Question

float a = 0;
while (true)
{
    a++;
    if (a > 16777216)
        break; // Will never break... a stops at 16777216
}

Can anyone explain this to me why a float value stops incrementing at 16777216 in this code?

Edit:

Or even more simple:

float a = 16777217; // a becomes 16777216

Answer 1

Short roundup of IEEE-754 floating point numbers (32-bit) off the top of my head:

• 1 bit sign (0 means positive number, 1 means negative number)
• 8 bit exponent (with -127 bias, not important here)
• 23 bits “mantissa”
• With exceptions for the exponent values 0 and 255, you can calculate the value as: (sign ? -1 : +1) * 2^exponent * (1.0 + mantissa)
  • The mantissa bits represent binary digits after the decimal separator, e.g. 1001 0000 0000 0000 0000 000 = 2^-1 + 2^-4 = .5 + .0625 = .5625 and the value in front of the decimal separator is not stored but implicitly assumed as 1 (if exponent is 255, 0 is assumed but that’s not important here), so for an exponent of 30, for instance, this mantissa example represents the value 1.5625

Now to your example:

16777216 is exactly 2²⁴, and would be represented as 32-bit float like so:

• sign = 0 (positive number)
• exponent = 24 (stored as 24+127=151=10010111)
• mantissa = .0
• As 32 bits floating-point representation: 0 10010111 00000000000000000000000
• Therefore: Value = (+1) * 2^24 * (1.0 + .0) = 2^24 = 16777216

Now let’s look at the number 16777217, or exactly 2²⁴+1:

• sign and exponent are the same
• mantissa would have to be exactly 2^-24 so that (+1) * 2^24 * (1.0 + 2^-24) = 2^24 + 1 = 16777217
• And here’s the problem. The mantissa cannot have the value 2^-24 because it only has 23 bits, so the number 16777217 just cannot be represented with the accuracy of 32-bit floating points numbers!