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//exmple 6 .4(b)//
clc
//clears the screen//
clear
//clears already existing variables//
a =0;
b =0;
q =0;
//bb=input(”Enter the first no (in decimal) : ” ) ;
//aa=input( ”Enter the number from which first no has to subtracted : ” ) ;
aa =8;
bb =1;
while ( aa >0)
//converting the inputs into binary numbers//
x = modulo ( aa ,2) ;
a = a + (10^ q ) * x ;
aa = aa /2;
aa = floor ( aa ) ;
q = q +1;
end
nn = a
q =0;
while ( bb >0)
x = modulo ( bb ,2) ;
b = b + (10^ q ) * x ;
bb = bb /2;
bb = floor ( bb ) ;
q = q +1;
end
printf ( ' \n The binary equivalent of first no is %f\n\n ' ,b ) ;
printf ( 'The binary equivalent of second no is %f\n\n' ,a ) ;
for i =1:40
a1 ( i ) = modulo (a ,10) ;
a = a /10;
a = round ( a ) ;
b1 ( i ) = modulo (b ,10) ;
b = b /10;
b = round ( b ) ;
end
bro (1) =0;
for i =1:40
c1 ( i ) = a1 ( i ) - b1 ( i ) - bro ( i ) ;
if c1 ( i ) == -1 then
bro ( i +1) = 1;
c1 ( i ) =1;
elseif c1 ( i ) == -2 then
bro ( i +1) = 1;
c1 ( i ) =0;
else
bro ( i +1) =0;
end
end
re =0;
format ( 'v' ,18) ;
for i =1:40
re = re +( c1 ( i ) *(10^( i -1) ) )
end
printf ( ' The difference of given two numbers is %f\n\n ' , re ) ;
q =1;
b =0;
f =0;
a = re ;
while (a >0)
//converting the binary result to decimal then 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 ( ' Difference in decimal notation is %d\n\n ' ,f ) ;
printf ( ' Difference in hexadecimal notation is %s \n ' , hex ) ;
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