C++ 向量 max_size();
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C++ vector max_size();
提问by Kundan
On 32 bit System.
在 32 位系统上。
std::vector<char>::max_size()returns 232-1, size ofchar— 1 bytestd::vector<int>::max_size()returns 230-1, size ofint— 4 bytestd::vector<double>::max_size()returns 229-1, size ofdouble— 8 byte
std::vector<char>::max_size()返回 2 32-1，大小为char— 1 个字节std::vector<int>::max_size()返回 2 30-1，大小为int— 4 字节std::vector<double>::max_size()返回 2 29-1，大小为double— 8 字节
can anyone tell me max_size()depends on what?
谁能告诉我max_size()取决于什么？
and what will be the return value of max_size()if it runs on 64 bit system.
max_size()如果它在 64 位系统上运行，返回值是什么。
采纳答案by rahul sinha
Simply get the answer by
只需通过以下方式获得答案
std::vector<dataType> v;
std::cout << v.max_size();
Or we can get the answer by (2^nativePointerBitWidth)/sizeof(dataType) - 1. For example, on a 64 bit system, long longis (typically) 8 bytes wide, so we have (2^64)/8 - 1 == 2305843009213693951.
或者我们可以通过 得到答案(2^nativePointerBitWidth)/sizeof(dataType) - 1。例如，在 64 位系统上，long long（通常）是 8 字节宽，所以我们有(2^64)/8 - 1 == 2305843009213693951.
回答by Anthony Williams
max_size()is the theoretical maximum number of items that could be put in your vector. On a 32-bit system, you could in theory allocate 4Gb == 2^32 which is 2^32 charvalues, 2^30 intvalues or 2^29 doublevalues. It would appear that your implementation is using that value, but subtracting 1.
max_size()是可以放入向量的理论最大项目数。在 32 位系统上，理论上您可以分配 4Gb == 2^32，即 2^32 个char值、2^30 个int值或 2^29 个double值。看起来您的实现正在使用该值，但减去 1。
Of course, you could never really allocate a vector that big on a 32-bit system; you'll run out of memory long before then.
当然，您永远不可能在 32 位系统上真正分配这么大的向量；在那之前你就会耗尽内存。
There is no requirement on what value max_size()returns other than that you cannot allocate a vector bigger than that. On a 64-bit system it might return 2^64-1 for char, or it might return a smaller value because the system only has a limited memory space. 64-bit PCs are often limited to a 48-bit address space anyway.
除了max_size()不能分配大于该值的向量之外，对返回什么值没有要求。在 64 位系统上，它可能会为 返回 2^64-1 char，或者它可能会返回较小的值，因为系统只有有限的内存空间。无论如何，64 位 PC 通常仅限于 48 位地址空间。
回答by Vladimir
max_size()returns
the maximum potential size the vector could reach due to system or library implementation limitations.
由于系统或库实施限制，向量可能达到的最大潜在大小。
so I suppose that the maximum value is implementation dependent. On my machine the following code
所以我认为最大值取决于实现。在我的机器上有以下代码
std::vector<int> v;
cout << v.max_size();
produces output:
产生输出：
4611686018427387903 // built as 64-bit target
1073741823 // built as 32-bit target
so the formula 2^(64-size(type))-1 looks correct for that case as well.
所以公式 2^(64-size(type))-1 对于这种情况看起来也是正确的。
