Why the loop becomes infinite?

Question

How to get out of an infinite loop?

unsigned char half_limit = 130;

for (unsigned char i = 0; i < 2 * half_limit; ++i)
{
    //smth is happening;
}

Help, please.

Thank you!

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the [How to Ask](https://stackoverflow.com/help/how-to-ask) page for help clarifying this question. — letsintegreat, Feb 19 '20 at 05:21

score 4 · Answer 1 · edited Feb 19 '20 at 05:32

4

Make your loop variable an int.

unsigned char can't exceed 255, so incrementing i past that will wrap it to 0.

2*130 is 260, because the type of literal 2 is int, and multiplying an int by unsigned char you get an int.

Thus, when i is an unsigned char, your loop termination condition will never be satisfied since i will always be less than 260, hence the infinite looping.

edited Feb 19 '20 at 05:32

Remy Lebeau

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answered Feb 19 '20 at 05:22

aparpara

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score 1 · Answer 2 · answered Feb 19 '20 at 05:18

1

How to get out of an infinite loop?

Paradoxically, you cannot. Because if you get out of a loop, then the loop was finite.

answered Feb 19 '20 at 05:18

eerorika

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1

Why y'all downvoting this answer. It is correct. Honestly, this question is bad. – Daniel Feb 19 '20 at 05:21
2

@Daniel this answer is technically correct, but misses the spirit of the question. The asker is obviously a beginner, and may be lacking in terminology, or may not be a native english speaker. This would be better as a clarifying comment or question under the main question. We can all take a moment to laugh at the cleverness of this non-answer, but it does little to help OP. – JohnFilleau Feb 19 '20 at 05:27
1

@John `it does little to help OP` I disagree. It is arguably more useful to learn the language, terminology and the philosophy around the puzzle than to be merely told the solution. – eerorika Feb 19 '20 at 13:19

score 0 · Answer 3 · answered Feb 19 '20 at 05:26

I suggest you take a look at this post:

What is an unsigned char?

Or you could try this approach:

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

int main(){
        int half_limit = 130;
        int i;
        for (i = 0; i < 2 * half_limit; ++i){
                printf("%d\n", i);
        }
        return 0;
}

score 0 · Answer 4 · answered Feb 19 '20 at 05:33

As i is of type unsigned char which is of size 1byte and can hold 0 to 255 values. If you increment i after 255 (i.e 0xFF) expectation is i to become 256 but i hold value 0 due overflow. (i.e in 256 is equivalent to 0x100 and only lower byte value will be available in i)

In order to run loop from 0 to 260 you need to change the type of i which can hold value beyond 255.

short int or int is preferable,

unsigned char half_limit = 130;

for (int i = 0; i < 2 * half_limit; ++i)
{
    //smth is happening;
}

score 0 · Answer 5 · answered Feb 19 '20 at 05:34

0

You're having an infinite loop because of unsigned char range which is from 0 to 255. So basically, i never reaches 2 * half_limit. i starts from 0 then goes to 255 then goes to 0 again and so on and so forth.

answered Feb 19 '20 at 05:34

Michaël Randria

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Why the loop becomes infinite?

5 Answers5