问题描述
我学习使用Xor运算符交换两个整数，例如：
int a = 21;
int b = 7;
a ^ = b ^ = a ^ = b;
我终于得到a = 7和b = 21。
我尝试使用xor操作符在数组上像这样：
int main（）
{
int a [] = { 7,21};
a [0] ^ = a [1] ^ = a [0] ^ = a [1];
cout<< a [0]<<'，'<< a [1]
return 0;
}
输出为 0,7
我在Xcode和g ++上编译代码，他们有同样的问题。
Xor交换数组工作正常，多行：
int main（）
{
int a [] = {7,21};
a [0] ^ = a [1];
a [1] ^ = a [0];
a [0] ^ = a [1];
cout<< a [0]<<'，'<< a [1]
return 0;
}
我会输出 21,7 / p>
这是我已经找到的信息：
- 问题是关于序列点：
- 即使对于简单整数，它们可能对此未定义行为有副作用：
- 有关xor交换的一些其他问题：
所以我应该避免使用xor交换，保证正确的结果。
但我还是不太清楚在上发生了什么a [0] ^ = a [1] ^ = a [0 ] ^ = a [1]; 它的序列点问题是什么？
我无法弄清楚编译器有什么不同 a [0] ^ = a [1] ^ = a [0] ^ = a [1]; 和 a ^ = b ^ = a ^ = b; ？
我的疑问是：
编译器输出 0,7 for a [0] ^ = a [1] ^ = a [0] ^ = a [1];
我知道这是序列指针问题，我可以理解为什么 printf d，％d，i ++，i ++）; 未定义为函数的一些编译器解析参数从左到右，有些从右到左。
但我不知道 a [0] ^ = a [1] ^ = a [0] ^ = a [1]; ，它看起来与 a ^ = b ^ = a ^ = b; 相同。所以我想知道它如何与数组一起使用。
You不能修改变量多于一次而没有中间序列点，如果这样做，则是未定义行为。
<$ c ^ c> a ^ = b ^ = a ^ = b;
尝试修改 a 上面的语句中的 b 打破了这个规则，你最终会得到一个未定义的行为。
注意，未定义的行为意味着任何行为都是可能的，获得任何输出。
良好阅读：
I learned using Xor operator to swap two integers,like:int a = 21;
int b = 7;
a^=b^=a^=b;
I would finally get a=7 and b=21.
I try to use xor operator on array like this way:
int main()
{
int a[] = {7,21};
a[0]^=a[1]^=a[0]^=a[1];
cout << a[0] <<',' <<a[1];
return 0;
}
The output is 0,7
I compile the code on Xcode and g++, they have the same issue.
Xor swap on array works fine with multiple lines:
int main()
{
int a[] = {7,21};
a[0]^=a[1];
a[1]^=a[0];
a[0]^=a[1];
cout << a[0] <<',' <<a[1];
return 0;
}
I would get output as 21,7
Here is the information what I already find:
- the issue is about sequence point: Array + XOR swap fails
- even for simple integers, they may have side affect to this undefined behavior: Why is this statement not working in java x ^= y ^= x ^= y;
- some other issue on xor swap: Weird XOR swap behavior while zeroing out data
So I should avoid using xor swap, instead, swap with temp would guarantee correct result.
But I still not very clear about what happen on a[0]^=a[1]^=a[0]^=a[1]; what is the sequence point issue with it?
I could not figure out what's the different on compiler between a[0]^=a[1]^=a[0]^=a[1]; and a^=b^=a^=b; ?
My doubt is:
" How does compiler output 0,7 for a[0]^=a[1]^=a[0]^=a[1];. "
I know this is sequence pointer issue, I could understand why printf("%d,%d",i++, i++); is undefined as some compiler parse parameter of function from left to right, and some do it from right to left.
But I do not know what is the problem on a[0]^=a[1]^=a[0]^=a[1];, it looks just the same as a^=b^=a^=b;. So I'd like to know how it works with array. So that I would know more about kind like "sequence pointer on index of array"
You cannot modify a variable more than once without an intervening sequence point, if you do so, it is Undefined Behavior.
a^=b^=a^=b;
Trying to modify the values of a and b in the above statement breaks this rule and you end up with an Undefined Behavior.
Note that Undefined Behavior means that any behavior is possible and you can get any output.
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