 cêntimos\n");
					troco-=0.50;
				}
				else if ((troco/0.20)>=1.00) {
					printf("1x 20 cêntimos\n");
					troco-=0.20;
					}
					else if ((troco/0.10)>=1.00) {
						printf("1x 10 cêntimos\n");
						troco-=0.10;
						}
						else if ((troco/0.05)>=1.00) {
							printf("1x 5 cêntimos\n");
							troco-=0.05;
							}
							else if ((troco/0.02)>=1.00) {
								printf("1x 2 cêntimos\n");
								troco-=0.02;
								}
								else if ((troco/0.01)>=1.00) {
									printf("1x 1 cêntimo\n");
									troco-=0.01;
									}
	}
	return 0;
}

The variable "troco" reaches 0.00, however the while cycle doesn't break on "troco>0" condition, so it enters in an infinite cycle. I can't tell why.

Here is your program with some printf statements that will output what troco is holding at each point in your program. Take a look at the values you are getting when you run it and see if you can figure anything out by this. This is called debugging, and as much as it can be a pain it will help you learn a lot.

Code:

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

int main() {

	float cafe=0.27, troco, d_int=2.00;
	troco=d_int-cafe;

	printf("Moedas do troco:\n");

	while (troco > 0) {

		if ((troco/2.00)>=1.00) {
			printf("1x 2 euros\n");
			printf("%d\n", troco);
			troco-=2.00;
		}
		else if ((troco/1.00)>=1.00) {
				printf("1x 1 euro\n");
				printf("%d\n", troco);
				troco-=1.00;
			}
			else if ((troco/0.50)>=1.00) {
					printf("1x 50 cêntimos\n");
					printf("%d\n", troco);
					troco-=0.50;
				}
				else if ((troco/0.20)>=1.00) {
					printf("1x 20 cêntimos\n");
					printf("%d\n", troco);
					troco-=0.20;
					}
					else if ((troco/0.10)>=1.00) {
						printf("1x 10 cêntimos\n");
						printf("%d\n", troco);
						troco-=0.10;
						}
						else if ((troco/0.05)>=1.00) {
							printf("1x 5 cêntimos\n");
							printf("%d\n", troco);
							troco-=0.05;
							}
							else if ((troco/0.02)>=1.00) {
								printf("1x 2 cêntimos\n");
								printf("%d\n", troco);
								troco-=0.02;
								}
								else if ((troco/0.01)>=1.00) {
									printf("1x 1 cêntimo\n");
									printf("%d\n", troco);
									troco-=0.01;
									}
	}
	system("PAUSE");
	return 0;
}

Using a non-integral value to control a while or for loop is dangerous. Your problem is that troco doesn't always reach exactly 0. In the floating-point representation it may reach something like 0.0000000011 which you see as 0 but is still greater than 0 for purposes of the boolean comparison.

Instead, try using a value that's small enough for purposes of your application but large enough to rule out tiny round-off differences. For example

Code:

while (troco > 0.000001) {

I did that, however I did not paste it here.

The last printf() gives troco=0. That's why I don't get it...

Originally Posted by R.Stiltskin

Using a non-integral value to control a while or for loop is dangerous. Your problem is that troco doesn't always reach exactly 0. In the floating-point representation it may reach something like 0.0000000011 which you see as 0 but is still greater than 0 for purposes of the boolean comparison.

Instead, try using a value that's small enough for purposes of your application but large enough to rule out tiny round-off differences. For example

Code:

while (troco > 0.000001) {

Oh, that's right, it worked. Thanks!

By the way, the reason your printf statements didn't reveal this is that you were printing troco as an integer ( "%d" ) or as a float but with only a few decimal places. Try printing it as a floating-point value, and specify the decimal precision, e.g.

Code:

printf("%.32f\n", troco);

and you will see more of what's actually there.