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convert floating point number to fixed point number
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#include <stdio.h> | |
#include <math.h> | |
#define FIXED_BIT 12 | |
unsigned short int float2fix(float n) | |
{ | |
unsigned short int int_part = 0, frac_part = 0; | |
int i; | |
float t; | |
int_part = (int)floor(n) << FIXED_BIT; | |
n -= (int)floor(n); | |
t = 0.5; | |
for (i = 0; i < FIXED_BIT; i++) { | |
if ((n - t) >= 0) { | |
n -= t; | |
frac_part += (1 << (FIXED_BIT - 1 - i)); | |
} | |
t = t /2; | |
} | |
return int_part + frac_part; | |
} | |
int main() | |
{ | |
float n; | |
n = 2.5; // 0d10240 | |
printf("%f => %d\n", n, float2fix(n)); | |
n = 2.625; // 0d10752 | |
printf("%f => %d\n", n, float2fix(n)); | |
n = 0.375; // 0d1536 | |
printf("%f => %d\n", n, float2fix(n)); | |
return 0; | |
} |
@len0rd can you explain what this does?
t = 0.5;
for (i = 0; i < FIXED_BIT; i++) {
if ((n - t) >= 0) {
n -= t;
frac_part += (1 << (FIXED_BIT - 1 - i));
}
t = t /2;
}
Its converting the fractional portion of the float to fixed point. The fractional portion of fixed point is represented by 1/(2^(i+1)).
ie: for a 4-bit fixed point number, where 3 bits are a fraction, 0.75 -> 0110.
Here's some more stuff for you to look at:
http://www-inst.eecs.berkeley.edu/~cs61c/sp06/handout/fixedpt.html
@len0rd Hello,
The article you linked is not accessible outside of berkeley, is it accessible somewhere else ?
Thank you for your help.
@sponthus looks like they made that change sometime in the last 5 years since I posted. Here’s a working copy on the wayback machine
Thanx that's perfect !
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Thanks! If you want this to handle negative numbers, you need to check for it before setting int_part.
lines 11-12 should be like this:
I also added some nan/inf checks to mine.