 2018-07-06 03:48

# c++针对二进制补码算术中整数的算术运算问题，请大神指点

c++ 二进制问题请教

01001和10010的二进制和是11011。

#include iostream>
#include cmath>
#include string>

using namespace std;

int main()
{
//Read in the bit pattern size
int L;
do {
cout << "Enter positive integer for the bit pattern size ";
cin >> L;
}while (L <= 0);

``````    //Read in two integers a and b
int a, b;
cout << "Enter an integer a ";
cin >> a;
cout << "Enter an integer b ";
cin >> b;

//Calculate the decimal arithmetic sum of a and b and print the result int
c1 = a + b;
cout << "In decimal " << a << " + " << b << " is " << c1 << endl;

//Compute the two's complement representations of a and b
//Each integer must be represented in L-bits pattern
//Also these two's complement representations must be returned as string data types
string A = decimalToTwocomplementString(a, L);
string B = decimalToTwocomplementString(b, L);

//Print the two's complement representations of a and b
cout << "The two's complement of " << a << " is\t " << A << endl;
cout << "The two's complement of " << b << " is\t " << B << endl;

//Compute the binary sum of the two's complement representations of a and b
//The result must be returned as L-bit pattern string data type

//Print the two's complement representation binary sum
cout << "The binary sum of " << A << " and " << B << " is " << C << endl;

//Convert the two's complement representation binary sum to decimal and print
int c2 = TwoComplementStringToDecimal(C);
cout << "In two's complement arithmetic, " << a << " + " << b << " is " << c2 << endl;

//Print some concluding results
if (c1 == c2) cout << c1 << " is equal to " << c2 << ". Good Job!" << endl;
else
``````

{
cout << c1 << " is not equal to " << c2 << endl;
cout << "Either " << c1 << " cannot be represented by the given bit pattern OR we have made some mistake!" << endl;
}
system("Pause");
return 0;
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#### 6条回答默认 最新

• 繁宵微梦 2018-07-06 14:01
已采纳
``````#include <iostream>
#include <string>
#include <cmath>

using namespace std;

string decimalToTwocomplementString(int num, int length) {
string binary;
int positive_num = abs(num);
while (positive_num != 0) {
if (num >= 0){
binary.insert(0,to_string(positive_num & 1));
positive_num >>= 1;
} else {
binary.insert(0,to_string(!(positive_num & 1)));
positive_num >>=1;
}
}
if (num < 0) {
for (reverse_iterator it = binary.rbegin(); it != binary.rend(); it++) {
if (*it == '1') {
*it = '0';
} else {
*it = '1';
break;
}
}
}
while (binary.length() < length) {
if (num >= 0) {
binary.insert(0, "0");
} else {
binary.insert(0, "1");
}
}
return binary;
}

string TwoComplementStringAddition (const string& a, const string& b) {
int carry = 0;
string result;
for (reverse_iterator ita = a.rbegin(), itb = b.rbegin(); ita != a.rend(); ita++, itb++) {
int num_a = *ita - '0';
int num_b = *itb - '0';
result.insert(0,to_string(num_a ^ num_b ^ carry));
if (num_a == 1 && num_b == 1) {
carry = 1;
} else if (num_a == 0 && num_b == 0) {
carry = 0;
}
}
return result;
}

int TwoComplementStringToDecimal(string binary) {
int result = 0;
string::iterator it = binary.begin();
bool negative = *it == '1';
if (negative) {
for (; it != binary.end(); it++) {
*it = *it == '1' ? '0' : '1';
}
for (reverse_iterator iterator = binary.rbegin(); iterator != binary.rend(); iterator++) {
if (*iterator == '1') {
*iterator = '0';
} else {
*iterator = '1';
break;
}
}
}
for (it = binary.begin(); it != binary.end(); it++) {
result += (*it - '0') * pow(2,distance(it,binary.end()) - 1);
}
return negative ? -result : result;
}

int main()
{
//Read in the bit pattern size
int L;
do {
cout << "Enter positive integer for the bit pattern size ";
cin >> L;
}while (L <= 0);

//Read in two integers a and b
int a, b;
cout << "Enter an integer a ";
cin >> a;
cout << "Enter an integer b ";
cin >> b;

//Calculate the decimal arithmetic sum of a and b and print the result int
int c1 = a + b;
cout << "In decimal " << a << " + " << b << " is " << c1 << endl;

//Compute the two's complement representations of a and b
//Each integer must be represented in L-bits pattern
//Also these two's complement representations must be returned as string data types
string A = decimalToTwocomplementString(a, L);
string B = decimalToTwocomplementString(b, L);

//Print the two's complement representations of a and b
cout << "The two's complement of " << a << " is\t " << A << endl;
cout << "The two's complement of " << b << " is\t " << B << endl;

//Compute the binary sum of the two's complement representations of a and b
//The result must be returned as L-bit pattern string data type

//Print the two's complement representation binary sum
cout << "The binary sum of " << A << " and " << B << " is " << C << endl;

//Convert the two's complement representation binary sum to decimal and print
int c2 = TwoComplementStringToDecimal(C);
cout << "In two's complement arithmetic, " << a << " + " << b << " is " << c2 << endl;

//Print some concluding results
if (c1 == c2) cout << c1 << " is equal to " << c2 << ". Good Job!" << endl;
else

{
cout << c1 << " is not equal to " << c2 << endl;
cout << "Either " << c1 << " cannot be represented by the given bit pattern OR we have made some mistake!" << endl;
}
return 0;
}

``````

这个是整个程序的代码，实现了那三个函数，其他的都没有改动。
虽然我实现的方法不算是很聪明，但是应该还是很好理解的

已采纳该答案
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• 毛线一团 2018-07-06 04:25

这个有点难的。首先decimal可以以不同类型表示，也可以理解成文件描述符（file descriptor)。其次需要做索引解耦（index decoupling), 效果好坏需要工业界的验证。

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• weixin_38472405 2018-07-06 09:28

11011是补码，反码就是11010，源码就是10101=-5，哪里不懂吗？

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• mosespaul 2018-07-06 10:54
``````#include <string>
#include <stdlib.h>

string decimalToTwocomplementString(int a, int L)
{
string result = "";
while(L > 0)
{
result += std::toString(a % 2);
a /= 2;
L--;
}
}

int TwoComplementStringToDecimal(string C)
{
return strtol(C.c_str(), NULL, 2);
}

{
return decimalToTwocomplementString(TwoComplementStringToDecimal(A) + TwoComplementStringToDecimal(B));
}

``````

PS:
1.题目中要求不能用数组变量，但没有规定不能用C函数，
所以函数TwoComplementStringToDecimal偷懒用了c_str()和strtol()。
如果不能用，你就得用String的枚举变量一个个算了，反正无论如何按这题的规定，运行效率都不会高。
2.要是支持C++11就更简单了，参考C++11 stol函数和fmt::FormatInt。

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• 繁宵微梦 2018-07-06 13:22

题主的意思应该是想实现这三个函数吧。我假设题主理解什么是原码补码反码，也知道怎么互相转换，但是难以用编程语言来描述这个过程。如果是这样，那可以看看我的答案。

看了题目，题目中只让用`string`类型，连数组都不让用。所以题目的意思让自己来实现这些算法。虽然调用C++库可以非常轻松的实现这个要求，那出这道题也失去了意义。

首先是实现十进制的整数类型转换为字符串形式的二进制格式

这个最常见的就是采用*"除2取余，逆序排列"*法

``````string decimalToTwocomplementString(int num, int length) {
string binary;
int positive_num = abs(num);
//判断正负，正数的补码就是原码，而负数的补码是在原码的基础上按位取反再加1
while (positive_num != 0) {
//如果是正数，直接取其原码
if (num >= 0){
binary.insert(0,to_string(positive_num & 1));
positive_num >>= 1;
//如果是负数，先取它的反码
} else {
binary.insert(0,to_string(!(positive_num & 1)));
positive_num >>=1;
}
}
//如果是负数，将之前获取到的反码加1，得到补码
//由于这个是字符串，所以没有真正意义上的加1，而是根据二进制数的性质，从右向左遍历
//找到第一个是0的数，把这个0变成1，再把之前遍历到的所有1改成0，就相当于加1了
if (num < 0) {
for (reverse_iterator it = binary.rbegin(); it != binary.rend(); it++) {
if (*it == '1') {
*it = '0';
} else {
*it = '1';
break;
}
}
}
//如果当前的字符串长度小于输入的二进制位数
//是正数则前面补0，负数则前面补1
while (binary.length() < length) {
if (num >= 0) {
binary.insert(0, "0");
} else {
binary.insert(0, "1");
}
}
return binary;
}

``````

接下来是两个字符串相加

``````string TwoComplementStringAddition (const string& a, const string& b) {
//代表是否有进位
int carry = 0;
string result;
//从右向左遍历这俩字符串
//因为这俩字符串长度一定是一样的，所以终止条件写谁都无所谓
for (reverse_iterator ita = a.rbegin(), itb = b.rbegin(); ita != a.rend(); ita++, itb++) {
//把当前遍历到的位的值转换成整型
int num_a = *ita - '0';
int num_b = *itb - '0';
//将当前这俩数值与进位数异或，获得当前位置相加后的结果
result.insert(0,to_string(num_a ^ num_b ^ carry));
//当这俩字符串当前数值都是1时，肯定发生进位
//而且在有进位的前提下，只要当前这俩数值不是0和0那么一定会再次发生进位
if (num_a == 1 && num_b == 1) {
carry = 1;
} else if (num_a == 0 && num_b == 0) {
carry = 0;
}
}
return result;
}
``````

最后是把字符串转为十进制的整数类型

其实和第一个函数大同小异

``````int TwoComplementStringToDecimal(string binary) {
int result = 0;
string::iterator it = binary.begin();
//判断这个二进制数的符号位是1还是0
//来决定这个数是正还是负
bool negative = *it == '1';
//如果是负数，要把补码转换为原码
//和原码转补码是同样的过程
//这里我为了偷懒，把符号位也给变成0了，但实际上负数的原码符号位是1
if (negative) {
for (; it != binary.end(); it++) {
*it = *it == '1' ? '0' : '1';
}
for (reverse_iterator iterator = binary.rbegin(); iterator != binary.rend(); iterator++) {
if (*iterator == '1') {
*iterator = '0';
} else {
*iterator = '1';
break;
}
}
}
//加权求值
for (it = binary.begin(); it != binary.end(); it++) {
result += (*it - '0') * pow(2,distance(it,binary.end()) - 1);
}
return negative ? -result : result;
}

``````
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• mosespaul 2018-07-06 18:53
``````#include <iostream>
#include <cmath>
#include <string>

using namespace std;

/*Implement code begin */
string decimalToTwocomplementString(int a, int L)
{
string result = "";

//In our computer, int type save as complement itself! So we can save it to string directly with enough length.
//For example in 32bit computer, -14 saved as 0xFFFFFFFFFFFFFFF2,
//We only need 5 length to record in this question which is 0x12 (= binrary10010)
//And as a positive number, 0x5(=01001) don't have sign bit.
while(L-- > 0) {
result += (a & 0x1) ? string("1"):string("0");
a >>= 1;
}
return result;
}

int TwoComplementStringToDecimal(string C)
{
int len = C.length();
int num = 0;

//Translate string to positive number
for(int i = 0; i < len; i++)
{
num <<= 1;
if('0' != C[i])
num |= 1;
}

if(len > 0 && '1' == C) {
//It's negative number, we can change it back.
//in this case, our number is 0x12, we need to change back to 0xFFFFFFFFFFFFFFF2
num |= (~0 << len);
}
cout << "transfer" << C << "to" << num;
return num;
}

{
return decimalToTwocomplementString(TwoComplementStringToDecimal(A) + TwoComplementStringToDecimal(B), A.length());
}

/*Implement code End */

int main()
{
//Read in the bit pattern size
int L;
do {
cout << "Enter positive integer for the bit pattern size ";
cin >> L;
}while (L <= 0);

//Read in two integers a and b
int a, b;
cout << "Enter an integer a ";
cin >> a;
cout << "Enter an integer b ";
cin >> b;

//Calculate the decimal arithmetic sum of a and b and print the result int
int c1 = a + b;
cout << "In decimal " << a << " + " << b << " is " << c1 << endl;

//Compute the two's complement representations of a and b
//Each integer must be represented in L-bits pattern
//Also these two's complement representations must be returned as string data types
string A = decimalToTwocomplementString(a, L);
string B = decimalToTwocomplementString(b, L);

//Print the two's complement representations of a and b
cout << "The two's complement of " << a << " is\t " << A << endl;
cout << "The two's complement of " << b << " is\t " << B << endl;

//Compute the binary sum of the two's complement representations of a and b
//The result must be returned as L-bit pattern string data type

//Print the two's complement representation binary sum
cout << "The binary sum of " << A << " and " << B << " is " << C << endl;

//Convert the two's complement representation binary sum to decimal and print
int c2 = TwoComplementStringToDecimal(C);
cout << "In two's complement arithmetic, " << a << " + " << b << " is " << c2 << endl;

//Print some concluding results
if (c1 == c2) cout << c1 << " is equal to " << c2 << ". Good Job!" << endl;
else
{
cout << c1 << " is not equal to " << c2 << endl;
cout << "Either " << c1 << " cannot be represented by the given bit pattern OR we have made some mistake!" << endl;
}

//system("Pause");
return 0;
}
``````

PS: 突然发现下午看题忘记填补码,而且include缺少stdlib.h，楼上代码应该是个解（不细看啦），先赞一个。
强迫症还是要把自己的解法改正写清楚，检查完符合题目，编译运行也过了.(linux g++ 5.4.0-6)
补码是原码反码后+1只是一种计算方法，带符号整形在内存中就是补码存在的，如题不需要这么大的数(32bit)去计算，那只要去掉符号扩展的位数就行了，反之加回即可。
如果有问题敬请读到的指出来看看，代码的乐趣在于能够聪明地使用它，换个思路简化些，Have Fun。
当然这个题目本身也有点小瑕疵啦，比如L取值如果小于能容纳的位数怎么办(大于等于是没问题的)。

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