3 ms·

Uh I'm confused but don't know c++. why doesn't that loop end instantly? I mean length - 1 < length should always be true, right? Or does it only terminate w

Uh I'm confused but don't know c++. why doesn't that loop end instantly? I mean length - 1 < length should always be true, right? Or does it only terminate when the number underflows? Terribly confused here

msk-lywenn 4y ago
It’s a condition to run, not a condition to stop
- wizofaus 4y ago
  That was my reaction - why should anyone think it might be the latter? Are there languages that do have such a syntax without explicit keywords ("do...until")?
  - msk-lywenn 4y ago
    I think lisp or scheme does. I was often confused by that when I was playing with it
bobthepanda 4y ago
It‘s an unsigned int, so past 0 it overflows back to the maximum
- 411111111111111 4y ago
  Ooh, i see. I wasn't aware that they under/overflow at 0 when they're unsigned. Thanks for broadening my horizon!
naavis 4y ago
It loops while the condition is true. When an underflow happens, it stops being true.
Jtsummers 4y ago
For loops are translatable from: for(initialize; condition; increment) { ... } to: initialize; while(condition) { ... increment } (more or less, some scoping things not encompassed by the above; this is also how pretty much every for loop in a C-syntax language works) The condition of a for loop is equivalent to a while loop's condition. So yes, length - 1 < length will be true on the first iteration, which is fine because the loop continues as long as that condition is true. What the above approach takes advantage of is that when underflow eventually happens you'll have this condition: MAXINT < length Which will terminate it for all possible values of length.
billforsternz 4y ago
The loop continues until i transitions from 0 to 0 minus 1. 0-1 in this case actually doesn't equal -1 since size_t is an unsigned type, instead it wraps around to be the largest possible positive integer instead. TLDR; yes as you speculate it terminates when the number underflows.

msk-lywenn 4y ago

It’s a condition to run, not a condition to stop

wizofaus 4y ago
That was my reaction - why should anyone think it might be the latter? Are there languages that do have such a syntax without explicit keywords ("do...until")?
- msk-lywenn 4y ago
  I think lisp or scheme does. I was often confused by that when I was playing with it

bobthepanda 4y ago

It‘s an unsigned int, so past 0 it overflows back to the maximum

411111111111111 4y ago
Ooh, i see. I wasn't aware that they under/overflow at 0 when they're unsigned. Thanks for broadening my horizon!

naavis 4y ago

It loops while the condition is true. When an underflow happens, it stops being true.

Jtsummers 4y ago

For loops are translatable from: for(initialize; condition; increment) { ... } to: initialize; while(condition) { ... increment } (more or less, some scoping things not encompassed by the above; this is also how pretty much every for loop in a C-syntax language works) The condition of a for loop is equivalent to a while loop's condition. So yes, length - 1 < length will be true on the first iteration, which is fine because the loop continues as long as that condition is true. What the above approach takes advantage of is that when underflow eventually happens you'll have this condition: MAXINT < length Which will terminate it for all possible values of length.

billforsternz 4y ago

The loop continues until i transitions from 0 to 0 minus 1. 0-1 in this case actually doesn't equal -1 since size_t is an unsigned type, instead it wraps around to be the largest possible positive integer instead. TLDR; yes as you speculate it terminates when the number underflows.