6 Apr
2019
6 Apr
'19
1:21 p.m.
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