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std::prev in C++

Last Updated : 22 Jul, 2022

std::prev returns an iterator pointing to the element after being advanced by certain number of positions in the reverse direction. It is defined inside the header file iterator. It returns a copy of the argument advanced by the specified amount in the backward direction. If it is a random-access iterator, the function uses just once operator + or operator - for advancing. Otherwise, the function uses repeatedly the increase or decrease operator (operator ++ or operator - -) on the copied iterator until n elements have been advanced. Syntax:

BidirectionalIterator prev (BidirectionalIterator it,
 typename iterator_traits::difference_type n = 1);
it: Iterator to the base position.
difference_type: It is the numerical type that represents 
distances between iterators of the BidirectionalIterator type.
n: Total no. of positions by which the
iterator has to be advanced. In the syntax, n is assigned
a default value 1 so it will atleast advance by 1 position.

Returns: It returns an iterator to the element 
n positions before it.

Output:

v1 = 1 2 3 4 5 6 7
v2 = 8 9 10 1 2 3 4

How can it be helpful ?

  • Moving iterator in Lists: Since, lists support bidirectional iterators, which can be incremented only by using ++ and - - operator. So, if we want to advance the iterator by more than one position, then std::next and if we want to decrement the iterator, then std::prev can be extremely useful. 
  • Output:
v1 = 1 2 3 7 8 9
v2 = 4 5 6 1 2 3 
  • Explanation: Here, just look how if we want copy only a selected portion of the list, then we can make use of std::prev, as otherwise we cannot use any +=, -= operators with bidirectional iterators supported by lists. So, we used std::prev and directly moved the iterator backwards by three positions from the end.

Can we use std::next in place of std::prev ?

One common query that may arise with std::prev is that can std::next be also used with a negative argument to move the iterator in backward direction. Well, the answer is yes

Output:

v1 = 1 2 3 4 5 6 7
v2 = 8 9 10 1 2 3 4

Explanation: So, we have just used std::next in place of std::prev and changed the second argument from 3 to -3 here and it still serves the same purpose. We can use std::next also, but there are two things that needs to be kept in mind:

  • Since, std::next in their syntax have an argument as forward iterator, so if we want to use negative no. for advancing that iterator, then it should be at least a bidirectional iterator.
  • Although, std::next can also be used, but std::prev() would be more readable when the intent is specifically to move backwards.
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