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#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <limits.h>
unsigned bit_pattern_A(int k) {
return (~0)<<k;
}
unsigned bit_pattern_B(int k, int j) {
return ~((~0)<<k) << j;
}
float fraction_bit_pattern(unsigned Y, unsigned k) {
return 0;
}
unsigned f2u(float f) {
union {
float f;
unsigned u;
} fu = { .f = f };
return fu.u;
}
/*
floating point number
The unsigned numbers can be compared.
The bits in the unsigned number can be mapped to bits in the floating point number in a monotonic way.
*/
int float_le(float x, float y) {
unsigned ux = f2u(x);
unsigned uy = f2u(y);
printf("ux: %x, uy: %x, ux<=uy: %d\n", ux, uy, ux<=uy);
/* Get the sign bits */
unsigned sx = ux >> 31;
unsigned sy = uy >> 31;
printf("sx: %d, sy: %d\n", sx, sy);
/* Give an expression using only ux, uy, sx, and sy */
printf("sx && !sy: %d, ((uy - sy) <= (ux - sx)): %d\n", sx && !sy, ux<=uy);
return (ux << 1 == 0 && uy << 1 == 0)
|| (sx && !sy)
|| (sx && sy) && (uy <= ux)
|| ((!sx && !sy) && (ux <= uy));
}
int main() {
/* 2.85
V=(2^E)*M
bias=2^(k-1) - 1
k=8
0111 1111
k bit exponent has values from (2^k)-bias
(A) V=7.0
E=(bias+2)-bias, has k-1 bit set plus 1, since (2^(k-1) -1)-bias == 0
M=1/2+1/4=3/4, first two bits
7.0=2^2(1+f)=4+4f=4+3
(B)
largest integer --> largest odd integer
*/
return 0;
/* 2.84 */
printf("%d\n", f2u(0.1f));
printf("%d\n", float_le(1, 0.1));
return 0;
/* 2.83
Y=B2U(y), y is k bit word
I couldn't solve this one. Was stuck looking for patterns with arrows and tables instead of algebra.
n = 0.yyyy..
n<<k = y.yyyy..
n<<k = Y + n
n<<k - n = Y
n(1<<k - 1) = Y
n = Y/(1<<k - 1)
B. (a) 1/3
(b) 6/15 -> 2/5
(c) 19/63
*/
return 0;
/*
32bit int and unsigned with arithmetic shift
*/
srand(time(NULL));
int x = random();
int y = random();
unsigned ux = (unsigned) x;
unsigned uy = (unsigned) y;
printf("x: %d, y: %d\n", x, y);
/*
x<y == -x>-y
+x +y -> always true
-x -y -> alwyas true
-x +y -> x<y is true, -x>-y is also true, since the sign changes
+x -y -> x<y is false, -x>-y is also false, since the sign changes
*/
printf("A: %d\n", (x<y) == (-x>-y));
/*
<=>
((x<<4) + (y<<4) + y - x) mod 2^w
(16x + 16y + y - x) mod 2^w
*/
printf("B: %d\n", ((x+y)<<4) + y-x == 17*y+15*x);
/*
~:
number --(~)--> -number -1
Using this observation we can do this (everything is under modulo 2^w, which is fine... right?):
~x + ~y + 1:
-x - 1 - y - 1 + 1
-x -y - 1
-(x+y) - 1
~(x+y)
*/
printf("C: %d\n", ~x + ~y + 1 == ~(x+y));
/*
number --(unsigned)--> number + msb * 2^32
x + (msbx * 2^32) - (y + msby * 2^32)
x - y + msbx * 2^32 - msby * 2^32
-(y-x) + msbx * 2^32 - msby * 2^32
-(unsigned)(y-x)
-((y-x) + (msbyx * 2^32))
-(y-x) - (msbyx * 2^32)
(msbx * 2^32 - msby * 2^32) = (msbyx * 2^32) = 0?
msbx = 0 and msby = 0 -> msbyx = 0 ?
0100 - 0101 = 1111 -> if x > y and x,y>0 this expression is not true
*/
y = 4;
x = 5;
printf("D: %d\n", (ux-uy) == -(unsigned)(y-x));
/*
number -->>2--> floor(number/4)
number --<<4--> number*4
So the first two bits are always lost. And that means for negative and postive
two's complement number that the number either equals or is less than before.
*/
printf("E: %d\n", ((x >> 2) << 2) <= x);
return 0;
/* 2.81 */
printf("%x\n", bit_pattern_A(16));
printf("%x\n", bit_pattern_B(8, 8));
return 0;
}
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