为（x ++，y）+（Y +，X）未定义或不确定的，如果没有指定，那是什么计算？ [英] Is (x++, y) + (y++, x) undefined or unspecified, and if unspecified, what can it compute?

问题描述

借助逗号序列算引入的顺序点在离pression 。我想知道这是否意味着，下面的程序避免不确定的行为。

The comma sequence operator introduces a sequence point in an expression. I am wondering whether this means that the program below avoids undefined behavior.

int x, y;

int main()
{
  return (x++, y) + (y++, x);
}

如果它确实避免不确定的行为，它可能仍然是不确定的，也就是说，返回几个可能的值之一。我认为，在C99，它只能计算 1 ，但实际上，GCC的各种版本编译这个程序进入返回可执行 2 。锵生成返回一个可执行 1 ，显然与我的直觉同意。

If it does avoid undefined behavior, it could still be unspecified, that is, return one of several possible values. I would think that in C99, it can only compute 1, but actually, various versions of GCC compile this program into an executable that returns 2. Clang generates an executable that returns 1, apparently agreeing with my intuition.

最后，就是这个东西，在C11改变？

Lastly, is this something that changed in C11?

推荐答案

就拿前pression：

Take the expression:

(x++, y) + (y++, x)

评估左到右：

x++  // yield 0, schedule increment of x
,    // sequence point: x definitely incremented now
y    // yield 0
y++  // yield 0, schedule increment of y
// explode because you just read from y and wrote to y
// with no intervening sequence point

有什么在禁止这一点，所以整个事情是未定义行为的标准。

There's nothing in the standard that forbids this, so the whole thing has undefined behavior.

对比度这个伪code：

Contrast this pseudocode:

f() { return x++, y; }
g() { return y++, x; }
f() + g()

Acoording到C99（5.1.2.3/2）的F的调用 和先按g 自己算的副作用和函数调用运算符包含它进入函数之前序列点。这意味着功能的执行不能交错。

Acoording to C99 (5.1.2.3/2) the calls to f and g themselves count as side effects, and the function call operator contains a sequence point just before it enters a function. This means function executions can't interleave.

在评价并行的东西的模式：

Under the "evaluate things in parallel" model:

f()  // arbitrarily start with f: sequence point; enter f
g()  // at the same time, start calling g: sequence point

由于 F的执行算作一个副作用本身，在克序列点（）暂停执行直到˚F又回来了。因此，没有未定义的行为。

Since the execution of f counts as a side effect itself, the sequence point in g() suspends execution until f has returned. Thus, no undefined behavior.

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