size_t,ptrdiff和std::向量::尺寸()

I thought that the correct type to use to store the difference between pointers was ptrdiff_t.

我认为用来存储指针之间差异的正确类型是ptrdiff_t。

As such, I'm confused by the way that my STL (msvc 2010) implements it's std::vector::size() function. The return type is size_t (this is mandated by the standard, as far as I understand it) and yet it's computed as the difference of pointers:

因此，我对我的STL (msvc 2010)实现它的方式感到困惑:std:::vector: size()函数。返回类型是size_t(根据我的理解，这是标准要求的)，但是它是作为指针的差值计算的:

// _Mylast, _Myfirst are of type pointer
// size_type, pointer are inherited from allocator<_Ty>
size_type size() const 
{
    return (this->_Mylast - this->_Myfirst);
}

Obviously, there's a bit of meta-magic that goes on in order to determine exactly what types size_type and pointer are. In order to be "sure" what types they are I checked this:

显然，要确定size_type和指针是什么类型，还需要进行一些元魔法。为了确定它们是什么类型，我检查了一下:

bool bs = std::is_same<size_t, std::vector<int>::size_type>::value;
bool bp = std::is_same<int * , std::vector<int>::pointer>::value;
// both bs and bp evaluate as true, therefore:
//   size_type is just size_t
//   pointer is just int*

Compiling the following with /Wall gives me a signed-to-unsigned mismatch for mysize2, but no warnings for mysize1:

使用/Wall编译下面的代码，我得到了mysize2的一个签名到未签名的不匹配，但是mysize1没有任何警告:

std::vector<int> myvector(100);
int *tail = &myvector[99];
int *head = &myvector[ 0];
size_t mysize1 = myvector.size();
size_t mysize2 = (tail - head + 1);

Changing the type of mysize2 to ptrdiff_t results in no warning. Changing the type of mysize1 to ptrdiff_t results in an unsigned-to-signed mismatch.

将mysize2类型更改为ptrdiff_t将导致无警告。将mysize1类型更改为ptrdiff_t会导致未签名到签名的不匹配。

Obviously I'm missing something...

显然我少了什么……

EDIT: I'm not asking how to suppress the warning, with a cast or a #pragma disable(xxx). The issue I'm concerned about is that size_t and ptrdiff_t may have different allowable ranges (they do on my machine).

编辑:我并不是问如何通过强制转换或#pragma disable(xxx)来抑制警告。我担心的问题是size_t和ptrdiff_t可能有不同的允许范围(它们在我的机器上有)。

Consider std::vector<char>::max_size(). My implementation returns a max_size equal to std::numeric_limits<size_t>::max(). Since vector::size() is creating an intermediate value of type ptrdiff_t before casting to size_t it seems that there could be problems here - ptrdiff_t is not big enough to hold vector<char>::max_size().

考虑std::向量< char >::max_size()。我的实现返回一个max_size，等于std::numeric_limits ::max()。由于向量::size()在转换到size_t之前创建了一个ptrdiff_t类型的中间值，看起来这里可能存在问题——ptrdiff_t不够大，不能容纳vector ::max_size()。

2 个解决方案

#1

Generally speaking, ptrdiff_t is a signed integral type of the same size as size_t. It must be signed so that it can represent both p1 - p2 and p2 - p1.

一般来说，ptrdiff_t是与size_t大小相同的有符号积分类型。它必须有符号，这样它就可以代表p1 - p2和p2 - p1。

In the specific case of the internals of std::vector, the implementor is effectively deriving size() from end() - begin(). Because of the guarantees of std::vector (contiguous, array based storage), the value of the end pointer will always be greater than the value of the begin pointer, and thus there is no risk of generating a negative value. In fact, size_t will always be able to represent a larger positive range than will ptrdiff_t, as it doesn't have to use half its range to represent negative values. Effectively, this means that the cast in this case from ptrdiff_t to size_t is a widening cast, which has well defined (and intuitively obvious) results.

在std::vector的内部情况中，实现者有效地从end() - begin()派生出size()。由于std::vector(连续的，基于数组的存储)的保证，结束指针的值总是大于开始指针的值，因此不存在生成负值的风险。实际上，size_t总是能够比ptrdiff_t表示一个更大的正值范围，因为它不需要用它的一半的值域来表示负值。有效地说，这意味着从ptrdiff_t到size_t的强制转换是一种扩展的强制转换，它具有很好的定义(直观地很明显)结果。

Also, note that this is not the only possible implementation of std::vector. It could just as easily be implemented as a single pointer and a size_t value holding the size, deriving end() as begin() + size(). That implementation would also resolve your max_size() concern. In reality, max_size is never actually attainable--it would require your program's entire address space to be allocated for the vector's buffer, leaving no room for the begin()/end() pointers, function call stack, etc.

另外，请注意，这并不是std::vector的唯一可能实现。它可以很容易地实现为一个指针和一个size_t值，包含大小，派生end()和begin() + size()。该实现还将解决您的max_size()问题。实际上，max_size实际上是不可能实现的——它需要您的程序的整个地址空间被分配给vector的缓冲区，而不为begin()/end()指针、函数调用堆栈等提供空间。

#2

There is nothing wrong with how std::vector::size() is implemented in STL. The this->_Mylast - this->_Myfirst == vector size is mere an coincidental fact which relies on how the vector is implemented.

std::vector::size()在STL中实现没有问题。这个->_Mylast ->_Myfirst =向量大小仅仅是一个巧合的事实，它依赖于向量是如何实现的。

Plus MSVC STL vector implementation has an #pragma warning(disable: 4244) which removes the warning.

加上MSVC STL向量实现有一个#pragma警告(禁用:4244)，它删除了警告。

#1