How to increment an iterator by 2

Question

Can anybody tell me how to increment the iterator by 2   iter   is available - do I have to do iter 2  How can I achieve this

User · Answer

You could use the  assignment by addition  operator  iter    2

User · Answer

std  advance  iter  2     This method will work for iterators that are not random-access iterators but it can still be specialized by the implementation to be no less efficient than iter    2 when used with random-access iterators

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If you don t know wether you have enough next elements in your container or not  you need to check against the end of your container between each increment  Neither    nor std  advance will do it for you   if    iter    collection end             stop  if    iter    collection end             stop   You may even roll your own bound-secure advance function   If you are sure that you will not go past the end  then std  advance  iter  2   is the best solution

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The very simple answer       iter   The long answer   You really should get used to writing   iter instead of iter     The latter must return  a copy of  the old value  which is different from the new value  this takes time and space   Note that prefix increment    iter  takes an lvalue and returns an lvalue  whereas postfix increment  iter    takes an lvalue and returns an rvalue

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http   www cplusplus com reference std iterator advance   std  advance it n     where n is 2 in your case   The beauty of this function is  that If  it  is an random access iterator  the fast  it    n   operation is used  i e  vector lt      iterator     Otherwise its rendered to  for int i   0  i  lt  n  i          it     i e  list lt      iterator

User · Answer

Assuming list size may not be an even multiple of step you must guard against overflow   static constexpr auto step   2      Guard against invalid initial iterator  if   list empty          for  auto it   list begin      nothing here    std  advance it  step                    do stuff                Guard against advance past end of iterator          if  std  distance it  list end     gt  step              break            Depending on the collection implementation  the distance computation may be very slow  Below is optimal and more readable  The closure could be changed to a utility template with the list end value passed by const reference   const auto advance     amp   list type  iterator amp  it  size t step        for  size t i   0  it    list end    amp  amp  i  lt  step  std  next it     i       static constexpr auto step   2   for  auto it   list begin    it    list end    advance it  step            do stuff        If there is no looping   static constexpr auto step   2  auto it   list begin     if  step  lt   list size          std  advance it  step

User · Answer

We can use both std  advance as well as std  next  but there s a difference between the two   advance modifies its argument and returns nothing  So it can be used as    vector lt int gt  v  v push back 1   v push back 2   auto itr   v begin    advance itr  1              modifies the itr cout  lt  lt   itr lt  lt endl          prints 2   next returns a modified copy of the iterator   vector lt int gt  v  v push back 1   v push back 2   cout  lt  lt   next v begin    1   lt  lt  endl       prints 2

User · Answer

If you don t have a modifiable lvalue of an iterator  or it is desired to get a copy of a given iterator  leaving the original one unchanged   then C  11 comes with new helper functions - std  next   std  prev   std  next iter  2               returns a copy of iter incremented by 2 std  next std  begin v   2      returns a copy of begin v  incremented by 2 std  prev iter  2               returns a copy of iter decremented by 2

[c++] How to increment an iterator by 2?

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