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c++pointerspointer-arithmetic
user69514
user69514

Reputation: 27629

C++ arithmetic with pointers

I am trying to add the following:

I have an array of double pointers call A. I have another array of double pointers call it B, and I have an unsigned int call it C.

So I want to do:

A[i] = B[i] - C;

how do I do it? I did:

A[i] = &B[i] - C;

I don't think I am doing this correctly.

Edit: What I want to do is, take the value at index i of the double pointer array and subtract an int from it, then store that result into a double pointer array at index i.

Upvotes: 0

Views: 310

Answers (3)

Draco Ater
Draco Ater

Reputation: 21226

What you want is:

&( *B[i] - C )

But i think you cannot assign it directly to A[i]. First you have to create a temporary (T) array of double.

for(int i=0; i< size_of_B; i++){
  T[i] = *B[i] - C;
}
for(int i=0; i< size_of_T; i++){
  A[i] = &T[i];
}

Upvotes: 0

kennytm
kennytm

Reputation: 523224

C++ doesn't have a simple syntax for mapping, you either

(1) Use a loop

for (int i = 0; i < 1482; ++ i)
  A[i] = B[i] - C;

(2) Use std::transform in <algorithm>:

#include <functional>
#include <algorithm>
...
std::transform(B, B+1482, A, std::bind2nd(std::minus<double>(), C));

(There may be a Boost library to simplify this.)

Upvotes: 0

pnt
pnt

Reputation: 1916

Your question is a bit unclear, but if A and B are arrays of pointers to double and you want to change each pointer with a fixed amount of exactly C, then, for each element in A:

A[i] = B[i] - C;

should do it. &B[i] takes the address of the pointer itself, so it is a completely different thing.

sample code:

for(int i = 0; i < size_of_A; ++i) A[i] = B[i] - C;

Upvotes: 4

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