STL 容器要求存储的值是可复制构造和可分配的.const T 显然不是任何 T 的可赋值类型，但我尝试使用它(只是出于好奇)并发现它可以编译，而且还表现为可赋值类型.
STL containers require the stored values to be copy constructible and assignable. const T is obviously not an assignable type for any T, but I tried to use it (just being curious) and found out that it compiles and, moreover, behaves as an assignable type.
vector<const int> v(1);
v[0] = 17;
这在 Visual Studio 2008 中成功运行并将 v[0] 分配给 17.
This successfully runs in Visual Studio 2008 and assigns v[0] to 17.
这不是其他人建议的实现中的错误.
This is not a bug in the implementation as others have suggested.
违反 C++ 标准库工具的要求不会使您的程序格式错误，而是会产生未定义的行为.
Violating the requirements of a C++ Standard Library facility does not render your program ill-formed, it yields undefined behavior.
您违反了存储在容器中的值类型必须是可复制构造和可分配的要求(const 类型显然不可分配)，因此您的程序表现出未定义的行为.
You have violated the requirement that the value type stored in a container must be copy constructible and assignable (const types are not assignable, obviously), so your program exhibits undefined behavior.
C++ 标准中的适用语言可以在 C++03 17.4.3.6 [lib.res.on.functions] 中找到:
The applicable language from the C++ Standard can be found in C++03 17.4.3.6 [lib.res.on.functions]:
在某些情况下(替换函数、处理函数、对用于实例化标准库模板组件的类型的操作)，C++ 标准库依赖于由 C++ 程序提供的组件.如果这些组件不满足他们的要求，则标准不会对实现提出要求.
In certain cases (replacement functions, handler functions, operations on types used to instantiate standard library template components), the C++ Standard Library depends on components supplied by a C++ program. If these components do not meet their requirements, the Standard places no requirements on the implementation.
特别是，在以下情况下效果未定义:
In particular, the effects are undefined in the following cases:
...
Visual C++ 标准库实现可以对这段代码做任何事情，包括静默删除或忽略 const 限定，它仍然符合标准.
The Visual C++ Standard Library implementation may do anything with this code, including silently removing or ignoring the const-qualification, and it is still standards-conforming.
这篇关于VC++ 允许对 STL 容器使用 const 类型.为什么?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！