Why does "for (i = 100; i <= 0; --i)" loop forever?

unsigned int i;
for (i = 100; i <= 0;开发者_开发技巧 --i)
    printf("%d\n",i);

Should be i >= 0 in the second condition in the loop if you want it to loop from 100 to 0.

That, and as others have pointed out, you'll need to change your definition of i to a signed integer (just int) because when the counter is meant to be -1, it will be some other positive number because you declared it an unsigned int.

Since i is unsigned, it will never be less than zero. Drop unsigned. Also, swap the <= for >=.

Since i is unsigned, the expression i <= 0 is suspicious and equivalent to i == 0.

And the code won't print anything, since the condition i <= 0 is false on its very first evaluation.

If the code is supposed to do nothing, nothing is wrong with it.

Assuming that you want it to print the loop index i from 100 to 1, you need to change i <= 0 to i > 0.

Because it is an unsigned int, you cant use i >= 0 because that will cause it to infinitely loop.

The loop checks for i <= 0;

i is never less-than-or-equal to zero. Its initial value is 100.

Technically nothing is wrong with that code. The test for i <= 0 is strange since i is unsigned, but it's technically valid, true when i is 0 and false otherwise. In your case i never happens to be 0.

I suspect you meant the test to be i > 0.

The <= 0 maybe? since it is false from the start

For loop

• init: i = 100
• test: i <= 0 // false on first pass

Change the test to i > 0 (100 times)

or i >= 0 (101 times) together with the declaration signed int i; so that it actually decreases down to -1. An unsigned int will go from 0 up to max-int (overflow).

If you want it to print all numbers from 100 down to 0 then you need

unsigned int i;
for (i = 100; i >= 0; --i)
    printf("%d\n",i);

The first time your loop ran in your original code, i was 100. The test '100 <= 0' failed and therefore nothing was showing.