Neater way to multiply many times

Question

I needed to find away to left shift a number on a platform with buggy << and >> bitwise operators. I came up with a rudimentary solution, but it looks ugly and inefficient. What's a better way to do this?

float leftShift8(int value) {
    return value * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2;
}

float rightShift8(int value) {
    return value / 2 / 2 / 2 / 2 / 2 / 2 / 2 / 2;
}

Answer 1

Why wouldn't you do something like

float rightShift(int value, int numberOfShift)
{
  float result = (float)value;
  while(numberOfShift-- > 0)
     value*=2.0f;
  return value;
}

float leftShift(int value, int numberOfShift)
{
  float result = (float)value;
  while(numberOfShift-- > 0)
     value/=2.0f;
  return value;
}

This would especially deal with the fact that you do not return int type, but float. Additionally, its a lot faster to do a 2*2*2 than pow(2,3)

Answer 2

If I understand well, you are using a computer that have "broken" bitwise operators? That is odd but maybe you can try this?

float leftShit8(int value)
{
    return pow(2, 8) * value;
}

This code simply do a power of 8 on the base that is 2. After that, you multiply by your value. It does the same thing as your code but I'm not sure why anyone would use this instead of bitwise since bitwise a much more faster I think.

You could make this more general and useful like this

float leftShit(int value, int numberOfShift)
{
    return pow(2, numberOfShift) * value;
}

EDIT: For the edited question about division, the is only a sligh change. Thanks to T.C for this (answer below), I include it in my answer to make it more complete

float rightShift(int value, int numberOfShift) 
{ 
    return value/pow(2, numberOfShift); 
} 

Anyway, I hope it will help.

Answer 3

Well, going on Jean-François's answer above, to solve the division:

float rightShift(int value, int numberOfShift)
{
    return value/pow(2, numberOfShift);
}

Again, there you go.