Unsigned int reverse iteration with for loops

Question

I want the iterator variable in a for loop to reverse iterate to 0 as an unsigned int, and I cannot think of a similar comparison to i > -1, as

Answer 1

I would use two variables:

unsigned int start = 10;
for (unsigned int j = 0, i = start; j <= start; ++ j, -- i) {
    // ...
}

You can also use a while loop:

unsigned int start = 10;
unsigned int i = start + 1;
while (i --) {
    // ...
}

Answer 2

One more way:

for(unsigned i = n-1; i < n ; --i) 
{       
    // Iterates from n-1 to 0
}

Simillarly for size_t (unsigned integer type) use the same trick

for(std::size_t i = n-1; i < n ; --i) 
{       
    // Iterates from n-1 to 0
}

Answer 3

You can use

for( unsigned int j = n; j-- > 0; ) { /*...*/ }

It iterates from n-1 down to 0.

Answer 4

Just:

int start = 10;
for(unsigned int iPlus1 = start + 1 ; iPlus1 > 0 ; iPlus1--) {
  // use iPlus1 - 1 if you need (say) an array index
  a[iPlus1 - 1] = 123; // ...
}

No?

Answer 5

for(unsigned i = x ; i != 0 ; i--){ ...

And if you want to execute the loop body when i == 0 and stop after that. Just start with i = x+1;

BTW, why i must be unsigned ?

Answer 6

The following does what you want:

for (unsigned i = 10; i != static_cast<unsigned>(-1); --i)
{
    // ...
}

This is perfectly defined and actually works. Arithmetic on signed types is accurately defined by the standard. Indeed:

From 4.7/2 (regarding casting to an unsigned type):

If the destination type is unsigned, the resulting value is the least unsigned integer congruent to the source integer (modulo 2^n where n is the number of bits used to represent the unsigned type)

and 3.9.1/4

Unsigned integers, declared unsigned, shall obey the laws of arithmetic modulo 2^n where n is the number of bits in the value representation of that particular size of integer