Infinite loop when using size_t in a count down for loop

Question

So I\'m using size_t instead of int in any indexing for loop to prevent negative indices. But when counting down, this leads to an overflow:

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Answer 1

A simplest way is to increase the upper value. For example

const size_t N = 10;

for (size_t i = N + 1; i != 0; --i) {
    // Do something, f.ex. array[i-1] = i-1
}

or

const size_t N = 10;

for (size_t i = N + 1; i-- != 0; ) {
    // Do something, f.ex. array[i] = i
}

In general case when i can be equal to the maximum value stored in an object of the type size_t you can use the following trick

#include <stdio.h>

int main( void )
{
    const size_t N = 10;

    for (size_t i = N, j = N; !( i == 0 && j == -1 ); j--)
    {
        i = j;
        printf( "%zu ", i );
    }

    printf( "\n" );
}

Otherwise you can use do-while loop. It is more suitable in this case. For example

size_t i = N;

do
{
    printf( "%zu ", i );
} while ( i-- != 0 );

Answer 2

It is not technically an overflow because size_t is an unsigned type, but it is definitely an infinite loop since the termination condition is always true.

Unsigned integers wrap around when decremented at 0. Note that your loop will run 11 times before the wrap around occurs, not 10.

You must check for the condition before decrementing the index. Starting the enumeration with an initial value one more than the maximum valid index improves visual consistency and simplifies the test.

Here is a corrected version where you can see that the initial value for i is the number of elements of the array:

int array[11];
for (size_t i = 11; i-- > 0; ) {
    // Do something, f.ex. array[i] = i
}

Answer 3

for (size_t i = 11; i-- > 0; ) {
    // Do something, f.ex. array[i] = i
}

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Note: The question starts the loop with value=10(which is strange, but not impossible). I start with 11, but the first time the loop body is enterered, it has already been decremented to 10.

Answer 4

The idiomatic, though not to everyone's taste way, is to use the slide operator:

for (size_t i = 10 + 1; i--> 0; )

It isn't really an operator but that's what it has become known as over the years.

Answer 5

size_t i = 10; i >= 0; is never false as size_t is some unsigned type and all values are greater than or equal to zero.

... size_t which is the unsigned integer type of the result of the sizeof operator; ...
C11 §7.19 2

A good compiler with warnings enabled would have warned about this.
Hopefully, that infinite loop would never had occurred as an investigation to the warning would have first rectified the problem.

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Best alternative depends on coding goals

Good code avoids magic numbers like this naked 10. Better if code derived that. In this simple case, it should have been 11.

#define A_SIZE 11
int array[A_SIZE];
...
for (size_t i = A_SIZE; i-- > 0; ) {
    // Do something, f.ex. array[i] = i
}

OTOH, code may have had break conditions in the loop and needs i in later code to indicate array[] usages

size_t i = A_SIZE;
while (i > 0) {
  if (...) break; 
  i--;
  // Do something, f.ex. array[i] = i
  if (...) break; 
}
// Do something with i

Code may have a contract requirement to use a 10 in various places.

// Contract says loop must handle indexes 0 to N, inclusive
#define N 10
int array[N + 1];

for (size_t i = N; i + 1 > 0; i--) {
  // Do something, f.ex. array[i] = i
}

Good optimizing compilers will not perform a +1 on each i + 1 > 0, but create equivalent efficient code.

Code is a fashion that best conveys the overall meaning of the code.

Answer 6

for (size_t i = 10; i <= 10; --i) // do something

When overflow do happens, it will round to the largest integer and thus the condition will fail.