为什么“for (i = 100; i <= 0; --i)”永远循环？
unsigned int i;
for (i = 100; i <= 0; --i)
printf("%d\n",i);
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unsigned int i;
for (i = 100; i <= 0; --i)
printf("%d\n",i);
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评论（9）
也许<= 0？因为从一开始它就是
falseFor 循环
将测试更改为
i >; 0（100 次）或
i >= 0（101 次）与声明signed int i;一起，使其实际上减少到 -1 。无符号 int 将从 0 到 max-int（溢出）。The <= 0 maybe? since it is
falsefrom the startFor loop
Change the test to
i > 0(100 times)or
i >= 0(101 times) together with the declarationsigned int i;so that it actually decreases down to -1. An unsigned int will go from 0 up to max-int (overflow).如果你希望它打印从 100 到 0 的所有数字，那么你需要
在原始代码中第一次运行循环时，i 是 100。测试“100 <= 0”失败，因此没有显示任何内容。
If you want it to print all numbers from 100 down to 0 then you need
The first time your loop ran in your original code, i was 100. The test '100 <= 0' failed and therefore nothing was showing.
如果你希望它从 100 循环到 0，则在循环的第二个条件中应该是
i >= 0。正如其他人指出的那样，你需要更改你的定义
i为有符号整数（只是int），因为当计数器为 -1 时，它将是其他正数，因为您将其声明为无符号int。Should be
i >= 0in 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
ito a signed integer (justint) because when the counter is meant to be -1, it will be some other positive number because you declared it anunsigned int.由于
i是无符号的，因此它永远不会小于零。删除未签名。另外，将<=替换为>=。Since
iis unsigned, it will never be less than zero. Dropunsigned. Also, swap the<=for>=.由于
i是无符号的，因此表达式i <= 0是可疑的，并且等价于i == 0。并且代码不会打印任何内容，因为条件
i <= 0在第一次计算时为 false。Since
iis unsigned, the expressioni <= 0is suspicious and equivalent toi == 0.And the code won't print anything, since the condition
i <= 0is false on its very first evaluation.如果代码不执行任何操作，那么它就没有问题。
假设您希望它打印循环索引
i从100到1，您需要将i <= 0更改为i > 0 。因为它是一个无符号整数，所以不能使用
i >= 0因为这会导致它无限循环。If the code is supposed to do nothing, nothing is wrong with it.
Assuming that you want it to print the loop index
ifrom 100 to 1, you need to changei <= 0toi > 0.Because it is an unsigned int, you cant use
i >= 0because that will cause it to infinitely loop.循环检查
i <= 0;i永远不会小于或等于零。其初始值为100。The loop checks for
i <= 0;iis never less-than-or-equal to zero. Its initial value is 100.从技术上讲，该代码没有任何问题。 i <= 0 的测试很奇怪，因为 i 是无符号的，但它在技术上是有效的，当 i 为 0 时为 true，否则为 false。在你的情况下，我永远不会碰巧是0。
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.
我怀疑你的意思是测试是
i > 0 。I suspect you meant the test to be
i > 0.