std::prev in C++

02 Aug, 2017
std::prev returns an iterator pointing to the element after being advanced by certain no. of positions in the reverse direction. It is defined inside the header file .
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.


// C++ program to demonstrate std::prev#include <iostream>#include <iterator>#include <deque>#include <algorithm>using namespace std;int main(){    // Declaring first container    deque<int> v1 = { 1, 2, 3, 4, 5, 6, 7 };      // Declaring another container    deque<int> v2 = { 8, 9, 10 };      // Declaring an iterator    deque<int>::iterator i1;      // i1 points to 1    i1 = v1.begin();      // Declaring another iterator to store return    // value and using std::prev    deque<int>::iterator i2;    i2 = std::prev(v1.end(), 3);      // Using std::copy    std::copy(i1, i2, std::back_inserter(v2));    // Remember, i1 stills points to 1    // and i2 points to 5    // v2 now contains 8 9 10 1 2 3 4      // Displaying v1 and v2    cout << "v1 = ";      int i;    for (i = 0; i < 7; ++i) {        cout << v1[i] << " ";    }      cout << "\nv2 = ";    for (i = 0; i < 7; ++i) {        cout << v2[i] << " ";    }      return 0;}
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.// C++ program to demonstrate std::prev#include <iostream>#include <iterator>#include <list>#include <algorithm>using namespace std;int main(){    // Declaring first container    list<int> v1 = { 1, 2, 3, 7, 8, 9 };      // Declaring second container    list<int> v2 = { 4, 5, 6 };      list<int>::iterator i1;    i1 = v1.begin();    // i1 points to 1 in v1      list<int>::iterator i2;    // i2 = v1.end() - 3;    // This cannot be used with lists    // so use std::prev for this      i2 = std::prev(v1.end(), 3);      // Using std::copy    std::copy(i1, i2, std::back_inserter(v2));    // v2 now contains 4 5 6 1 2 3      // Displaying v1 and v2    cout << "v1 = ";      int i;    for (i1 = v1.begin(); i1 != v1.end(); ++i1) {        cout << *i1 << " ";    }      cout << "\nv2 = ";    for (i1 = v2.begin(); i1 != v2.end(); ++i1) {        cout << *i1 << " ";    }      return 0;}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.
// C++ program to demonstrate std::prev#include <iostream>#include <iterator>#include <list>#include <algorithm>using namespace std;int main(){    // Declaring first container    list<int> v1 = { 1, 2, 3, 7, 8, 9 };      // Declaring second container    list<int> v2 = { 4, 5, 6 };      list<int>::iterator i1;    i1 = v1.begin();    // i1 points to 1 in v1      list<int>::iterator i2;    // i2 = v1.end() - 3;    // This cannot be used with lists    // so use std::prev for this      i2 = std::prev(v1.end(), 3);      // Using std::copy    std::copy(i1, i2, std::back_inserter(v2));    // v2 now contains 4 5 6 1 2 3      // Displaying v1 and v2    cout << "v1 = ";      int i;    for (i1 = v1.begin(); i1 != v1.end(); ++i1) {        cout << *i1 << " ";    }      cout << "\nv2 = ";    for (i1 = v2.begin(); i1 != v2.end(); ++i1) {        cout << *i1 << " ";    }      return 0;}
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.
// C++ program to demonstrate std::next#include <iostream>#include <iterator>#include <deque>#include <algorithm>using namespace std;int main(){    // Declaring first container    deque<int> v1 = { 1, 2, 3, 4, 5, 6, 7 };      // Declaring another container    deque<int> v2 = { 8, 9, 10 };      // Declaring an iterator    deque<int>::iterator i1;      // i1 points to 1    i1 = v1.begin();      // Declaring another iterator to store return    // value and using std::next    deque<int>::iterator i2;    i2 = std::next(v1.end(), -3);      // Using std::copy    std::copy(i1, i2, std::back_inserter(v2));    // Remember, i1 stills points to 1    // and i2 points to 5    // v2 now contains 8 9 10 1 2 3 4      // Displaying v1 and v2    cout << "v1 = ";      int i;    for (i = 0; i < 7; ++i) {        cout << v1[i] << " ";    }      cout << "\nv2 = ";    for (i = 0; i < 7; ++i) {        cout << v2[i] << " ";    }      return 0;}
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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