How to make a dynamic array? Two parameters

Posted 13 Nov '12 - 2:23

Edited 20 Nov '12 - 21:09

Solution 1

You may use the new operator, the documentation[^] provides sample code.

You could also use a STL container, like the std::vector (see, for instance here[^]).

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Posted 13 Nov '12 - 2:44

CPallini246K

Comments

sajad abbasi - 13 Nov '12 - 16:33

Thanks for the help.

CPallini - 13 Nov '12 - 16:48

You are welcome.

Albert Holguin - 14 Nov '12 - 17:32

For beginners... I'd recommend learning to use the heap (new/delete) first before moving on to containers. +5

CPallini - 15 Nov '12 - 3:23

I agree with you, beginners should learn to master new/delete. By the way, thank you.

Espen Harlinn - 15 Nov '12 - 10:59

a 5, std::vector became immensely more useful after data() was added to the interface, removing the need for a separate dynamic_array container - which was discussed at some point.

Solution 3

Perhaps this snippet can help:

 
    vector<int> x;
 
    int size;
    cin >> size;
 
    while (size--)
    {
        int temp; cin >> temp; x.push_back(temp);
    }
/*
    Better is, but the compiler dislikes it:
    while (size--)
        cin >> x.push_back();
    or
    while (size--)
        x.push_back(cin>>);
*/ 
 
    // processing...

    for (vector<int>::iterator it = x.begin(); it != x.end(); it++)
        cout << *it << endl;
 
    x.clear();

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Posted 15 Nov '12 - 21:56

Member 9586495591

Edited 15 Nov '12 - 21:57

Solution 4

# include <stdio.h>
# include <malloc.h>
# include <stdlib.h>
 
int main()
{
	int size;
	scanf("%d",&size);
	int* p = (int *)malloc(sizeof(int)*size);
	if(NULL == p)
		exit(-1);
	for(int i = 0;i < size;++ i)
		scanf("%d",&p[i]);
	for(int i = 0;i < size;++ i)
		printf("%3d",p[i]);
	return 0;
}</stdlib.h></malloc.h></stdio.h>

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Posted 17 Nov '12 - 0:05

ccww552010264

Solution 6

int main()
{
  int size;
  int *ptr_arr;          //pointer delaration
  cin>>size;             //input size from user
  ptr=new int[size];     //array declared in heap or dynamic memory
  for(int i=0;i<size;i++)>
     cin>>ptr_arr[i];        //takes input in dynamic array
  for(int i=0;i<size;i++)>
     cout<<ptr_arr[i];
 
return 0;
}

I hope it helps

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Posted 19 Nov '12 - 2:27

saad_lah657

Edited 19 Nov '12 - 2:34

Solution 7

The only way to make a two dimension without the use of the advanced C++ features is the use an array of pointers, note the two for loops, one is for initialization, and the second is for cleanup purposes.

Example:

#include <iostream>
using namespace std;
 
int main()
{
	int **a;
	int i = 0, j = 0;
	int k;
 
	cout << "input dimension 1:";
	cin >> i;
	cout << "input dimension 2:";
	cin >> j;
 
	a = new int *[i];
	for (k = 0; k < i; k++)
	{
		a[k] = new int[j];
	}
	
	cout << "a[0][0]: ";
	cin >> a[0][0];
	
	a[0][0] = 1024;
	cout << "value: " << a[0][0] << endl;
 
	for (k = 0; k < i; k++)
	{
		delete a[k];
	}
	delete  a;
 
	return 0;
}

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Posted 21 Nov '12 - 12:12

The Santos Nerd278

Edited 21 Nov '12 - 12:23

Solution 2

Array declaration in C/C++ will not be success if you don't provide a constant size. For the beginners who do not familiar with pointers the following may give a good understanding.

1. Declare an array with a maximum size; when your input exceeds this limit you can always show an error message.
int x[100];

2. when you deal with an array size within the maximum size, keep the dynamic size in a variable and use the variable as the size of your array instead of the maximum size.

       int main(){
          int x[100];  //maximum size is 100
          int size;
          cin>>size;   //to accept the size
          if (size>=100){
             cout<<"Array size exceeds the maximum allocation";
             return 0;
          }
          //when inputting numbers
          for(int i=0; i<size; i++){             
             cin>>x[i];
          }
          //when displaying the result
          for(int i=0; i<size; i++){             
             cout<<x[i]<<endl;
          }
          return 0;
       }

Hope this will help.

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Posted 14 Nov '12 - 9:48

Shan Veerakutty390

Edited 15 Nov '12 - 4:30

Comments

Maximilien - 15 Nov '12 - 10:49

IMO, You answer will not help at all. How is declaring a static array answering the original question? There are multiple ways, even in plain old C to allocate dynamic arrays(malloc and realloc) , and in C++ to have better and more foolproof ways of doing it (new xor std::vector).

Solution 5

have a good time dear mr sajjad
you're from iran?me too.

look you have to declare a dynamic array like above:

int n=0,*a=0;
cout<<"enter the length of array :\n";
cin>>n;
a=new int[n];

after that you can control the array with an index , like above :

int j=0;
j=a[7];

good lock and god with you

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Posted 17 Nov '12 - 0:10

Mahdi Nejadsahebi5.5K

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

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How to make a dynamic array? Two parameters

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How to make a dynamic array? Two parameters

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