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Hi
I get a number from dynamic array and then print, I can not.
What do I do?
input:2
3
3
0
0
5
1
2
3

code:

#include<iostream>
using namespace std;
void s(int f[],int d)
{
	cout<<"matris  element i,j   :\ni=";
        cin>>f[0,0];cout<<"j=";cin>>f[0,1];
        f[0,2]=d;
        cout<<"add number not 0 ";
	for(int k=1;k<d+1;k++)
	{   
		cout<<"\n array i="; cin>>f[k,0];
		cout<<"array j=";cin>>f[k,1];
		cout<<"value array ij=";cin>>f[k,2];
		cout<<f[k,2];
	}
	for(int i=1;i<d+1;i++)
	 {
		for(int j=0;j<3;j++)
		cout<<f[i,j];
	    cout<<"\n";
	 }
}
void main()
{
    int w;
	cout<<"number not 0 :";
	cin>>w;
    int *b=new int [w+1,3];
    s(b,w);
}
Posted 20-Nov-12 10:16am
Updated 20-Nov-12 10:18am
v2
Comments
joshrduncan2012 20-Nov-12 16:55pm
   
I don't understand your question. Can you explain more, please?

1 solution

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Solution 1

The comma (,) is a list operator. The array operation (operator [] (int index);) has only one argument. The list operator will work in following way: all operations (separated by comma) will been executed but still the last is the operand.
For example you can write int* px = new int [n=calcsize(),n+=8,push(n),42]; will finally be the same like: n = calcsize(); n+=8; push(n); int* px = new int [42];. So at your code all your array operations are using the parameter behind the last comma:
int *b=new int [w+1,3]; is equal to int *b=new int [3];
cin>>f[k,0]; is equal to cin>>f[0];
and so on.
Perhaps you can explan what to achive with your expressions.
Best regards.
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