sort (arr, arr   n)

Why do we write arr n in sort function in (function in algorithm library) C . What does it mean arr n?

CodePudding user response：

1. std::sort accepts iterators to beginning and end of some range (end points to first element beyond range).
2. A pointer can be an iterator
3. In C an array of type sometype[n] decays to a pointer of type: sometype*. So arr is treated as a pointer and arr n advances this pointer by n elements (so it point to first element beyond array).

Now alternative ways to write this code to make it more clear and less bug prone:

std::sort(std::begin(arr), std::end(arr));

// or using C  20 ranges:
std::ranges::sort(arr);

CodePudding user response：

std::sort takes a pair of iterators and them uses std::swap to sort the elements in that range. The iterators must provide implementations for operator* and operator . These requirements are defined as "named requirements" here.

If you thing about it, any pointer fulfills these criteria. In other words, you can think of iterators as a generalization of pointers. By passing a pointer to the first &arr[0] and a pointer to one past the last element &arr[n] you are providing the begin and end iterators. arr and arr n are fancy abbreviations for &arr[0] and &arr[n].

CodePudding user response：

Given the declaration of type arr[], the expression arr n is evaluated as the address of the nth element in arr.

In other words (hoping that it helps clarifying), arr n is equivalent to &arr[n].