CODEWITHSUNDEEP
c++vectormax-size
Kundan
Kundan

Reputation: 321

C++ vector max_size();

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.

Upvotes: 25

Views: 96629

Answers (3)

rahul sinha
rahul sinha

Reputation: 166

Simply get the answer by 

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 long is (typically) 8 bytes wide, so we have (2^64)/8 - 1 == 2305843009213693951.

Upvotes: 15

Anthony Williams
Anthony Williams

Reputation: 68691

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 on a 32-bit system; 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.

Upvotes: 35

Vladimir
Vladimir

Reputation: 170849

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.

Upvotes: 14

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