swap rows in a two dimensional array

Do this

int b[3];
int a[3][3]={0,0,0,1,1,1,2,2,2};

memcpy(b, a[2], sizeof(int)*3);
memcpy(a[2], a[0], sizeof(int)*3);
memcpy(a[0], b, sizeof(int)*3);

>> int *b;
>> int a[3][3]={0,0,0,1,1,1,2,2,2};
You cannot use a pointer to swap the rows. Since two-dimensional array is allocated consecutively.
You need to explicitly swap the data in it.

Thank you.
However, do you think there is a pointer to a row in the two dimensional array?
int *b;
     int a[3][3]={0,0,0,1,1,1,2,2,2};
     b = a[2];    Is a[2] a pointer to the last row of a[][]?

>Is a[2] a pointer to the last row of a[][]?

Kind of.  a[][] is an array of integers, not pointers.


a[2] is the first element of the last row.


Kent

Thanks a lot.

I can do
     int *b;
     int a[3][3]={0,0,0,1,1,1,2,2,2};
     b = a[2];
without any error meanning a[2] is a pointer to the last row? If it is, why we cannot use it to point something else: a[2] = b or a[2] = a[1]?

Hi learn,

     int *b;
     int a[3][3]={0,0,0,1,1,1,2,2,2};
     int *row[3] = {&a[0][0],&a[1][0],&a[2][0]};
     b = row[0]; row[0] = row[1]; row[1]=b;

It doesnt actually swap the data, it just moves the goalposts. :-)

Paul

OK......at least we swap the rows by pointer in C++ using "new", can we do the similar job using mallocate?

Kdo's method will work nicely with malloc.

Paul

Paul,

How Kdo's method will work with malloc?

if you just want to swap the row then rajeev_devin's method will do.

for the question of using pointers to swap the row, then theres a catch,
if you dynamically allocated the array like this

int* arr[3];

for(int i=0; i < 3; i++)
   arr[i]=new int[3];

//then you can swap row like this
int *temp;

temp=arr[1];
arr[1]=arr[2];
arr[2]=temp;

but if declared like this

int arr[3][3];

you cant swap the rows using pointers. cause the compiler actually stores the elements in a one dimentional form.
so when you just want to access a value at position a[i][j], the compiler calculates the position using this

arr[(i*3)+j]

a generalized formula would be a[(i*columnSize)+j]

but if you want to access it in that way in your code, the compiler will throw a subscript error, but you can try this in the following way

int *a=(int*)arr;
i=0,j=2;
printf("%d", a[(i*3)+j]);


so if you try to just swap the pointers it wont actually change the rows.
hope you understand this, you can ask more if you want.

Nafis

Hi learn,

>>How Kdo's method will work with malloc?

typedef struct Row
{
 int cell[3];
} Row;

typedef Row [3] Matrix;

Matrix m = (Matrix*)malloc(sizeof(Matrix));

// Swap row 0 and 1.
Row temp = m[1];
m[1] = m[0];
m[0] = temp;

I havent tested this. You may have to tinker with it but you get the idea.

Paul

Paul,

the structs method is not Kent's method, its mine, the author is interested in Kent's method.

Regards,
Siddhesh

Siddesh,

You are correct. Using structs was your way. My apologies. I think asker meant your method as Kent posted about another side of the question, my previous post was misleading. :-(

Paul

nafis_devlpr,

Are you talking "new" in C++?

>>Are you talking "new" in C++?

I don't understand that.

nafis_devlpr,

In your code
   for(int i=0; i < 3; i++)
   arr[i]=new int[3];

I think we cannot use "new" in C:

you can replace

arr[i] = new int[3];

with

arr[i] = (int ) malloc(sizeof(int)*3));