C++ vector max_size();

问题:

On 32 bit System.

1. std::vector<char>::max_size() returns 2³²-1, size of char — 1 byte
2. std::vector<int>::max_size() returns 2³⁰-1, size of int — 4 byte
3. std::vector<double>::max_size() returns 2²⁹-1, size of double — 8 byte

can anyone tell me max_size() depends on what?

and what will be the return value of max_size() if it runs on 64 bit system.

回答1:

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 char values, 2^30 int values or 2^29 double values. It would appear that your implementation is using that value, but subtracting 1.

Of course, you could never really allocate a vector that big; you'll run out of memory long before then.

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.

回答2:

max_size() returns

the maximum potential size the vector could reach due to system or library implementation limitations.

so I suppose that the maximum value is implementation dependent. On my machine the following code

std::vector<int> v;
cout << v.max_size();

produces output:

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.

回答3:

Simply get the answer by

Vector<data-type>v;
Cout<< v.max_size();

Or we can get the answer by

(2^bit of system)/sizeof(datatype) -1
Eg for 64 bit system
Iong long datatype has size =8 bit
Ans =(2^64)/8 -1
2305843009213693951