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diff --git a/45/CH6/EX6.10/example_6_10.sce b/45/CH6/EX6.10/example_6_10.sce
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+//example  6.10
+clc;
+clear;
+format('v',18);
+//bb=input('enter the first number(in decimal):' );
+//aaa=input('enter the second number(negative) :');
+aaa=-12618
+bb=18357;
+aa=-1*aaa;
+a=0;
+q=0;
+while(aa>0)             // finding the binary equivalents 
+    x=modulo(aa,2);
+    a= a + (10^q)*x;
+    aa=aa/2;
+    aa=floor(aa);
+    q=q+1; 
+end
+r=0;
+b=0;
+while(bb>0)             
+    x=modulo(bb,2);
+    b= b + (10^r)*x;
+    bb=bb/2;
+    bb=floor(bb);
+    r=r+1; 
+end
+m=b
+for i=1:16
+    a1(i)=modulo(a,10);
+    a=a/10;
+   a=round(a);
+    p1(i)=0;
+    b1(i)=modulo(b,10);
+    b=b/10;
+   b=round(b);
+end
+p1(1)=1;    
+for i=1:16             // finding the 2's compliment of second number 
+        a1(i)=bitcmp(a1(i),1);
+        end
+    car(1)=0;
+for i=1:16
+    c1(i)=car(i)+a1(i)+ p1(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;
+re=0;
+    for i=1:16
+        re=re+(c1(i)*(10^(i-1)))   
+    end;
+    printf(' The binary representation of first number is ');
+    disp(m);
+         printf(' The 2''s compliment of second nmber is');
+         disp(re);
+a1=c1;
+ar(1)=0;
+for i=1:8
+    c1(i)=ar(i)+a1(i)+ b1(i);   // addin both the nmbers (binary addition)
+    if c1(i)== 2 then           // lower byte 
+        ar(i+1)= 1;
+        c1(i)=0;
+    elseif c1(i)==3 then 
+         ar(i+1)= 1;
+        c1(i)=1;
+    else 
+        ar(i+1)=0;
+    end
+end
+c1(9)=ar(9)
+re=0;
+format('v',18);
+for i=1:8
+    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',ar(9));
+for i=9:16
+    c1(i)=ar(i)+a1(i)+ b1(i);// upper byte 
+    if c1(i)== 2 then
+        ar(i+1)= 1;
+        c1(i)=0;
+    elseif c1(i)==3 then 
+         ar(i+1)= 1;
+        c1(i)=1;
+    else 
+        ar(i+1)=0;
+    end
+end
+c1(17)=ar(17);
+format('v',25);
+ree=0;
+for i=9:16
+    ree=ree+(c1(i)*(10^(i-9)));   
+end
+for i=9:16
+    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',ar(17));//displaying results 
+printf(' The total sum is ' );
+disp(re);
+printf(' with a carry %d',ar(17));