在新表达式中分配内存后是否对初始化程序求值? [英] Is initializer evaluated after memory allocation in new expression?

问题描述

考虑代码

auto p = new T( U(std::move(v)) );

则初始化器为U(std::move(v)).假设T( U(std::move(v)) )没有抛出.如果在基础内存分配之后对初始化程序进行了评估，则该代码将是强异常安全的.否则，事实并非如此.如果抛出了内存分配，则v应该已经被移动了.因此，我对内存分配和初始化程序评估之间的相对顺序感兴趣.它是定义的，未指定的还是什么?

The initializer is then U(std::move(v)). Let's assume that T( U(std::move(v)) ) does not throw. If the initializer is evaluated after the underlying memory allocation, the code is then strong-exception-safe. Otherwise, it is not. Had memory allocation thrown, v would have already been moved. I'm therefore interested in the relative order between memory allocation and initializer evaluation. Is it defined, unspecified, or what?

推荐答案

是的，初始化是在分配后进行评估的.引用C ++ 17(N4659)[expr.new] 8.3.4/19:

Yes, the initialisation is evaluated after the allocation. Quoting C++17 (N4659) [expr.new] 8.3.4/19:

分配函数的调用在 new-initializer 中的表达式求值之前进行排序. 分配的对象的初始化在 new-expression 的值计算之前进行排序.

The invocation of the allocation function is sequenced before the evaluations of expressions in the new-initializer. Initialization of the allocated object is sequenced before the value computation of the new-expression.

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