C program of booth’s multiplication algorithm : **Booth’s multiplication algorithm**is a multiplication algorithm that multiplies two signed binary **numbers** in two’s complement notation.

(source : Wikipedia) **Booth’s algorithm** can be implemented by repeatedly adding (with ordinary unsigned binary addition) one of two predetermined values *A* and *S* to a product *P*, then performing a rightward arithmetic shift on *P*. Let **m** and **r** be the multiplicand and multiplier, respectively; and let *x* and *y* represent the number of bits in **a** and **b**.

For more clear picture, say we wish to multiply:

3 × (−4), with **a** = 3 and **b** = −4, and *x* = 4 and *y*= 4 then :

- a = 0011, -a = 1101, b = 1100
- A = 0011 0000 0
- S = 1101 0000 0
- P = 0000 1100 0

**Now lets check Booth’s Standard Algorithm Implementation**:

- Determine the values of
*A*and*S*, and the initial value of*P*. All of these numbers should have a length equal to (*x*+*y*+ 1).- A: Fill the most significant (leftmost) bits with the value of
**a**. Fill the remaining (*y*+ 1) bits with zeros. - S: Fill the most significant bits with the value of (−
**a**) in two’s complement notation. Fill the remaining (*y*+ 1) bits with zeros. - P: Fill the most significant
*x*bits with zeros. To the right of this, append the value of**b**. Fill the least significant (rightmost) bit with a zero.

- A: Fill the most significant (leftmost) bits with the value of
- Determine the two least significant (rightmost) bits of
*P*.- If they are 01, find the value of
*P*+*A*. Ignore any overflow. - If they are 10, find the value of
*P*+*S*. Ignore any overflow. - If they are 00, do nothing. Use
*P*directly in the next step. - If they are 11, do nothing. Use
*P*directly in the next step.

- If they are 01, find the value of
- Arithmetically shift the value obtained in the 2nd step by a single place to the right. Let
*P*now equal this new value. - Repeat steps 2 and 3 until they have been done
*y*times. - Drop the least significant (rightmost) bit from
*P*. This is the product of**a**and**b**.

Now friends let us see how to implement **Booth’s multiplication in C**. I have written a sample code on Booth’s multiplication **algorithms in C** which accepts numbers from range -7 to 7 and displays the binary equivalents and complete booth operations in the tabular form and then the output of multiplication in both binary and decimal form.

### C program of Booth’s Multiplication Algorithm :

#include<stdio.h> #include<conio.h> #include<process.h> #include<math.h> int get(int a) { char ch='B'; int flag=0; if(a==1) ch='A'; do { printf("ENTER VALUE OF %c: ",ch); scanf("%d",&a); if(a< 0) { a = a * -1; flag = 1; } if(8<=a) printf("\n\t!INVALID NUMBER.ENTER VALUE (-8 < A < 8)!\n"); }while(8<=a); if(flag) a = a *-1; return(a); } void add(int *a,int *b) { int x,i,c=0; for(i=3;i>=0;i--) { x=a[i]; a[i]=c^x^b[i]; if(((c==1)&&(x==1))||((x==1)&&(b[i]==1))||((b[i]==1)&&(c==1))) c = 1; else c = 0; } } void binary(int x,int*arr) { int i,p=x,c[4]={0,0,0,1}; for(i=0;i<4;i++) arr[i] = 0; if(x < 0) x = x *-1; i = 3; do { arr[i]=x%2; x = x/2; i--; }while(x!=0); if(p<0) { for(i=0;i<4;i++) arr[i]=1-arr[i]; add(arr,c); } printf("\n\nTHE BINARY EQUIVALENT OF %d IS : ",p); for(i=0;i<4;i++) printf("%d",arr[i]); } void rshift(int x,int *y) { int i; for(i=3;i>0;i--) y[i] = y[i-1]; y[0] = x; } int main() { int q=0,i,j,a,b,A[4]={0,0,0,0},C[4]={0,0,0,1},C1[8]={0,0,0,0,0,0,0,1}; int s=0,z=0,Q[4],M[4],temp,temp1[4],ans[8],y,x=0,c=0; printf("----------------------------------------------------------------------\n"); printf("-------------------made by C code champ ------------------------------\n"); printf("----------------------------------------------------------------------\n"); printf("\n\n\t\tBOOTHS MULTIPLICATION ALGORITHM \n"); printf("\n----------------------------------------------------\n"); a = get(1); b=get(0); printf("\n---------------------------------------------------\n"); binary(a,M); binary(b,Q); printf("\n---------------------------------------------------\n"); printf("OPERATION\t\t A\t Q\tQ'\t M"); printf("\n\n INITIAL\t\t"); for(i=0;i<4;i++) printf("%d",A[i]); printf("\t"); for(i=0;i<4;i++) printf("%d",Q[i]); printf("\t"); printf("%d\t",q); for(i=0;i<4;i++) printf("%d",M[i]); for(j=0;j<4;j++) { if((Q[3]==0)&&(q==1)) { printf("\n A:=A+M \t\t"); add(A,M); for(i=0;i< 4;i++) printf("%d",A[i]); printf("\t"); for(i=0;i< 4;i++) printf("%d",Q[i]); printf("\t%d\t",q); for(i=0;i< 4;i++) printf("%d",M[i]); } if((Q[3]==1)&&(q==0)) { printf("\n A:=A-M \t\t"); for(i=0;i<4;i++) temp1[i] = 1-M[i]; add(temp1,C); add(A,temp1); for(i=0;i< 4;i++) printf("%d",A[i]); printf("\t"); for(i=0;i< 4;i++) printf("%d",Q[i]); printf("\t%d\t",q); for(i=0;i< 4;i++) printf("%d",M[i]); } printf("\n Shift \t\t\t"); y = A[3]; q = Q[3]; rshift(A[0],A); rshift(y,Q); for(i=0;i<4;i++) printf("%d",A[i]); printf("\t"); for(i=0;i<4;i++) printf("%d",Q[i]); printf("\t"); printf("%d\t",q); for(i=0;i<4;i++) printf("%d",M[i]); } printf("\n\n---------------------------------------------------\n"); printf("\nTHE ANSWER IN BINARY IS : "); for(i=0;i<4;i++) ans[i]=A[i]; for(i=0;i<4;i++) ans[i+4]=Q[i]; if(((a< 0)&&(b>0))||((a>0)&&(b< 0))) { for(i=0;i<8;i++) ans[i]=1-ans[i]; for(i=7;i>=0;i--) { x = ans[i]; ans[i]=c^x^C1[i]; if(((c==1)&&(x==1))||((x==1)&&(C1[i]==1))||((C1[i]==1)&&(c==1))) c=1; else c=0; } } for(i=0;i<8;i++) printf("%d",ans[i]); for(i=7;i>=0;i--) { s = s + (pow(2,z) * ans[i]); z = z+1; } if(((a< 0)&&(b>0))||((a>0)&&(b< 0))) printf("\nTHE ANSWER IN DECIMAL IS : -%d\n",s); else printf("\nTHE ANSWER IN DECIMAL IS : %d\n",s); getch(); }

We hope you all have enjoyed the **C program** of Booth’s multiplication algorithm. If you have any issues with code or its logic, feel free to contact us in form of comments.

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