C++ 在不使用第三个变量的情况下交换两个变量值
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原文地址: http://stackoverflow.com/questions/1826159/
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Swapping two variable value without using third variable
提问by Muhammad Akhtar
One of the very tricky questions asked in an interview.
面试中提出的非常棘手的问题之一。
Swap the values of two variables like a=10and b=15.
交换两个变量的值，如a=10和b=15。
Generally to swap two variables values, we need 3rd variable like:
通常要交换两个变量值，我们需要第三个变量，例如：
temp=a;
a=b;
b=temp;
Now the requirement is, swap values of two variables without using 3rd variable.
现在的要求是，在不使用第三个变量的情况下交换两个变量的值。
回答by Yacoby
Using the xor swap algorithm
使用异或交换算法
void xorSwap (int* x, int* y) {
if (x != y) { //ensure that memory locations are different
*x ^= *y;
*y ^= *x;
*x ^= *y;
}
}
Why the test?
为什么要考试？
The test is to ensure that x and y have different memory locations (rather than different values). This is because (p xor p) = 0and if both x and y share the same memory location, when one is set to 0, both are set to 0.
When both *x and *y are 0, all other xor operations on *x and *y will equal 0 (as they are the same), which means that the function will set both *x and *y set to 0.
测试是为了确保 x 和 y 具有不同的内存位置（而不是不同的值）。这是因为(p xor p) = 0如果 x 和 y 共享相同的内存位置，当一个设置为 0 时，两者都设置为 0。当 *x 和 *y 都为 0 时，*x 和 *y 上的所有其他异或运算将等于0（因为它们是相同的），这意味着该函数会将 *x 和 *y 设置为 0。
If they have the same values but not the same memory location, everything works as expected
如果它们具有相同的值但不相同的内存位置，则一切都按预期进行
*x = 0011
*y = 0011
//Note, x and y do not share an address. x != y
*x = *x xor *y //*x = 0011 xor 0011
//So *x is 0000
*y = *x xor *y //*y = 0000 xor 0011
//So *y is 0011
*x = *x xor *y //*x = 0000 xor 0011
//So *x is 0011
Should this be used?
应该使用这个吗？
In general cases, no. The compiler will optimize away the temporary variable and given that swapping is a common procedure it should output the optimum machine code for your platform.
在一般情况下，没有。编译器将优化掉临时变量，并且鉴于交换是一个常见过程，它应该为您的平台输出最佳机器代码。
Take for example this quick test program written in C.
以这个用 C 编写的快速测试程序为例。
#include <stdlib.h>
#include <math.h>
#define USE_XOR
void xorSwap(int* x, int *y){
if ( x != y ){
*x ^= *y;
*y ^= *x;
*x ^= *y;
}
}
void tempSwap(int* x, int* y){
int t;
t = *y;
*y = *x;
*x = t;
}
int main(int argc, char* argv[]){
int x = 4;
int y = 5;
int z = pow(2,28);
while ( z-- ){
# ifdef USE_XOR
xorSwap(&x,&y);
# else
tempSwap(&x, &y);
# endif
}
return x + y;
}
Compiled using:
编译使用：
gcc -Os main.c -o swap
The xor version takes
异或版本需要
real 0m2.068s
user 0m2.048s
sys 0m0.000s
Where as the version with the temporary variable takes:
带有临时变量的版本需要：
real 0m0.543s
user 0m0.540s
sys 0m0.000s
回答by jk.
the general form is:
一般形式是：
A = A operation B
B = A inverse-operation B
A = A inverse-operation B
however you have to potentially watch out for overflows and also not all operations have an inverse that is well defined for all values that the operation is defined. e.g. * and / work until A or B is 0
但是，您必须潜在地注意溢出，并且并非所有操作都具有为定义操作的所有值良好定义的逆。例如 * 和 / 工作直到 A 或 B 为 0
xor is particularly pleasing as it is defined for all ints and is its own inverse
xor 特别令人愉快，因为它是为所有整数定义的，并且是它自己的逆
回答by Dor
a = a + b
b = a - b // b = a
a = a - b
回答by CB Bailey
No-one has suggested using std::swap, yet.
没有人建议使用std::swap, 。
std::swap(a, b);
Idon't use any temporary variables and depending on the type of aand bthe implementation may have a specalization that doesn't either. The implementation should be written knowing whether a 'trick' is appropriate or not. There's no point in trying to second guess.
我不使用任何临时变量，并且根据类型a和b实现可能有一个也没有的专业化。编写实现时应该知道“技巧”是否合适。尝试第二次猜测是没有意义的。
More generally, I'd probably want to do something like this, as it would work for class types enabling ADL to find a better overload if possible.
更一般地说，我可能想要做这样的事情，因为它适用于类类型，如果可能的话，它使 ADL 能够找到更好的重载。
using std::swap;
swap(a, b);
Of course, the interviewer's reaction to this answer might say a lot about the vacancy.
当然，面试官对这个答案的反应可能会说很多关于空缺的信息。
回答by Vilx-
As already noted by manu, XOR algorithm is a popular one which works for all integer values (that includes pointers then, with some luck and casting). For the sake of completeness I would like to mention another less powerful algorithm with addition/subtraction:
正如 manu 已经指出的那样，XOR 算法是一种流行的算法，适用于所有整数值（包括指针，有一些运气和转换）。为了完整起见，我想提及另一种功能较弱的加法/减法算法：
A = A + B
B = A - B
A = A - B
Here you have to be careful of overflows/underflows, but otherwise it works just as fine. You might even try this on floats/doubles in the case XOR isn't allowed on those.
在这里你必须小心上溢/下溢，否则它也能正常工作。在不允许 XOR 的情况下，您甚至可以在浮点数/双精度数上尝试此操作。
回答by MSalters
Stupid questions deserve appropriate answers:
愚蠢的问题应该得到适当的回答：
void sw2ap(int& a, int& b) {
register int temp = a; // !
a = b;
b = temp;
}
The only good use of the registerkeyword.
唯一好用的register关键字。
回答by vikas
#include<iostream.h>
#include<conio.h>
void main()
{
int a,b;
clrscr();
cout<<"\n==========Vikas==========";
cout<<"\n\nEnter the two no=:";
cin>>a>>b;
cout<<"\na"<<a<<"\nb"<<b;
a=a+b;
b=a-b;
a=a-b;
cout<<"\n\na="<<a<<"\nb="<<b;
getch();
}
回答by user3258202
Since the original solution is:
由于原始解决方案是：
temp = x; y = x; x = temp;
You can make it a two liner by using:
您可以使用以下方法使其成为两个班轮：
temp = x; y = y + temp -(x=y);
Then make it a one liner by using:
然后使用以下方法使其成为单衬里：
x = x + y -(y=x);
回答by Ashwini
Swapping two numbers using third variable be like this,
使用第三个变量交换两个数字是这样的，
int temp;
int a=10;
int b=20;
temp = a;
a = b;
b = temp;
printf ("Value of a", %a);
printf ("Value of b", %b);
Swapping two numbers without using third variable
在不使用第三个变量的情况下交换两个数字
int a = 10;
int b = 20;
a = a+b;
b = a-b;
a = a-b;
printf ("value of a=", %a);
printf ("value of b=", %b);
回答by Naeem Ul Hassan
#include <iostream>
using namespace std;
int main(void)
{
int a,b;
cout<<"Enter a integer" <<endl;
cin>>a;
cout<<"\n Enter b integer"<<endl;
cin>>b;
a = a^b;
b = a^b;
a = a^b;
cout<<" a= "<<a <<" b="<<b<<endl;
return 0;
}
Update:In this we are taking input of two integers from user. Then we are using the bitwise XORoperation to swap them.
更新：在这里，我们从用户那里输入两个整数。然后我们使用按位异或操作来交换它们。
Say we have two integers a=4and b=9and then:
假设我们有两个整数a=4和b=9，然后：
a=a^b --> 13=4^9
b=a^b --> 4=13^9
a=a^b --> 9=13^9
