问题
unsigned int i;
for (i = 100; i <= 0; --i)
printf("%d\n",i);
回答1:
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.
回答2:
Since i is unsigned, it will never be less than zero. Drop unsigned. Also, swap the <= for >=.
回答3:
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.
回答4:
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.
回答5:
The loop checks for i <= 0;
i is never less-than-or-equal to zero. Its initial value is 100.
回答6:
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.
回答7:
I suspect you meant the test to be i > 0.
回答8:
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).
回答9:
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.
来源：https://stackoverflow.com/questions/4788298/why-does-for-i-100-i-0-i-loop-forever