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法律研究C++ 程序实现布斯乘法算法用于乘以 2 个带符号数
布斯算法是一种乘法算法，它用 2 的补码表示法对两个带符号二进制数进行相乘。布斯使用了比加法更快的移位台式计算器，并创建了一种算法以提高它们的计算速度。
算法
Begin Put multiplicand in BR and multiplier in QR and then the algorithm works as per the following conditions: 1. If Qn and Qn+1 are same i.e. 00 or 11 perform arithmetic shift by 1 bit. 2. If Qn Qn+1 = 10 do A= A + BR and perform arithmetic shift by 1 bit. 3. If Qn Qn+1 = 01 do A= A – BR and perform arithmetic shift by 1 bit. End
代码示例
#include<iostream>
using namespace std;
void add(int a[], int x[], int q);
void complement(int a[], int n) {
int i;
int x[8] = { NULL };
x[0] = 1;
for (i = 0; i < n; i++) {
a[i] = (a[i] + 1) % 2;
}
add(a, x, n);
}
void add(int ac[], int x[], int q) {
int i, c = 0;
for (i = 0; i < q; i++) {
ac[i] = ac[i] + x[i] + c;
if (ac[i] > 1) {
ac[i] = ac[i] % 2;
c = 1;
}else
c = 0;
}
}
void ashr(int ac[], int qr[], int &qn, int q) {
int temp, i;
temp = ac[0];
qn = qr[0];
cout << "\t\tashr\t\t";
for (i = 0; i < q - 1; i++) {
ac[i] = ac[i + 1];
qr[i] = qr[i + 1];
}
qr[q - 1] = temp;
}
void display(int ac[], int qr[], int qrn) {
int i;
for (i = qrn - 1; i >= 0; i--)
cout << ac[i];
cout << " ";
for (i = qrn - 1; i >= 0; i--)
cout << qr[i];
}
int main(int argc, char **argv) {
int mt[10], br[10], qr[10], sc, ac[10] = { 0 };
int brn, qrn, i, qn, temp;
cout << "\n--Enter the multiplicand and multipier in signed 2's
complement form if negative--";
cout << "\n Number of multiplicand bit=";
cin >> brn;
cout << "\nmultiplicand=";
for (i = brn - 1; i >= 0; i--)
cin >> br[i]; //multiplicand
for (i = brn - 1; i >= 0; i--)
mt[i] = br[i];
complement(mt, brn);
cout << "\nNo. of multiplier bit=";
cin >> qrn;
sc = qrn;
cout << "Multiplier=";
for (i = qrn - 1; i >= 0; i--)
cin >> qr[i];
qn = 0;
temp = 0;
cout << "qn\tq[n+1]\t\tBR\t\tAC\tQR\t\tsc\n";
cout << "\t\t\tinitial\t\t";
display(ac, qr, qrn);
cout << "\t\t" << sc << "\n";
while (sc != 0) {
cout << qr[0] << "\t" << qn;
if ((qn + qr[0]) == 1) {
if (temp == 0) {
add(ac, mt, qrn);
cout << "\t\tsubtracting BR\t";
for (i = qrn - 1; i >= 0; i--)
cout << ac[i];
temp = 1;
}
else if (temp == 1) {
add(ac, br, qrn);
cout << "\t\tadding BR\t";
for (i = qrn - 1; i >= 0; i--)
cout << ac[i];
temp = 0;
}
cout << "\n\t";
ashr(ac, qr, qn, qrn);
}
else if (qn - qr[0] == 0)
ashr(ac, qr, qn, qrn);
display(ac, qr, qrn);
cout << "\t";
sc--;
cout << "\t" << sc << "\n";
}
cout << "Result=";
display(ac, qr, qrn);
}输出
--Enter the multiplicand and multipier in signed 2's complement form if negative-- Number of multiplicand bit=5 multiplicand=0 1 1 1 1 No. of multiplier bit=5 Multiplier=1 0 1 1 1 qn q[n+1] BR AC QR sc initial 00000 10111 5 1 0 subtracting BR 10001 ashr 11000 11011 4 1 1 ashr 11100 01101 3 1 1 ashr 11110 00110 2 0 1 adding BR 01101 ashr 00110 10011 1 1 0 subtracting BR 10111 ashr 11011 11001 0 Result=11011 11001
更新于： 30-Jul-2019
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