Calcuate single line of 2d array

Hey guys.

None of my books tell me how to do this so im wondering if its even possible, but say i have an array like this:

Code:

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int myArray[4][4] = {
{ 1,2,3,4 },
{ 5, 6, 7, 8 },
{ 7, 10, 11, 12 },
{ 13, 14, 15, 16 }
};

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Its it possible to calculate and find the total of only the top row?

I know I can do it via an index like this:

Code:

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int totalRow = myArray[0][0] + myArray[0][1] ...etc

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But is there a simpler way to do that, as if I have a large array with say over 10
columns in size I really do not want to create a huge expression per row.

Any advice appreiciated guys :)

Um, use a for loop?

Quote:

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Originally Posted by laserlight

Um, use a for loop?

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Thanks Laserlight I nearly have this sussed. Say I want to calculate the top row of the above array, I did this:

Code:

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for ( int k = myArray[0][0]; k <= myArray[0][3]; k++ )
{
  rowTotal += k;
}

cout << "Row total is: " << rowTotal << endl;

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But its giving me the output of 9 when it should be 10. Im starting at the position [0][0]
as I should do, and im certain my ending loop index is also correct. Row[0] column[3]

More likely:

Code:

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for ( int k = 0; k < sizeof(myArray[0]); k++ )
{
    rowTotal += myArray[0][k];
}

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I thought sizeof returned the bit-size of a variable ( eg: integer is 2 bit ). Im getting the answer of 133.

Its surely not as hard as im making it out ot be.

Quote:

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I thought sizeof returned the bit-size of a variable ( eg: integer is 2 bit ).

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No, it returns the number of bytes, not bits.

Quote:

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Im getting the answer of 133.

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That's because I forgot to divide by sizeof(int), thus the array is accessed out of bounds.

Try:

Code:

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#include <iostream>

int main()
{
    int myArray[4][4] = {
        { 1,2,3,4 },
        { 5, 6, 7, 8 },
        { 7, 10, 11, 12 },
        { 13, 14, 15, 16 }
        };
    int rowTotal = 0;
    unsigned int size = sizeof(myArray[0]) / sizeof(int);
    for (unsigned int k = 0; k < size; ++k)
    {
        rowTotal += myArray[0][k];
    }
    std::cout << rowTotal << std::endl;
}

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Code:

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for ( int k = 0; k < sizeof(myArray[0]) / sizeof(myArray[0][0]); k++ )
{
    rowTotal += myArray[0][k];
}

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sizeof gives the BYTE-size [1] of the type you give sizeof.

[1] Technically, the smalles addressable unit, which is what sizeof returns, could be something other than an 8-bit byte, but that's only on rare and unusual architectures, not your common PC or workstation system.

--
Mats

Quote:

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Technically, the smalles addressable unit, which is what sizeof returns, could be something other than an 8-bit byte, but that's only on rare and unusual architectures, not your common PC or workstation system.

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True on a technicality, yet also by technicality such a unit is still called a byte by the C++ Standard.