blob: 04ad4b7c9c9ddb571b03e0d38541dd52f7c73ea6 (
plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
|
//example 6.3
clc;
clear;
//a=input("enter the first 8 bit number :");
//b=input("enter the second 8 bit number :");
a=0000111110101100;
b=0011100001111111;
for i=1:16
a1(i)=modulo(a,10);
a=a/10;
a=round(a);
b1(i)=modulo(b,10);
b=b/10;
b=round(b);
end
car(1)=0;
for i=1:8
c1(i)=car(i)+a1(i)+ b1(i); // adding the Higher bytes
if c1(i)== 2 then
car(i+1)= 1;
c1(i)=0;
elseif c1(i)==3 then
car(i+1)= 1;
c1(i)=1;
else
car(i+1)=0;
end
end
c1(9)=car(9)
re=0;
format('v',18);
for i=1:9
re=re+(c1(i)*(10^(i-1)))
end
printf('The sum of lower bytes of two binary numbers is %d\n',re );
printf(' with a carry is %d\n',car(9));
re=re-(c1(9)*(10^(8)))
for i=9:16 // adding the Higher bytes
c1(i)=car(i)+a1(i)+ b1(i);
if c1(i)== 2 then
car(i+1)= 1;
c1(i)=0;
elseif c1(i)==3 then
car(i+1)= 1;
c1(i)=1;
else
car(i+1)=0;
end
end
c1(17)=car(17);
format('v',25);
ree=0;
for i=9:17
ree=ree+(c1(i)*(10^(i-9)));
end
for i=9:17
re=re+(c1(i)*(10^(i-1)))
end
printf(' The sum of upper bytes of the given numbers is %d\n',ree);
printf(' with a carry is %d\n',car(17)); // displaying results
printf(' The total sum is %f',re );
|
