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c++c++11valarray
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Reputation: 10614

valarray divide its element

When I divide a valarray by its first element, only the first element becomes 1 and others keep their original value.

#include <iostream>
#include <valarray>
using namespace std;

int main() {
    valarray<double> arr({5,10,15,20,25});
    arr=arr/arr[0]; // or arr/=arr[0];

    for(int value:arr)cout << value << ' ';
    return 0;
}

The actual output is:

1 10 15 20 25

The expected output is:

1 2 3 4 5

Why is the actual output not as expected?

I use g++(4.8.1) with -std=c++11

Upvotes: 5

Views: 607

Answers (2)

Jonathan Wakely
Jonathan Wakely

Reputation: 171413

The details of why this happens are due to implementation tricks used in valarray to improve performance. Both libstdc++ and libc++ use expression templates for the results of valarray operations, rather than performing the operations immediately. This is explicitly allowed by [valarray.syn] p3 in the C++ standard:

Any function returning a valarray<T> is permitted to return an object of another type, provided all the const member functions of valarray<T> are also applicable to this type.

What happens in your example is that arr/arr[0] doesn't perform the division immediately, but instead it returns an object like _Expr<__divide, _Valarray, _Constant, valarray<double>, double> which has a reference to arr and a reference to arr[0]. When that object is assigned to another valarray the division operation is performed and the result stored directly into the left-hand side of the assignment (this avoids creating a temporary valarray to store the result and then copying it into the left-hand side).

Because in your example the left-hand side is the same object, arr, it means that the reference to arr[0] stored in the expression template refers to a different value once the first element in arr has been updated with the result.

In other words, the end result is something like this:

valarray<double> arr{5, 10, 15, 20, 25};
struct DivisionExpr {
  const std::valarray<double>& lhs;
  const double& rhs;
};
DivisionExpr divexpr = { arr, arr[0] };
for (int i = 0; i < size(); ++i)
  arr[i] = divexpr.lhs[i] / divexpr.rhs;

The first iteration of the for-loop will set arr[0] to arr[0] / arr[0] i.e. arr[0] = 1, and then all subsequent iterations will set arr[i] = arr[i] / 1 which means the values don't change.

I'm considering making a change to the libstdc++ implementation so that the expression template will store a double directly instead of holding a reference. This would mean arr[i] / divexpr.rhs will always evaluate arr[i] / 5 and not use the updated value of arr[i].

Upvotes: 2

skypjack
skypjack

Reputation: 50568

This one works:

#include <iostream>
#include <valarray>
using namespace std;

int main() {
    valarray<double> arr({5,10,15,20,25});
    auto v = arr[0];
    arr=arr/v; // or arr/=arr[0];

    for(int value:arr)cout << value << ' ';
    return 0;
}

The problem is that you are trying to use a value (arr[0]) from an array that you are modifying at the same time (arr).
Intuitively, once you have updated arr[0] by doing arr[0]/arr[0], what value does it contain?
Well, that's the value that will be used from now on to divide the other values...
Please, note that the same applies for arr/=arr[0] (first of all, arr[0]/arr[0] takes place, than all the others, in a for loop or something like that).
Also note from the documentation that operator[] of a std::valarray returns a T&. This confirms the assumption above: it is turned to 1 as the first step of your iteration, then all the other operations are useless.
By simply copying it solves the issue, as in the example code.

Upvotes: 4

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