未定义的行为。 .. [英] Undefined Behavior. ..
问题描述
以下语句是否会调用未定义的行为：
a ^ = b，b ^ = a，a ^ = b;
给出a和b是int型？
要小心，我还没写过^ = b ^ = a ^ = b;当然，这是
undefined。我对前一个陈述感到困惑！
另外，说明声明未定义的原因！
Will the following statement invoke undefined behavior :
a^=b,b^=a,a^=b ;
given that a and b are of int-type ??
Be cautious, I have not written a^=b^=a^=b ; which, of course, is
undefined. I am having some confusion with the former statement!
Also, state the reason for the statement being undefined!
推荐答案
co ************** @ gmail。 com 写道：
以下语句是否会调用未定义的行为：
a ^ = b，b ^ = a，a ^ = b;
假设a和b是int型？
Will the following statement invoke undefined behavior :
a^=b,b^=a,a^=b ;
given that a and b are of int-type ??
否，因为逗号运算符构成序列点
No, because the comma operator constitutes a sequence point
要小心，我还没写过^ = b ^ = a ^ = b;当然，这是
undefined。我对前一个陈述感到困惑！
另外，说明声明未定义的原因！
Be cautious, I have not written a^=b^=a^=b ; which, of course, is
undefined. I am having some confusion with the former statement!
Also, state the reason for the statement being undefined!
6.5.17.1
逗号运算符的左操作数被计算为void表达式;
那里是评估后的一个序列点。
再见，Jojo
6.5.17.1
The left operand of a comma operator is evaluated as a void expression;
there is a sequence point after its evaluation.
Bye, Jojo
<<谨慎，我还没写过^ = b ^ = a ^ = b;当然，这是
undefined。我对前一个陈述感到困惑！>>
我认为这个表达式是有效的，因为赋值运算符被评估
从右到左所以它被解析为
（a ^ =（b ^ =（a ^ = b）））
以上表达式交换两个数字
please如果我错了，请纠正我...
<<Be cautious, I have not written a^=b^=a^=b ; which, of course, is
undefined. I am having some confusion with the former statement!>>
I think this expression is valid as assignment operator is evaluated
from right to left so it is parsed as
(a^=(b^=(a^=b)))
above expression swaps two numbers
please correct me if i am wrong...
c.***********@gmail.com 写道：
<<谨慎，我还没有写a ^ = b ^ = a ^ = b;当然，这是
undefined。我对前一个陈述感到困惑！>>
我认为这个表达式是有效的，因为赋值运算符被评估
从右到左所以它被解析为
（a ^ =（b ^ =（a ^ = b）））
以上表达式交换两个数字
please如果我错了，请纠正我...
<<Be cautious, I have not written a^=b^=a^=b ; which, of course, is
undefined. I am having some confusion with the former statement!>>
I think this expression is valid as assignment operator is evaluated
from right to left so it is parsed as
(a^=(b^=(a^=b)))
above expression swaps two numbers
please correct me if i am wrong...
这是在没有中间序列点的情况下修改''a'两次 - >
未定义的行为
再见，Jojo
This is modifying ''a'' twice without an intermediate sequence point ->
undefined behavoir
Bye, Jojo
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