read in array into another array C++

Question

after reading inputs to an array:

int * inputs;

the "inputs" is 1 dimensional array: inputs[6], then reading this array out, the values are:

inputs[0]=1
inputs[1]=2
inputs[2]=3
inputs[3]=4
inputs[4]=5
inputs[5]=6

I would like to read this array into another one dimensional array:

int counter=0;
int * allElements = new int[6];

for(int i=0; i<6; i++)
{
      allElements[counter++] = inputs[i];
}

That is a traditional way of reading all of the elements into one dimensional array and I believe if I read the elements of "allElements" this way:

for(int i=0; i<6; i++)
   printf("%d ", allElements[i]);

and it should be: 1 2 3 4 5 6

However, I would like to read all elements of that array into the 1 dimensional array such that when I do it like this:

for(int i=0; i<6; i++)
   printf("%d ", allElements[i]);

It should be: 1 3 5 2 4 6

How can I achieve this way?

You could write a function which calculates the index of allElements from the index of inputs. — kol
– kol, Commented Nov 14, 2011 at 23:33
I don't think any of this deserves to be called "C++"... why are you using pointers and new? Use the standard library containers for arrays (vector or array). — Kerrek SB
– Kerrek SB, Commented Nov 14, 2011 at 23:34
possible duplicate of read array 2 dimensions in another way C++ — aschepler
– aschepler, Commented Nov 14, 2011 at 23:35
@KerrekSB probably, some codes I wrote it in C# way, but I would like to do it in C++ — olidev
– olidev, Commented Nov 14, 2011 at 23:37
@JeremyFriesner: "C++" is as much about idiom as it is about language features. — Lightness Races in Orbit
– Lightness Races in Orbit, Commented Nov 14, 2011 at 23:41

ildjarn · Accepted Answer · 2011-11-15 00:36:20Z

3

int * allElements = new int[6];

for(int i=0; i<6; i+=2)
{
    allElements[i/2] = inputs[i];
    allElements[3+i/2] = inputs[i+1];
}

edited Nov 15, 2011 at 0:36

ildjarn

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answered Nov 14, 2011 at 23:57

ComfortablyNumb

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1 Comment

Joe McGrath Over a year ago

Might as well unroll it if it is going to be for only 1 case.

rodrigo · Accepted Answer · 2011-11-15 08:58:56Z

2

What about:

for(int i=0; i<6; i++)
{
    allElements[i] = inputs[2*(i%3) + (i/3)];
}

Imagine inputs is a two-dimension array, of 3x2, then i%3 is one coordinate and i/3 the other. Just transpose it into a 2x3 matrix, and done!

edited Nov 15, 2011 at 8:58

answered Nov 14, 2011 at 23:42

rodrigo

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1 Comment

rodrigo Over a year ago

@JoeMcGrath Right, I made it backwards. Corrected.

Joe McGrath · Accepted Answer · 2011-11-15 00:41:01Z

0

const int size =6;
int inputs[size]={1,2,3,4,5,6};
int allElements[size];

...

  if (size % 2)
  {
      int middle=(size/2) +1;
      for(int i=0; i<size; i+=2)
      {   
          allElements[i/2] = inputs[i];
          if (i < size-2)
            allElements[middle+i/2] = inputs[i+1];          
      }      

  }
  else
  {
    int middle=size/2;
    for(int i=0; i<size; i+=2)
    { 
      allElements[i/2] = inputs[i];
      allElements[middle+i/2] = inputs[i+1];
    }      
  }

Logic can definitely be simplified. Just first attempt and left it alone.

If want it to work for the 1 case

allElements[0]=inputs[0];      
allElements[1]=inputs[2];
allElements[2]=inputs[4];
allElements[3]=inputs[1];
allElements[4]=inputs[3];
allElements[5]=inputs[5];

edited Nov 15, 2011 at 0:41

answered Nov 15, 2011 at 0:22

Joe McGrath

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