为什么大小不是std :: initializer_list的模板参数？ [英] Why is the size not a template argument of std::initializer_list?

问题描述

std :: initializer_list 由编译器从括号括起的init列表构建，并且此列表的大小必须是编译时常量。

std::initializer_list is constructed by the compiler from a brace-enclosed init list and the size of this list must be a compile time constant.

那么为什么委员会决定从模板参数中忽略大小呢？这可能会阻止一些优化，并使某些事情变得不可能（从 std :: initializer_list 中初始化 std :: array ）。

So why did the committee decide to omit the size from the template arguments? This possibly prevents some optimizations and makes some things impossible (initializing std::array from a std::initializer_list).

推荐答案

现有系统的一个优点是，您可以导出需要 initializer_list 从DLL。如果它是在模板上的大小，他们必须作为来源。

One upside of the existing system is that you can export functions which take an initializer_list from a DLL. If it were templated on the size, they would have to be shipped as source.

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