Trying to pass a matrix by reference

Question

Sorry i am new to c, i am trying to pass this matrix to a procedure by reference so i don't have to copy it in memory. I couldn't find any explanation. This is the closest i could get but it doesn't work. The point of the program is only for that, i made it to test it.

#include <stdio.h>

typedef int tmatrix[5][5];

void print (tmatrix*mtr)
{

    int l , m;  

    l=0;

    while (l<=4)

    {
        m=0;

        while (m<=4)
        {
            printf("%d", *mtr[l][m]);

            m = m+1;
        }

        printf("\n");

        l=l+1;

    }
}
//-----------------------------------
int main()
{
int i , j;

tmatrix matrix;


i=0;

while (i <= 4)
{

    j=0;

    while (j<=4)
    {

        matrix[i][j] = 3;

        j = j+1;

    }

    i = i+1;

}

print(&matrix);

return 0;

}

It should print:

But it prints:

33333
54198992041990930
1977890592-1961670060002752492
03232520
664-21479789407743168

I know it may be something to do with the pointers because i think those are addresses, but i got no clues.

The expression `*mtr[l][m]` does not do what you think it does because of [*operator precedence*](http://en.cppreference.com/w/c/language/operator_precedence). — Some programmer dude, May 19 '16 at 07:00
You also have to remember that arrays naturally decays to pointers to their first element, so there's no need to pass arrays by reference (or rather *emulate* passing by reference, since C doesn't have it). — Some programmer dude, May 19 '16 at 07:02
Thank you very much guys , i've been stuck with this for hours. — Nico Baldelli, May 19 '16 at 07:30

score 3 · Accepted Answer · edited May 19 '16 at 09:06

3

matrix is a 2D array of integers. You can directly pass it to the function. Passing an array will be the same as passing it by reference.
The printf inside the function will also be changed to directly access it. *mtr[l][m] --> mtr[l][m]
While calling it, you can call it directly as shown. print(&matrix) --> print(matrix)

The modified code is below.

#include <stdio.h>

typedef int tmatrix[5][5];

void print (tmatrix mtr)
{

    int l , m;  

    l=0;

    while (l<=4)

    {
        m=0;

        while (m<=4)
        {
            printf("%d", mtr[l][m]);

            m = m+1;
        }

        printf("\n");

        l=l+1;

    }
}
//-----------------------------------
int main()
{
int i , j;

tmatrix matrix;


i=0;

while (i <= 4)
{

    j=0;

    while (j<=4)
    {

        matrix[i][j] = 3;

        j = j+1;

    }

    i = i+1;

}

print(matrix);

return 0;

}

edited May 19 '16 at 09:06

J. Chomel

8,193
15
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answered May 19 '16 at 07:09

Rishikesh Raje

8,556
2
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31

Thank you very much Rishikesh Raje. But can i ask you, what if i want to pass it by value? – Nico Baldelli May 19 '16 at 07:34
@NicoBaldelli in C, arrays are always passed by reference. You cannot pass it by value. – Rishikesh Raje May 19 '16 at 07:41
@RishikeshRaje, wrong, all is passed by value in C, but an array decays into a pointer – David Ranieri May 19 '16 at 07:48
So, effectively, you cannot pass an array by value, right? – Rishikesh Raje May 19 '16 at 07:50
Not directly, the only way is embedding the array into a `struct` and passing a pointer to this `struct` – David Ranieri May 19 '16 at 08:22

Trying to pass a matrix by reference

1 Answers1