Vector index variable declaration (size_t or std::vector<DATATYPE>::size_type)

vector c++
vector c++ example
vector of vectors - c++
c vector constructor
c++ vector initialization
std::size_t vs size_t
c++ vector::push_back
std::vector find

What is the best practice for declaring a new variable for comparison with the size of a vector? which of the following should we favor (i.e. vector of doubles)?

  1. uint compareVar;
  2. std::uint64_t compareVar;
  3. std::size_t compareVar;
  4. std::vector<double>::size_type compareVar; // how is this different than size_t?

and why?

The one that you must use is std::vector::size_type. it is usually std::size_t but standard doesn't specify it. it is possible one implementation uses another type. You should use it whenever you want to refer to std::vector size.

uint, uint64_t and even size_t must not be used because in every implementation and platform the underlying vector implementation may use different types.

Iteration over std::vector: unsigned vs signed index variable, Some people tell you to use std::size_t as the index variable type. for(std:: vector<T>::iterator it = v.begin(); it != v.end(); ++it) { /* std::cout << *it; . before 'it' error: 'it' was not declared in this scope In member function 'void MyClass<T>:: sort()� vector<int>::size_type is a type that is guaranteed to hold the size of the biggest vector you may have, and thus it's guaranteed to let you index all the elements of the vector (since indexes go from 0 to size-1); it is the type used for indexes and sizes in all the vector methods.

If you need an equality or ordinal comparison only, the best type would be the one used by vector's implementation, meaning

std::vector<MyType>::size_type compareVar

This guarantees that the type you use matches the type of vector's implementation regardless of the platform.

Note that since the type is unsigned, computing the difference between vector's size and compareVar would require some caution to avoid incorrect results when subtracting a larger value from a smaller one.

std::vector, 1) std::vector is a sequence container that encapsulates dynamic size This means that a pointer to an element of a vector may be passed to� The type of the size of an std::vector<T> is std::vector<T>::size_type. That's it. On some architecture and for some compilers it might be the same as size_t, in some others it might not. The assumption that a variable of type size_t can hold the same values than one of type std::vector<T>::size_type can fail.

In order of "goodness": 4 3 1 2.

std::vector::size_type is an implementation defined unsigned integer type that is guaranteed to be able to index all elements in any vector. It is good to use this, because it is guaranteed to be at least as large as is needed in all cases, but is not required to be any larger than that.

std::size_t is an implementation defined unsigned integer type that is guaranteed to be able to represent the size of any object. std::vector::size_type is typically same as this, but that is not guaranteed in theory.

uint There is no such type in C++, but I assume that you mean unsigned. This is not guaranteed to be able to represent all indices of a vector. In fact, on 64 bit systems which are ubiquitous, largest representable value of unsigned is often much much less than largest possible vector. Still, if you know that the upper limit is much less in your program, then using this is probably fine... as long as that assumption never changes.

std::uint64_t is an unsigned type that is not guaranteed to exist on all systems, and is not guaranteed to be sufficiently large to represent all indexes of a vector, although it is likely be sufficient on 64 bit systems and those with even smaller address space, which essentially covers nearly all computers at the moment.

std::size_t, std::size_t is commonly used for array indexing and loop counting. Programs that use other types, such as unsigned int, for array indexing may� To access any element in vector by index vector provides two member functions i.e. at() operator[] Let’s discuss them in detail, Access an element in vector using operator [] std::vector provides [] operator i.e. element_reference operator[] (size_type n); It returns the reference of element in vector at index n. Suppose we have a vector of

vector - C++ Reference, Vectors are sequence containers representing arrays that can change in size. thus, vectors do not reallocate each time an element is added to the container. reference, allocator_type::reference, for the default allocator: value_type& any non-negative value of difference_type, usually the same as size_t <stdexcept>. Returns a reference to the element at position n in the vector container. A similar member function, vector::at, has the same behavior as this operator function, except that vector::at is bound-checked and signals if the requested position is out of range by throwing an out_of_range exception.

std::vectors, not dynamic arrays! - C++ Articles, std::vector<value_type> variable_name(number_of_elements); They support the [] operator for element access much like arrays do (and their 1 2, for (size_t i = 0; i < a_vector.size(); ++i) std::cout << a_vector[i] << std::endl;� Of course, each access to the vector also puts its management content into the cache as well, but as has been debated many times (notably here and here), the difference in performance for iterating over a std::vector compared to a raw array is negligible. So the same principle of efficiency for raw arrays in C also applies for C++'s std::vector.

C++ std::vector : declare, initialize, functions of vector, etc, Unlike std::array whose length is specified at the time of declaration and remains constant till compile time, we can change the length of std::vector dynamically� Initialization vector can be done in many ways. 1) Initialize a vector by push_back() method Algorithm Begin Declare v of vector type. Call push_back() function to insert values into vector v.

  • size_t depends on the environment (32/64 bits, compiler, etc). And it's how the standard vector::size is implemented.
  • Yes, it is, but there is no guarantee that std::vector::size_type is always equal to size_t.
  • Thanks. But doesn't size_t do the same? isn't it platform independent as well?
  • @Bu_ali size_t is indeed platform-independent, but std::vector's implementation is under no obligation to use it.