c++ - Why does std::accumulate behave like this with standard arrays? -


I'm just coming in C ++ and I think I have handle pointer, but std: : Accumulate () has confused me.

Looking at the array: 

  int a [3] = {5, 6, 7};

I want to add values ​​of the array to std :: accumulate () , so I give it an indicator for the first element, then the last, Then the initial value of the receiver. 

  std :: cached (a, a + 2, 0); Std :: deposits (& amp; a [0], & a [2], 0); Oh: None of these gives the sum of the first two elements:  11 . 
 
 On the other hand, if the second argument is a nonsensical indicator, just outside the limit ... 
 
  std :: deposits (a, a + 3, 0); Std :: deposits (& amp; a [0], & a [3], 0); The correct value of  
 
 ...  18  has been returned. 
 
 Can anyone please explain this? I know that I can not use simple arrays, but it's next to the point.

C ++ categories are defined [first, last] , And all STL algorithms work that way. In this case, all the elements behind the std :: accumulate defined class, starting with first and without actually last Ends denote it.

As such, calling it like std :: accumulate (a, a + 3, 0) is really correct and calling it with std :: Is equal to (start (a), end (a), 0) .

Also note that this is a typical exception to the pointer behind the last element, "There is no sign outside the allocated array".


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