Integer Underflow when initializing for loop variable with size_t

Question

I was running the code example below, initially trying to print a character array backwards where count represented the number of characters in the array. Everytime I ran the for loop, i was not initalized to count - 1 instead it was initialized to some very large number. But outside the loop, i is properly initalized to 4. I can't explain why this is happening.

size_t count {5};
size_t i {count - 1UL}; // i is initalized to 4

for(size_t i {count - 1}; i >= 0UL; --i) {
    std::cout << i << std::endl;
} // infinite loop, i is initalized to 18446744073709141874

So I should only ever use loop counters with integers when iterating backwards? — Mutating Algorithm, Sep 14 '19 at 13:28
Terminology update, [there is no such thing as "integer underflow"](https://en.m.wikipedia.org/wiki/Arithmetic_underflow). — n. m. could be an AI, Sep 14 '19 at 13:54

L. F. · Accepted Answer · 2019-09-15T01:54:49.030

An "unsigned" integer does not have a sign and is always nonnegative. If you do --i on i = 0 then you will get the maximum value of std::size_t.

Do this instead:

// iterates from (count - 1) to 0
for (std::size_t i = count; i--;)

So how does this work? i-- is the postfix decrement operator, which returns the prior value of i. This way, the loop condition is actually comparing count, ..., 1 to zero, while in the loop we get count - 1, ..., 0.

A more generalized version of this idiom is sometimes called the --> operator:

// iterates from (high - 1) to low
for (std::size_t i = high; i-- > low;)

Reads perfect now :-) – cmaster - reinstate monica Sep 15 '19 at 09:55 — cmaster - reinstate monica, Sep 15 '19 at 09:55

score 1 · Answer 2 · edited Sep 14 '19 at 13:49

1

std::size_t is unsigned integer type. So values of the type never can be negative.

Rewrite the loop like

std::size_t i {count - 1};

do
{
    std::cout << i << std::endl;
} while ( i-- );

edited Sep 14 '19 at 13:49

Aykhan Hagverdili

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answered Sep 14 '19 at 13:31

Vlad from Moscow

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Aykhan Hagverdili · Answer 3 · 2019-09-14T13:48:22.117

Firstly, i in the outer scope is unused and shaded, secondly an unsigned integer is always greater than or equal to 0. You could have your compiler detect these if only you turned on your compiler warnings:

<source>:7:38: error: comparison of unsigned expression >= 0 is always true [-Werror=type-limits]

    7 |     for(std::size_t i {count - 1}; i >= 0UL; --i) {

      |                                    ~~^~~~~~

<source>:5:17: error: unused variable 'i' [-Werror=unused-variable]

    5 |     std::size_t i {count - 1UL}; // i is initalized to 4

      |

Solution:

for (unsigned long long i = 4ull; i != -1ull; --i) {
    std::cout << i << '\n';
}

Integer Underflow when initializing for loop variable with size_t

3 Answers3