If you are using integer texture format, not a fixed point normalized texture format, you can directly do the appropriate integer texture fetches using integer samplers (usampler* or isampler*, the former could come handy in your case) as that way you get the actual integer texel values without the need of ALU instructions to scale it back up.
If you do use normalized texture formats (maybe because you are using a GPU generation that does not support integer texture formats), then your idea may work as well, however AFAIK there is no absolute guarantee about precision for pre-GL3 hardware.
But, the cast from uint8 to float introduces error
No error is introduced casting ints or uints to floats unless an int’s or uint’s magnitude exceeds the maximum value representable with the number of bits in the float’s mantissa. No uint8 can ever do that, even for a float16.
A way to find the mantissa and exponent from an int is:
log2(34) ~= 5.0875
exponent = ceil(5.0875) = 6
mantissa = 34 / ( 2^(exponent) ) ~= 0.5313If I map back I get
0.5313 * 2^6 = 34.0032
Does IEEE754 use other method much more accurate or exactly accurate?
Yes, IEEE 754 uses another, exactly accurate (and faster and simpler) method. In most cases, IEEE 754 floating point numbers are encoded in normalized form. The log2 value of a number is never calculated, making the normalized encoding faster, simpler, and exact (as long as the magnitude of the number fits within the mantissa’s bits).
IEEE 754 normalization means that the binary point position of a value is shifted one bit (and the exponent incremented or decremented by one) until the absolute value of the value is one or between one and two (except for the special values of zero, infinity, NaN, etc.). The exponent indicates how many positions the binary point was moved and the mantissa is not changed (except that the leading bit, which after normalization will be one, is removed to allow an extra bit of precision in the least-significant bit of the mantissa).
So, in your example, the binary value of 34 is 100010. Normalized, that is represented as 1.00010 with an exponent of five (added to a bias to allow both positive and negative exponents). After the leading one is removed the mantissa will be .00010 with however many trailing zeroes as are required to fill the remaining less significant bits of the mantissa for the particular floating point size (16, 32, 64, 80 bits).
Reconstructing the value, you concatenate the implicit leading one to the mantissa, for 1.00010, then shift the binary point five bits to the right from its position to the right of the leading one, for a value of 100010.0, which equals 34 exactly.
This can be implemented in hardware very efficiently and is very fast.