On Sat, 2019-04-06 at 11:28 +0200, cyrille henry wrote: > > > An other solution is to use bitwise operator : > > my_float = (byte1 << 24) & (byte2 << 16) & (byte3 << 8) & byte4;
From what I understand, the result is not a 32 bit floating point number, but a 32 bit unsigned int (assuming that each byte is an unsigned 8 bit int). Roman
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