C语言 不使用“-”运算符减去两个数字
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Subtracting two numbers without using '-' operator
提问by pranay
i tried with the following code , but i can't understand why it's giving me wrong answer. i am computing the 2's complement and adding with another no.
我尝试使用以下代码，但我不明白为什么它给了我错误的答案。我正在计算 2 的补码并加上另一个数字。
#include <stdio.h>
int add(int a, int b) {
while (a) {
a = (a & b) << 1;
b = a^b;
}
return b;
}
int sub(int a, int b) // add a with b's 2's complement.
{
return (add(a, add(~b, 1)));
}
int main() {
int a, b, res;
a = 3, b = 1;
res = sub(a, b);
printf("%d\n", res);
return 0;
}
采纳答案by pranay
i used a different add() function as suggested by NullUserException, it works now:
我使用了 NullUserException 建议的不同 add() 函数，它现在可以工作：
int add(int a,int b)
{
int x;
x = a^b;
while(a&b)
{
b = ((a&b)<<1);
a = x;
x = a^b;
//b=(a^b);
}
return x;
}
回答by ShinTakezou
Considering how negative numbers are represented, the following will compute a - b:
考虑如何表示负数，以下将计算 a - b：
int a, b, c;
// assign to a and b
c = a + (~b + 1); // () not needed, just to show the point
as the OP already noted:) This moves the attention to your add implementation, that is of course wrong. The following is an odd way to do it (just since other better ways are already given)
正如 OP 已经指出的那样:) 这将注意力转移到您的添加实现上，这当然是错误的。以下是一种奇怪的方法（因为已经给出了其他更好的方法）
int add1(int a, int b, int *c)
{
int r = *c & 1;
a &= 1; b &= 1;
*c = a&b | a&r | b&r;
return a^b^r;
}
int inv(int a)
{
int i, r = 0;
for(i = 0; i < sizeof(int)*8; i++)
{
r = r<<1 | (a&1);
a >>= 1;
}
return r<<1;
}
int add(int a, int b)
{
int r = 0, i;
int c = 0;
for(i=0; i < sizeof(int)*8; i++)
{
r |= add1(a>>i, b>>i, &c);
r <<= 1;
}
return inv(r);
}
int sub(int a, int b)
{
return add(a, add(~b, 1));
}
(keeping the same idea the code can be made better, just too tired to do it finer)
（保持同样的想法，代码可以做得更好，只是太累了，不能做得更好）
回答by Max Kapsecker
You also can implement this recursively. In C this might look like:
您也可以递归地实现这一点。在 C 中，这可能如下所示：
#include <stdio.h>
int add(int a, int b){
if(b == 0) return a;
int sum = a ^ b;
int carry = (a & b) << 1;
return add(sum, carry);
}
int subtract(int a, int b){
return add(a, add(~b, 1));
}
int main(){
int a = 3;
int b = 1;
int sum = add(a, b);
printf("%i + %i = %i \n", a, b, sum);
int difference = subtract(a, b);
printf("%i - %i = %i \n", a, b, difference);
return 0;
}
回答by Mohammad Adnan
add method implementation is incorrect. do like this -> A java way of this.
add 方法实现不正确。这样做 -> 一种 java 方式。
public int add(int a, int b){
do {
a = a & b; //carry
b = a ^ b; //addition
a = a << 1; //carry shift to one bit left
}while(a != 0); //exit
return b; //addition result
}
public int sub(int a, int b){
return add(a, add(~b, 1));
}
回答by Akash Das
import java.util.List;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.Hashtable;
import java.util.LinkedList;
import java.util.Scanner;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
import javax.xml.soap.Node;
public class mainone {
public static void main(String args[]){
int a=12;
int b=4;
Integer c=new Integer(b);
String d=Integer.toString(c);
String e="-";
String f=e.concat(d);
Integer g=Integer.parseInt(f);
System.out.println(a+g);
}
}
