how do you swap two rows in a matrix (in C)?

Question

for example, given a matrix:

1 2 3

4 5 6

7 8 9

if you are goint to swap row[0] and row[1], resulting matrix would be:

4 5 6

1 2 3

7 8 9

can you guys help me get a code in C for this?

Answer 1

temprow = row[1];
row[1] = row[0];
row[0] = temprow;

Answer 2

The answer depends entirely on how your "matrix" is implemented, because the c language has no notion of such a thing.

Are you using two dimensional arrays?

double m[3][3];

Or something else?

Two dimensional arrays

You will have to move individual elements by hand.

for (i=0; i<ROWLENGTH; ++i){
  double temp;
  temp = m[r2][i];
  m[r2][i] = m[r1][i];
  m[r1][i] = temp;
}

(here r1 and r2 are ints that have been set to the two row you want to swap) or see James' memcpy implementation which may well be faster but requires a whole rows worth of temporary memeory.

Ragged Arrays

If this operation is very common and profiling reveals that it is consuming a lot of time, you might consider using a ragged array implementation of the matrix. Something like this:

double **m;
m = malloc(sizeof(double*)*NUMROWS);
/* put error checking here */
for (i=0; i<NUMROWS; ++i){
  m[i] = malloc(sizeof(double)*ROWLENGTH);
  /* error checking again */
}

The fun part about this structure is that you can still access it with the [][] notation, but the row swap operation becomes

double *temp;
temp = m[r2];
m[r2] = m[r1];
m[r1] = temp;

Ragged arrays have two disadvantages from your point of view (well, three 'cause of the memory management hassle): they require extra storage for the row pointers, and you can't use inline initialization.

Row-as-astructure

C does not support array assignments of the form;

double r[3], q[3] = { 1, 2, 3 };
r = q; /* ERROR */

but it does support by-value assignment semantics for structures. Which gives you the implementation that several people have suggested without explaining:

typedef struct { double r[ROWLENGTH] } row;
row m[NUMROWS] = { {1, 2, 3}, {4, 5, 6}, {7, 8 9}};

row temp = m[2];
m[2] = m[1];
m[1] = temp;

which is slick. It requires a whole row of memory, but if the compiler is any good is probably fast. The big disadvantage is that you can not address individual matrix elements with the [][] syntax anymore. Rather you write m[i].r[j];

Others

There are many, many other ways to implement a "matrix" in c, but they are mostly much more complicated and useful only in specialized situations. By the time you need them you'll be able to answer this questions for yourself in the context of each one.

Answer 3

typedef int Row[3];
Row Matrix[3];

Row Temp;

memcpy(Temp, Matrix[0], sizeof(Row));
memcpy(Matrix[0], Matrix[1], sizeof(Row));
memcpy(Matrix[1], Temp, sizeof(Row));

Answer 4

I'd probably swap one element at a time to avoid using a lot of extra storage. If you're working primarily with things like graphics transforms where the matrices are typically 3x3 or 4x4, James Curran's approach is probably a bit better. If you are (or might be) working with really large matrices, this will save memory, and quite possibly run faster:

int x[3][3] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};

for (int i=0; i<3; i++) {
    int temp = x[0][i];
    x[0][i] = x[1][i];
    x[1][i] = temp;
}

Answer 5

solve this your homework?

typedef struct {int m[3];} Row;
typedef int RowAccess[3];

main()
{
  Row tmp,row[]={{1,2,3},{4,5,6},{7,8,9}};
  RowAccess *rowa=row;
  tmp=row[0];
  row[0]=row[1];
  row[1]=tmp;
  /* easy access to matrix here: (is this what you want?) */
  rowa[0][0]=0;
  rowa[0][1]=1;
  ...
  return 0;
}

Answer 6

Hy! this is my first post on stack overflow, I know it's pretty long, hope I won't get banned!

Probably one of the most elegant approaches would be using a function that swaps the two received arguments - using it to swap matrix components. Let's say somethig like swap(a,b). As many have already said, we should consider using a auxiliary variable

auxiliary = a ;
a = b ;
b = auxiliary ; 

Recently, I picked up a new method, which I found impressing, using bitwise XOR operation (http://en.wikipedia.org/wiki/Xor) thus a auxiliary is not needed

 a ^= b ;
 b ^= a ;
 a ^= b ;

You can easily use this operation to swap two elements ( a and b ) - I belive this is off topic, but I insisted on this idea because I found it pretty interesting. Finally, answering your question, you could use let's say

int swap (int *a , int *b){
    (*a)^=(*b);
    (*b)^=(*a);
    (*a)^=(*b);
    return 0; 
}

while having a matrix declared as

#define ROW_COUNT 5
#define COLUMN_COUNT 5
....
int a[ROW_COUNT][COLUMN_COUNT];

You can use your the XOR way the swap the rows, firstly identifyng the elements needed to be swapped ( according to row index, as you already said )

printf("\nSwap Row: "); scanf("%d", &swp1) ; // first row index
printf("With Row: "); scanf("%d", &swp2);    // second row index

for (j = 0 ; j < COLUMN_COUNT ; j++){
     swap(  &a[swp1][j] , &a[swp2][j] );   
}

I hope this will be usefull in your further practice.

Also try this example, I'm sure you'll understand the whole idea much better afterwards (don't forget matrix index starts at 0 !)

#include "stdio.h"
#include "conio.h"

#define ROW_COUNT 5
#define COLUMN_COUNT 5



int swap (int *a , int *b){
        (*a)^=(*b);
        (*b)^=(*a);
        (*a)^=(*b);
        return 0; 

}        

int main(){
    int i, j ;
    int swp1, swp2 ; 
    int a[ROW_COUNT][COLUMN_COUNT];

    // Create ( ROW_COUNT X COLUMN_COUNT ) random matrix

    for (i = 0 ; i < ROW_COUNT ; i++ ) 
        for (j = 0 ; j < COLUMN_COUNT ; j++ )  a[i][j] = rand();

    // Display matrix before row swap

    for (i = 0 ; i < ROW_COUNT ; i++ ){ 
        for (j = 0 ; j < COLUMN_COUNT ; j++ )  printf("%d\t",a[i][j]);
        printf("\n");     
    }

    // Elements to be swapped

    printf("\nSwap Row: "); scanf("%d", &swp1) ;  // first row index
    printf("With Row: "); scanf("%d", &swp2);     // second row index

    // Swapping right here

    for (j = 0 ; j < COLUMN_COUNT ; j++){
         swap(  &a[swp1][j] , &a[swp2][j] );   
    }


    // Display once again   

    printf("\n");
    for (i = 0 ; i < ROW_COUNT ; i++ ){ 
        for (j = 0 ; j < COLUMN_COUNT ; j++ )  printf("%d\t",a[i][j]);
        printf("\n");     
    }   



    getch();            
 return 0;   
}