VC++允许对 STL 容器使用 const 类型。为什么？
STL 容器要求存储的值是可复制构造和可赋值的。 const T 显然不是任何 T 的可分配类型，但我尝试使用它（只是出于好奇）并发现它可以编译，而且表现为可分配类型。
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。
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;
This successfully runs in Visual Studio 2008 and assigns v[0] to 17.
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正如其他人所建议的那样，这并不是实施中的错误。
违反 C++ 标准库设施的要求不会导致程序格式错误，而是会产生未定义的行为。
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（显然，const 类型不可分配），因此您的程序表现出未定义的行为。
C++ 标准中适用的语言可以在 C++03 17.4.3.6 [lib.res.on.functions] 中找到：
Visual C++ 标准库实现可以对此代码执行任何操作，包括静默删除或忽略 const 限定，并且它仍然符合标准。
This is not a bug in the implementation as others have suggested.
Violating the requirements of a C++ Standard Library facility does not render your program ill-formed, it yields undefined behavior.
You have violated the requirement that the value type stored in a container must be copy constructible and assignable (
consttypes are not assignable, obviously), so your program exhibits undefined behavior.The applicable language from the C++ Standard can be found in C++03 17.4.3.6 [lib.res.on.functions]:
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.
这根本不应该起作用。正如您所说，在第 §23.1 ¶ 3 节中，指定了存储在容器中的对象必须是
CopyConstructible（如第 §20.1.3 节中所指定）和可分配。类型
T的Assignable要求是，t和u类型为T< /code>，您可以这样做：将
T&作为返回值，将t等价于u作为后置条件。 (§23.1 ¶4)因此，
const类型显然不可可赋值，因为执行t = u会引发编译错误（§7.1. 5.1¶5)。我认为这是微软实施中的一个错误。如果您尝试实例化
vector（使用和不使用-std=c++0x进行测试，Linux 上的 g++ 会发出典型的 25 kajillion-lines 模板错误）代码> 标志，以防万一）。顺便说一句，这也在 IBM 常见问题解答。
理论上，正如 @James McNellis 所说，编译器不需要在向量实例化上崩溃（如果它是未定义的行为，任何事情都可能发生 - 包括一切正常）；但是，在赋值语句中违反了标准，应该会产生编译错误。
事实上，
operator[]成员返回一个vector::reference ；该值必须是T的左值（§23.1 ¶5 表 66）；由于T是const类型，因此它将是const左值。因此，我们陷入 (§7.1.5.1 ¶5)，它将尝试对const元素执行赋值的代码定义为“格式错误”，这需要一个编译错误或至少警告，因为赋值给const是一个可诊断的规则（§1.4 ¶1-2）（未指定“无需诊断”语句）。最终编辑
实际上，@James McNellis 是对的；一旦您通过实例化
vector调用了未定义的行为，通常的规则就不再有价值，因此无论它做什么，实现仍然符合标准 - 包括删除const 从元素类型或生成通常的鼻恶魔。This simply shouldn't work. At §23.1 ¶ 3 it's specified, as you said, that objects stored in a container must be
CopyConstructible(as specified at §20.1.3) andAssignable.The
Assignablerequirements for a typeTare that, beingtanduof typeT, you can do:having a
T&as return value andtequivalent touas postcondition. (§23.1 ¶4)Thus,
consttypes are clearly notAssignable, since doingt = uwould raise a compilation error (§7.1.5.1 ¶5).I suppose that this is a bug in Microsoft implementation. g++ on Linux emits the typical 25 kajillion-lines template error if you even try to instantiate a
vector<const int>(tested both with and without the-std=c++0xflag, just in case).By the way, this is also explained in detail in this IBM FAQ.
In theory, as @James McNellis said, the compiler is not required to blow up on the vector instantiation (if it's undefined behavior anything can happen - including everything working fine); however, on the assignation statement there's a violation of the standard that should produce a compilation error.
In facts, the
operator[]member returns avector<const int>::reference; that one is required to be an lvalue ofT(§23.1 ¶5 table 66); sinceTis of aconsttype, it will be aconstlvalue. So we fall down to (§7.1.5.1 ¶5), which defines code that tries to perform an assignment to aconstelement as "ill-formed", and this demands a compilation error or at least a warning, because assignment-to-constis a diagnosable rule (§1.4 ¶1-2) (no "no diagnostic is required" statement is specified).Final edit
Actually, @James McNellis is right; once you've invoked undefined behavior by instantiating
vector<const int>, the usual rules stop having value, so the implementation is still standard-conforming whatever it does - including removing theconstfrom the element type or generating the usual nasal demons.