使用初始化列表声明动态分配数组的大小

Question

使用初始化列表声明动态分配数组的大小

c++gccvisual-c++language-lawyerdynamic-arrays

6

我试图使用初始化列表初始化一个动态声明的数组，但我注意到在使用GCC时必须提供数组大小，否则会出现错误。如果省略数组大小并尝试使用MSVC做同样的操作，则不会出现任何错误。当使用初始化列表与动态数组时是否必须提供数组大小？这是一些实现定义的东西吗，这就是为什么两个编译器不同的原因吗？

int *array { new int [3] {0, 1, 2} }; // Works with both MSVC and GCC.
int *array { new int [] {0, 1, 2} }; // Works only with MSVC, not GCC.

- Mohammad Ali

Clang也接受两个版本演示。 - Jarod42

（initializer-list 标签是指 std::initializer_list） - 463035818_is_not_a_number

虽然我不能百分之百确定并详细说明，但我认为g++在这里是错误的，第二个版本应该被接受。这个参考链接也给出了double* p = new double[]{1,2,3};作为可行的例子。 - Lukas-T

https://github.com/cplusplus/draft/commit/44d91596bf91ddbaada724ceda4d893a853dd444 - Language Lawyer

1个回答

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可以查看英文原文，
原文链接

- dfrib · Accepted Answer

这是C++20中实现的新表达式中的数组大小推导，请参阅P1009R2。

Bjarne Stroustrup pointed out the following inconsistency in the C++ language:
double a[]{1,2,3}; // this declaration is OK, ...
double* p = new double[]{1,2,3}; // ...but this one is ill-formed!
Jens Maurer provided the explanation why it doesn’t work: For a new-expression, the expression inside the square brackets is currently mandatory according to the C++ grammar. When uniform initialization was introduced for C++11, the rule about deducing the size of the array from the number of initializers was never extended to the new-expression case. Presumably this was simply overlooked. There is no fundamental reason why we cannot make this work [...]

Proposed wording

The reported issue is intended as a defect report with the proposed resolution as follows. The effect of the wording changes should be applied in implementations of all previous versions of C++ where they apply. [...]

从GCC的C++标准支持页面可以看出，GCC将P1009R2列为在GCC 11中已实现，并且我们可以验证GCC 11已将实现向后移植，以接受OP的示例作为符合C++11的规范。

演示（GCC 11 / -std=c++11）。