Decimal itself converts to hexadecimal

Question

I have small issue. I create 2D int array. When I cout it I get hexadecimals numbers insted of decimals. I'm using Dev C++.

#include <cstdlib>
#include <iostream>

using namespace std;

int main()
{
    const int max=9;
    int ar[max][max]={
        {0,6,0,0,2,0,0,4,0},
        {5,0,0,3,0,0,0,0,0},
        {0,8,0,0,1,0,0,0,0},
        {6,0,0,0,0,7,0,0,0},
        {0,3,7,0,0,0,2,8,0},
        {0,2,0,8,0,0,0,3,0},
        {0,0,0,0,0,0,0,0,0},
        {7,0,0,4,0,0,0,0,1},
        {0,0,0,0,6,0,0,2,0}};

    for (int i=0;i<max;i++){
        for(int j=0;j<max;j++){
            cout<<ar[i,j]<<" ";
        }
        cout<<"\n";
    }

    system("pause");
    return 0;
}

In return I get this http://www.dropmocks.com/mf8wl

Don't use Dev C++: http://stackoverflow.com/tags/dev-c%2b%2b/info — Fred Larson, Nov 02 '11 at 13:15
`system("pause")` is a nasty habit, too: http://stackoverflow.com/q/1107705/10077 — Fred Larson, Nov 02 '11 at 13:16

score 9 · Accepted Answer · answered Nov 02 '11 at 13:12

That's because you've fallen into the cunning trap of operator ,!

First of all, C doesn't have multi-dimensional arrays as a primitive. As you've declared here, a 2D array is just "an array of arrays". Therefore, it makes no sense to access an array with a[i,j]. You should first get the "row" with a[i] then index the "column" with [j], whence a[i][j].

So, why does your code compile? Because the , is an operator in C. The evaluation of a,b is effectively the evaluation of expression a, then b, returning the result of evaluating b. Hence you're actually printing a[j] here, which is an int[], printed in hex as the address of the array.

Why have an operator , at all, you ask? Other than to be confusing, it's mostly for constructs like for, where you might want multiple expressions in an initialiser or incrementation construct, i.e. for (j = 0, i = 0; i + j < k; i++, j += 2) or similar.

score 4 · Answer 2 · answered Nov 02 '11 at 13:10

cout << ar[i,j] << " ";

This should read

cout << ar[i][j] << " ";

The comma operator , does (from here)

The comma operator (,) is used to separate two or more expressions that are included where only one expression is expected. When the set of expressions has to be evaluated for a value, only the rightmost expression is considered.

So this is printing out ar[j] (after evaluating i and discarding the result) which is a pointer and hence hexadecimal.

score 2 · Answer 3 · answered Nov 02 '11 at 13:10

2

The C++ syntax for accessing array members is:

ar[i][j]

You coming from a Pascal background? :P

answered Nov 02 '11 at 13:10

Luchian Grigore

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score 0 · Answer 4 · answered Nov 02 '11 at 13:11

0

You want to do

cout<<ar[i][j]<<" ";

http://ideone.com/G5n4Z with GNU says its undefined behavior.

answered Nov 02 '11 at 13:11

Daniel A. White

187,200
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score 0 · Answer 5 · answered Nov 02 '11 at 13:11

0

cout<<ar[i,j]<<" ";

Change this line to:

cout<<ar[i][j]<<" ";

answered Nov 02 '11 at 13:11

Kevin

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Decimal itself converts to hexadecimal

5 Answers5