Following up on the excellent answer davido came up with, and using the code he gave, 1.9041105342991877e+258 gives the binary value 0100001110111101100010001110100001001000001011011111000110101110

Going the other way, the binary string just obtained is displayed as 1.90411053429919e+258

Increasing the least significant bytes to see when we see a change, 

0100001110111101100010001110100001001000001011011111000110101110 - sti
+ll 1.90411053429919e+258

1100001110111101100010001110100001001000001011011111000110101110 - sti
+ll 1.90411053429919e+258

1110001110111101100010001110100001001000001011011111000110101110 - sti
+ll 1.90411053429919e+258

1111001110111101100010001110100001001000001011011111000110101110 - sti
+ll 1.90411053429919e+258

1111101110111101100010001110100001001000001011011111000110101110 - sti
+ll 1.90411053429919e+258

1111111110111101100010001110100001001000001011011111000110101110 - fin
+ally 1.9041105342992e+258

[download]

Why did we have to go in 6 bits from the end to see a change in the floating point representation?

There are 53 bits in the mantissa, and if the last 6 don't show, that means that there are only 47 being used for the display in floating point. log(2^47)/log(10) gives us about 14 significant digits. This is what the answers above seem to indicate.

I think that the answer is going to depend on how many digits the machine displays for a given IEEE754 value. For my machine, I need to change the value by roughly about 10^-14 from its original value to see a change.

P.S. I hope I transcribed everything correctly here. As always, I have to use another machine to run the perl.

In reply to Re^2: Determining the minimum representable increment/decrement possible? by ExReg
in thread Determining the minimum representable increment/decrement possible? by BrowserUk

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