[Solved] Why does the following loop execute

Question

Since tmp.size() is 1, subtracting 3 from it produces a negative value. That is, subtraction would produce a negative value if it weren’t for tmp.size() being size_t, which is unsigned.

When an unsigned is about to go negative on subtraction, it “wraps around” to a very large value. That is why your loop fails to stop. It also explains the “signed to unsigned comparison” warning produced by the compiler^*, which should have put you on high alert.

To avoid this problem, move subtraction to the other side, and make it addition:

for (size_t i = 0 ; i+3 <= tmp.size() ; i += 2) {
    ...
}

^* If you did not get that warning, change the settings on your compiler to warn you about everything. This saves a lot of time spent in debugging many common problems that are easily detectable by compilers.

Accepted Answer

Since tmp.size() is 1, subtracting 3 from it produces a negative value. That is, subtraction would produce a negative value if it weren’t for tmp.size() being size_t, which is unsigned.

When an unsigned is about to go negative on subtraction, it “wraps around” to a very large value. That is why your loop fails to stop. It also explains the “signed to unsigned comparison” warning produced by the compiler^*, which should have put you on high alert.

To avoid this problem, move subtraction to the other side, and make it addition:

for (size_t i = 0 ; i+3 <= tmp.size() ; i += 2) {
    ...
}

^* If you did not get that warning, change the settings on your compiler to warn you about everything. This saves a lot of time spent in debugging many common problems that are easily detectable by compilers.