What's the wrong with my code ?

Question

1.00/5 (1 vote)

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Hi.. Can someone please tell me what's the wrong with this code ?

C++

int getSum(int*, int, int);

void main(void)
{
	int myArray[2][5]{{ 10, 10, 10, 10, 10 }, { 2, 2, 2, 2, 2 }};
	cout << getSum(*myArray, 2, 5) << endl;
}

int getSum(int* arr, int firstLength, int seconLength)
{
	int sum{};

	for (int i{}; i < firstLength; i++)
		for (int j{}; j < seconLength; j++)
			sum += *((arr+i)+j);
	return sum;
}

All I want to do is get the summary of this two dimensional array using pointers.. But this function returns 92 even if the correct answer is 60.

And please just don't say to use arr[i][j] expression.. I want to do this with pointers..
Thanks in advance !

Posted 20-May-14 4:01am

Mark

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Comments

chandanadhikari 20-May-14 10:36am

People at CP would love to help but why not debug and find it out yourself ?

Sergey Alexandrovich Kryukov 20-May-14 11:00am

Debug what? Why do you think this code could compile? :-)
—SA

Mark 20-May-14 10:41am

It debugs without any errors.. So how am I suppose to find the mistake ? o_O

Legor 21-May-14 3:37am

Why do you think your code is wrong?

5 solutions

Solution 6

please check this code and you anderstand the your wrong

C++

#include "stdafx.h"
#include <iostream>

using namespace std;


int getSum(int*, int, int);
 
void main(void)
{
	int myArray[2][5]={{ 10, 10, 10, 10, 10 }, { 2, 2, 2, 2, 2 }};
			cout<<"before sent to func"<<endl;
	for (int i=0; i<2;i++)
		for(int j=0;j<5;j++)
		cout<<"myArray["<<i<<"]["<<j<<"]="<<myArray[i][j]<<endl;
	cout <<"SUM="<< getSum(*myArray, 2, 5) << endl;
}
 
int getSum(int* arr, int firstLength, int seconLength)
{
	int sum=0;

 	cout<<"after sent to func"<<endl;
	for (int i=0; i < firstLength; i++)
		for (int j=0; j < seconLength; j++)
		{
			cout<<"myArray["<<i<<"]["<<j<<"]="<<*((arr+i)+j)<<endl;
			sum += *((arr+i)+j);
		}
	return sum;
}

before sent to func

myArray[0][0]=10
myArray[0][1]=10
myArray[0][2]=10
myArray[0][3]=10
myArray[0][4]=10
myArray[1][0]=2
myArray[1][1]=2
myArray[1][2]=2
myArray[1][3]=2
myArray[1][4]=2

the elements to sent SUM

myArray[0][0]=10
myArray[0][1]=10
myArray[0][2]=10
myArray[0][3]=10
myArray[0][4]=10
myArray[1][0]=10
myArray[1][1]=10
myArray[1][2]=10
myArray[1][3]=10
myArray[1][4]=2
SUM=92

Posted 20-May-14 5:53am

f.sattari

Updated 20-May-14 6:07am

v2

Comments

Mark 20-May-14 12:12pm

Thanks.. I didn't know that elements from multidimensional arrays can be retrieved as from single dimensional arrays.. :)

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OriginalGriff · Accepted Answer · 2014-05-20T04:47:00

Solution 2

Well, think about what you are doing.
A 3 x 2 array would be indexed as:

0,0   0,1   0,2
1,0   1,1   1,2

and as a single contiguous block memory would be:

C#

 0     1     2     3     4     5
0,0   0,1   0,2   1,0   1,1   1,2

So when you use the indexes to add to the block start address, you can't just add the indexes together:

C++

sum += *((arr+i)+j);

because that will never access the last two elements!
Instead, you need to multiply one of the indexes by the number of elements in the row:

C++

sum += *(arr + (i * 3) + j);

(For your array, it won't be "3" by the way - but I'm very confident you can work out what it should be!)

Posted 20-May-14 4:47am

OriginalGriff

Comments

Richard MacCutchan 20-May-14 11:08am

I hope your confidence isn't misplaced :)

Mark 20-May-14 11:18am

Of course the confidence is misplaced :/ I still don't get that.

Richard MacCutchan 20-May-14 11:21am

I don't think he could have made it any clearer. And Griff is well known round here for the clarity of his answers.

OriginalGriff 20-May-14 11:29am

:laugh:
What don't you understand?

Mark 20-May-14 11:45am

How the "*(arr + (i * 3) + j);" retrieve the elements of the second dimension..

OriginalGriff 20-May-14 11:53am

well, arr is the address of the first element, yes?
and each row has 3 elements, yes?
So the address of the first element of the first row is arr + 0
And the address of the first element of the second row is arr + 3
And the address of the first element of the third row would be arr + 6
And so on: the address of the first element of row "n" will be arr + (n * 3)
(Bear in mind that all sensible languages are zero based, not one)

Also the address of the second element of the first row is arr + 1
The second element of the second row is (arr + 3) + 1
And so on.

Which means that the address of the "y"th element of the "x"th row is
arr + (x * 3) + y

Make sense now?

Mark 20-May-14 12:12pm

Thanks.. I didn't know that elements from multidimensional arrays can be retrieved as from single dimensional arrays.. :)

OriginalGriff 20-May-14 12:17pm

:laugh: Memory is just one *BIG* single dimensional array in reality: all other storage designs are basically just layers over the top of that!

Mark 20-May-14 12:35pm

It is.. :/ Everything we see are just illusions :/ I thought they are real :/

Richard MacCutchan · Accepted Answer · 2014-05-20T05:01:00

Solution 3

Some strange syntax in your code; it would not even compile with Microsoft's C compiler. I finally got it to work with the following:

C++

int getSum(int* arr, int firstLength, int seconLength)
{
    int sum = 0;
    
    for (int i = 0; i < firstLength; i++)
        for (int j = 0; j < seconLength; j++)
            sum += *(arr+i*seconLength+j);
    return sum;
}

int main()
{
    int myArray[2][5] = {{ 10, 10, 10, 10, 10 }, { 2, 2, 2, 2, 2 }};
    cout << getSum((int*)myArray, 2, 5) << endl;
    
}

Posted 20-May-14 5:01am

Richard MacCutchan

Comments

Sergey Alexandrovich Kryukov 20-May-14 11:24am

It's actually much simpler than that. :-)
—SA

Richard MacCutchan 20-May-14 11:31am

That all depends how you define "simple" :)

Sergey Alexandrovich Kryukov 20-May-14 13:46pm

:-)

Mark 20-May-14 11:42am

"sum += *(arr+i*seconLength+j);"

This solved the problem.. Thanks... But I still dont get what part i*seconlength+j takes in this code.

As I understand, in the first round of the first loop and, the first round of the second loop..

arr+i = arr[0] element
arr+i*seconLength = arr[0] element
arr+i*seconLength+j = arr[0] element..

As I see, this line doesn't care about the 2nd dimension of the array..
Can u please explain some further how this retrieve values from the arr[0] and arr[1] array ?

Thanks..

Richard MacCutchan 20-May-14 11:55am

First time round when i is 0, i * seconLength = 0 * 5 = 0 so adding j (from 0 to 4) gets you the first 5 elements (0 to 4).
Second time round when i is 1, i * seconLength = 1 * 5 = 5 so adding j (from 0 to 4) gets you the second 5 elements (i.e 5+0, 5+1 ...).

f.sattari 20-May-14 11:58am

arr=address for a[0]
if arr[m][n]
address for first element in i row = arr+i*n
and address for element in m,n=arr+i*n+j

Mark 20-May-14 12:10pm

Oh. I got it.. Thanks :) I didn't knew that elements of multi dimensional arrays can be retrieved like from a single dimensional array...

Sergey Alexandrovich Kryukov · Accepted Answer · 2014-05-20T05:22:00

Solution 4

You really need to show the code you were dealing with, not anything else. What you show, with all those sum{}, j{} and i{} is just the gibberish which could not compile.

It would be enough to write this:

C++

int getSum(int* arr, int firstLength, int seconLength)
{
	int sum = 0;
	for (int index = 0; index < firstLength * seconLength; ++index)
	   sum += arr[index];
	return sum;
}

Why? Because such multidimensional arrays are actually the single-dimensional ones, please see:
Please see: http://www.cplusplus.com/doc/tutorial/arrays[^].

—SA

Posted 20-May-14 5:22am

Sergey Alexandrovich Kryukov

Comments

Mark 20-May-14 12:12pm

Thanks.. I didn't know that elements from multidimensional arrays can be retrieved as from single dimensional arrays.. :)

Sergey Alexandrovich Kryukov 20-May-14 13:43pm

You are welcome. Now you know this simple and fundamental fact.
Good luck, call again.
—SA

Stefan_Lang 21-May-14 2:05am

I considered commenting about those {} bits, too. But I did experience some odd residuals in postings due to HTML getting in the way, so I didn't know if that was in the original code at all! It sure looked odd, and as you say it shouldn't compile, so I strongly believe that is not the literal code the author has in front of him.

Stefan_Lang 21-May-14 2:15am

P.S.: You're right, this version is really much better. I was considering changing the arr argument to the actual type, but frankly I don't like passing around multidimensional array types - you always end up either with changed dimensionality requirements, or worse, with requirements to use your function for varying array sizes. Your solution avoids that problem altogether!

That said, maybe the sum function really should only take two arguments: the array and the total size. The caller knows the total size after all, and the sum function really doesn't care what the original dimensionality is!

Aescleal 22-May-14 10:03am

T t{}; means "t is a T that's default initialised."

It's not gibberish - it's standard C++.

Stefan_Lang · Accepted Answer · 2014-05-20T04:41:00

Solution 1

try changing

C++

sum += *((arr+i)+j);

to

C++

sum += *((arr+i*firstLength)+j);

Posted 20-May-14 4:41am

Stefan_Lang

Comments

Mark 20-May-14 10:46am

Sorry :/ it returns 84 not the correct summary..

Richard MacCutchan 20-May-14 11:02am

i * seconLength is required, to get past the first group of 5. And it took me a few goes to work that out.

Stefan_Lang 21-May-14 2:09am

Yeah, somehow I always mix up rows and columns. :-( Next time I should simply point out the reasonable arr[*][*] way rather than fix odd memory pointer arithmetic ;-)

Richard MacCutchan 21-May-14 3:31am

Me too, but OriginalGriff's pictures and explanation make it all nice and clear.

Mark 20-May-14 12:12pm

Thanks.. I didn't know that elements from multidimensional arrays can be retrieved as from single dimensional arrays.. :)

Sergey Alexandrovich Kryukov 20-May-14 13:45pm

The solution is way simpler than that... :-)
—SA