How 0x3333 0x41DF could be 27.9 value?
I use hexadecimal to decimal value but it was 1B.E6666666666666666666.
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2https://www.h-schmidt.net/FloatConverter/IEEE754.html and https://en.wikipedia.org/wiki/Single-precision_floating-point_format – Michael Petch Nov 08 '22 at 12:30
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https://www-sop.inria.fr/sinus/Softs/Vigie/HHDB/ieee.html – matt Nov 08 '22 at 12:30
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There's no obvious standard for "16 bits decimal floating-point" numbers. I don't blame you for the confusion, whoever wrote this does not understand the concept. "real" is redundant with "floating" here. – MSalters Nov 08 '22 at 12:31
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1@MSalters : it is actually a 32-bit single precision value, I think the OP is confused. – Michael Petch Nov 08 '22 at 12:33
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2@MSalters It is 32-bit IEEE-754. It is simply split to 2 16-bit integer registers, because Modbus protocol is ancient. – user694733 Nov 08 '22 at 12:35
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Example: 0x41df3333 as a single precision 32-bit ieee-754 number is 27.9 – Michael Petch Nov 08 '22 at 12:37
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Oh help. This is middle-endian? `0x999A41E9` becomes `0x9A99E941` = 29.2? I can see why that would be confusing. – MSalters Nov 08 '22 at 12:37
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@MSalters The two 16-bit word values combine to form: 0x41e9999a which is 29.2 – Michael Petch Nov 08 '22 at 12:40
1 Answers
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0x3333 0x41DF combined = 0x41DF3333. As a 32 bit IEEE float, this is evaluated as follows:
Convert to binary:
0x41DF3333 =
4 1 D F 3 3 3 3 =
0100 0001 1101 1111 0011 0011 0011 0011 =
01000001110111110011001100110011
Evaluate as a 32 bit IEEE float:
01000001110111110011001100110011 =
Sign (s) Exponent (e) Significand (S)
11111111112222
12345678901234567890123
---- -------------------- -----------------------
0 10000011=1+2+128=131 10111110011001100110011
s e assumed 1 S1 S2 S3 S21 S22 S23
(-1)^0 x 2^(131-127) x (1 + 1 x 1/2^1 + 0 x 1/2^2 + 1 x 1/2^3 + ... + 0 x 1/2^21 + 1 x 1/2^22 + 1 x 1/2^23) =
1 x 2^( 4 ) x (1 x 1/2^0 + 1 x 1/2^1 + 0 x 1/2^2 + 1 x 1/2^3 + ... + 0 x 1/2^21 + 1 x 1/2^22 + 1 x 1/2^23) =
(1 x 2^4 + 1 x 2^3 + 0 x 2^2 + 1 x 2^1 + ... + 0 x 1/2^17 + 1 x 1/2^18 + 1 x 1/2^19) =
(1 x 2^23 + 1 x 2^22 + 0 x 2^21 + 1 x 2^20 + ... + 0 x 1/2^2 + 1 x 1/2^1 + 1 x 1/2^0 ) / 2^19 =
(1 1 0 1 0 1 1) / 524288 =
(110111110011001100110011 ) / 524288 =
(1101 1111 0011 0011 0011 0011 ) / 524288 =
(14299955 ) / 524288 =
27.899999618530273
The last calculation used a calculator. Noting that tha last non-zero digit after the decimal point, 3, is not 5, we know this is not the exact value. The following code gets the exact float value:
Mac_3.2.57$cat floatGet.c
#include <stdio.h>
int main(void){
union Data {
int i;
float f;
}data;
data.i = 0x41DF3333;
printf("%100.100f\n", data.f);
return 0;
}
Mac_3.2.57$cc floatGet.c
Mac_3.2.57$./a.out
27.8999996185302734375000000000000000000000000000000000000000000000000000000000000000000000000000000000
Mac_3.2.57$
Andrew
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