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SE user DU Jiaen provided the following answer, edited by Martin Zabel and found here. The expression i >= 0 is always true if i is of an unsigned integer type. The alternative and simple way...
Answer
#1: Initial revision
by
aura-lsprog-86
·
2025-02-15T07:52:19Z (about 1 month ago)
_SE user [DU Jiaen](https://stackoverflow.com/users/3003290) provided the following answer, edited by [Martin Zabel](https://stackoverflow.com/users/5466118) and found [here](https://stackoverflow.com/a/35570713/5397930)._
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The expression `i >= 0` is always true if `i` is of an unsigned integer type. The alternative and simple way is to change it into:
uint64_t i;
for (i = 15; i != -1; i--) { ... }
This works no matter if i is a signed or an unsigned integer. In the `uint64_t` case, `-1` will first get converted into `0xFFFFFFFFFFFFFFFF` and then compared with `i`.
If you want to get rid of the compile warning, change it into:
i != (uint64_t)-1
but you need to ensure the `uint64_t` is exactly the type of `i`.