Mathematical operation to keep number not less than zero

Question

In programming, modulus is useful to keep numbers in range not exceeding an upper bound limit.

For example:

int value = 0;
for (int x=0; x<100; x++)
    cout << value++%8 << " "; //Keeps number in range 0-7

Output: 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7...

---

Now consider this situation:

int value = 5;
for (int x=0; x<100; x++)
    cout << value-- << " ";

Output: 5 4 3 2 1 0 -1 -2 -3 -4 -5 -6 -7...

My question is: How to set the lower bound limit to 0 WITHOUT using any conditional statement like if or switch case?

Output I want: 5 4 3 2 1 0 0 0 0 0 0 0 ...

Answer 1

How about std::max?

int value = 5;
for (int x=0; x<100; x++) {
    cout << value << " ";
    value = std::max(0, --value);
}

Answer 2

Use std::max(value,0) or the ternary operator. Both will compile to code without conditional jumps.

Answer 3

Perhaps it can be done with

typedef unsigned u32b;

u32b sat_subu32b(u32b x, u32b y)
{
    u32b res = x - y;
    res &= -(res <= x); 
    return res;
}

which is taken from here. More specifically it seems to implement saturated subtraction; the saturated decrement would be

u32b sat_dec(u32bx)
{
    u32b res = x-1;
    res &= -(res <= x); 
    return res;
}

and seems to be branch-free.

Answer 4

The two situations are different. As you've shown, the % will make the numbers loop around from 0 to 7, whereas in the negative case you seem to be asking for the max.

Depending on whether you count the ternary operator as a conditional, you could do:

int value = 5;
for (int x=0; x<100; x++)
    cout << value > 0 ? value-- : 0 << " ";

This has the side effect that value will no longer be decremented once it has reached 0.

Alternatively, you could use a std::max:

int value = 5;
for (int x=0; x<100; x++)
    cout << std::max(0, value--) << " ";

Answer 5

using max function in std.
max function detail:

template <class T> const T& max (const T& a, const T& b) {
  return (a<b)?b:a;     
}