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//example 6.9
clc;
clear;
//aaa=input('enter the number(in decimal) :');
aaa=-19750 // given input
aa=-1*aaa;
format('v',18);
a=0;
q=0;
while(aa>0) // converting it to binary
x=modulo(aa,2);
a= a + (10^q)*x;
aa=aa/2;
aa=floor(aa);
q=q+1;
end
for i=1:16
a1(i)=modulo(a,10);
a=a/10;
a=round(a);
b1(i)=0;
end
b1(1)=1;
for i=1:16 /// finding the 2's compliment
a1(i)=bitcmp(a1(i),1);
end
car(1)=0;
for i=1:16
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);
re=0;
for i=1:17
re=re+(c1(i)*(10^(i-1)))
end;
printf('\n The 2''s compliment is');
disp(re);
q=1;
b=0;
f=0;
a=re;
while(a>0) // converting to hexadecimal
r=modulo(a,10);
b(1,q)=r;
a=a/10;
a=floor(a);
q=q+1;
end
for m=1:q-1
c=m-1
f = f + b(1,m)*(2^c);
end
hex=dec2hex(f);
printf('\n In Hexadecimal notation is %sH\n\n',hex) ;// displaying the result
disp('As the memory of a first generation microcumputer is orgnised in bytes . The lower byte is stored in 2000 address and the higher byte is stored in 2001 address.');
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