Sorted indexes of a random numbers - arrays

In which way i can find indexes of my random numbers and store them.
Example:
300, 2, 43, 12, 0, 1, 90
Values -> 0 1 2 12 43 90 300
Indexes -> 0 1 2 3 4 5 6
So. Can i store instead of my values their indexes?
Like This
300 2 43 12 0 1 90
6 2 4 3 0 1 5
And will it possible for negative numbers also?

EDIT: (correction to my previously posted incorrect solution)
#include <stdio.h>
#include <stdlib.h>
typedef struct {
int val;
int in0;
int in1;
} pair_t;
int cmpVal( const void *a, const void *b ) { return ((pair_t*)a)->val - ((pair_t*)b)->val; }
int cmpOrg( const void *a, const void *b ) { return ((pair_t*)a)->in0 - ((pair_t*)b)->in0; }
int main() {
int i;
int unsort[] = { 300, 2, 43, 12, 0, 1, 90 };
const int n = sizeof unsort/sizeof unsort[0];
// Make a copy in unsorted order including orginal sequence.
pair_t *worken = malloc( n * sizeof *worken );
for( i = 0; i < n; i++ )
worken[i].val = unsort[i], worken[i].in0 = i;
// Sort by value ascending
qsort( worken, n, sizeof pair_t, cmpVal );
// Register this sequence with each element
for( i = 0; i < n; i++ )
worken[i].in1 = i;
// Restore original sequence
qsort( worken, n, sizeof pair_t, cmpOrg );
// Copy the indices (of sorted version) to 'persistant' array.
int sorted[n] = { 0 };
for( i = 0; i < n; i++ )
sorted[i] = worken[i].in1;
// Toss 'working' buffer.
free( worken );
// List original sequence
for( i = 0; i < n; i++ )
printf( "%4d", unsort[ i ] );
putchar( '\n' );
// List corresponding indices (as if sorted)
for( i = 0; i < n; i++ )
printf( "%4d", sorted[ i ] );
putchar( '\n' );
return 0;
}
Output
300 2 43 12 0 1 90
6 2 4 3 0 1 5
Trivial assignment loop to "replace values with indices" in original array left out for clarity...
EDIT #2:
The OP suggests the unsorted array is to have its values replaced(!) with indices indicating the sort order.
This following does as much with the proviso that array values are not near the top end of values for ints.
#include <stdio.h>
#include <limits.h>
void show( int u[], size_t cnt ) { // Show current array values
for( size_t i = 0; i < cnt; i++ )
printf( "%4d", u[ i ] );
putchar( '\n' );
}
void oddSort( int u[], size_t cnt ) {
show( u, cnt );
// Succesively find and replace highest values with decreasing large int values.
int peak = INT_MAX;
for( size_t set = 0; set < cnt; set++ ) {
int maxID = 0;
while( u[maxID] >= peak ) maxID++; // find first non-replaced value
for( size_t i = maxID + 1; i < cnt; i++ )
if( u[i] < peak && u[i] > u[maxID] )
maxID = i;
u[maxID] = peak--;
}
// transpose down to 0, 1, 2...
for( size_t i = 0; i < cnt; i++ )
u[i] -= peak + 1;
show( u, cnt );
}
int main() {
{
int u[] = { 300, 2, 43, 12, 0, 1, 90 };
oddSort( u, sizeof u/sizeof u[0] );
}
putchar( '\n' );
{
// Test with negatives (coincidentally lowest value in first pos)
int u[] = { -256, 300, 2, 43, 12, 0, 1, 90 };
oddSort( u, sizeof u/sizeof u[0] );
}
return 0;
}
Output:
300 2 43 12 0 1 90
6 2 4 3 0 1 5
-256 300 2 43 12 0 1 90
0 7 3 5 4 1 2 6

Related

I want to insert in a vector multiple values in a location

When I create a vector , let's say size of 5 , elements are 1,2,3,4,5 and I want to add at the location( for example index 2) the numbers 200 and 300 , the vector should look like 1 ,2 ,200,300,3,4,5.
#include <stdio.h>
#include <stdlib.h>
int main(){
int v[] ={1,2,3,4,5,6,7};
int n = sizeof(v)/sizeof(int);
int location,element;
printf("Enter location:");
scanf("%d",&location);
printf("Enter element:");
scanf("%d",&element);
for(int i = n - 1; i >= location;i--){
v[i + 1] = v[i];
}
v[location] = element;
for(int i = 0; i <= n;i++){
printf("%d ",v[i]);
}
}
You need to loop like this
for(;;){
int location,element;
printf("Enter location:");
scanf("%d",&location);
if(location == -1)
break;
printf("Enter element:");
scanf("%d",&element);
for(int i = n - 1; i >= location;i--){
v[i + 1] = v[i];
}
v[location] = element;
}
BUT you code is severely broken, you are adding to a vector thats of fixed size. you cannot do that
In this code
for (int i = n - 1; i >= location; i--) {
v[i + 1] = v[i];
}
n is the count of number of elements, so on the first loop you do
v[n] = v[n-1]
ie
v[7] = v[6]
well v[7] is off the end of the array (array indexes are 0 to 6) Thats very bad
You declared a fixed size array
int v[] ={1,2,3,4,5,6,7};
You can not enlarge it.
Thus this loop
for(int i = 0; i <= n;i++){
printf("%d ",v[i]);
}
invokes undefined behavior due to using the index with the value n that is outside the valid range of indices [0, n) for the source array.
You need to allocate the array dynamically and when you are going to add one more element you will need to reallocate the array.
Here is a demonstration program.
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
int main( void )
{
size_t n = 7;
int *v = malloc( n * sizeof( int ) );
memcpy( v, ( int [7] ){ 1, 2, 3, 4, 5, 6, 7 }, n * sizeof( int ) );
for ( size_t i = 0; i < n; i++ )
{
printf( "%d ", v[i] );
}
putchar( '\n' );
size_t location = 0;
printf( "Enter location: " );
scanf( "%zu", &location );
if ( n < location ) location = n;
int element = 0;
printf( "Enter element: " );
scanf( "%d",&element );
int *tmp = realloc( v, ( n + 1 ) * sizeof( int ) );
if ( tmp != NULL )
{
v = tmp;
memmove( v + location + 1, v + location, ( n - location ) * sizeof( int ) );
v[location] = element;
++n;
}
for ( size_t i = 0; i < n; i++ )
{
printf( "%d ", v[i] );
}
putchar( '\n' );
free( v );
}
The program output might look like
1 2 3 4 5 6 7
Enter location: 2
Enter element: 200
1 2 200 3 4 5 6 7

function which arranges the elements of an array in ascending order

I try to create a function which arranges the elements of an array in ascending order. But the result isn't good. I think I have forgotten one thing... thanks for your help.
#include <stdio.h>
void ft_sort_int_tab(int *tab, int size) {
int i;
int j;
int temp;
size -= 1;
i = 0;
while (i < size) {
if (tab[i] > tab[i + 1]) {
temp = tab[i];
tab[i] = tab[i + 1];
tab[i + 1] = temp;
}
i++;
}
}
int main(void) {
int tab[9] = {9, 5, 2, 3, 8, 4, 16, 20, 24};
ft_sort_int_tab(tab, 9);
for (int i = 0; i < 9; i++) {
printf("%d ", tab[i]);
}
}
The result : 5 2 3 8 4 9 16 20 24
You've coded the inner-sweep loop of a bubblesort, but you apparently missed the point of the algorithm, as that's only half the algorithm.
Each pass of bubblesort swaps adjacent elements if they're out of order relative to each other (not the entire sequence). When a pass is finished, you're guaranteed the extreme value (largest or smallest, depending on your comparator choice) has found its home at the end of the swept partition. At that point, it should not be visited again. The next sweep run up to, but not include that element. Each sweep gets more and more elements closer to their proper homes, and at least one (the last one of that partition length) in its permanent home). As an optimization, you can track whether the current sweep every actually swapped any values.
Optionally, if the sweep completes with no swaps, you can eject entirely; the sequence is sorted. This attribute is what gives bubblesort its only redeeming quality. It is O(n) in best case, when the input sequence is already sorted when utilizing swap-detection. But it is also O(n^2) in general and worst case, making it a poor sorting choice in general.
Regardless
#include <stdio.h>
void ft_sort_int_tab(int *tab, int size)
{
while (size-- > 0) // note descending ceiling
{
int swapped = 0;
for (int i=0; i<size; ++i) // partition sweep
{
if (tab[i] > tab[i + 1])
{
int temp = tab[i];
tab[i] = tab[i + 1];
tab[i + 1] = temp;
swapped = 1;
}
}
if (!swapped) // early exit on no-swap sweep
break;
}
}
int main(void)
{
int tab[9] = {9, 5, 2, 3, 8, 4, 16, 20, 24};
ft_sort_int_tab(tab, 9);
for (int i = 0; i < 9; i++)
printf("%d ", tab[i]);
fputc('\n', stdout);
}
Output
2 3 4 5 8 9 16 20 24
It seems you are trying to implement the bubble sort algorithm.
Your function uses only one loop to move the maximum element in its target position. But you need to repeat the process for all other elements of the array.
The function can look the following way as it is shown in the demonstrative program below.
#include <stdio.h>
void ft_sort_int_tab( int *tab, size_t size )
{
for ( size_t last = size; !( size < 2 ); size = last )
{
for ( size_t i = last = 1; i < size; ++i )
{
if ( tab[i] < tab[i-1] )
{
int tmp = tab[i];
tab[i] = tab[i - 1];
tab[i - 1] = tmp;
last = i;
}
}
}
}
int main(void)
{
int tab[] = { 9, 5, 2, 3, 8, 4, 16, 20, 24 };
const size_t N = sizeof( tab ) / sizeof( *tab );
for ( size_t i = 0; i < N; i++ )
{
printf( "%d ", tab[i] );
}
putchar( '\n' );
ft_sort_int_tab( tab, N );
for ( size_t i = 0; i < N; i++ )
{
printf( "%d ", tab[i] );
}
putchar( '\n' );
return 0;
}
The program output is
9 5 2 3 8 4 16 20 24
2 3 4 5 8 9 16 20 24
Pay attention to that in general the number of elements in an array should be specified by an object of the type size_t. It is the type of the value returned by the operator sizeof. An object of the type int can be not enough large to store the possible number of elements in an array.
A more general approach is to add one more parameter to the function that will specify the criteria of sorting an array. With this approach you can for example sort an array either in the ascending order or in the descending order using the same one function.
Here is a demonstrative program.
#include <stdio.h>
int ascending( int x, int y )
{
return x < y;
}
int descending( int x, int y )
{
return y < x;
}
void ft_sort_int_tab( int *tab, size_t size, int cmp( int, int ) )
{
for ( size_t last = size; !( size < 2 ); size = last )
{
for ( size_t i = last = 1; i < size; ++i )
{
if ( cmp( tab[i], tab[i-1] ) )
{
int tmp = tab[i];
tab[i] = tab[i - 1];
tab[i - 1] = tmp;
last = i;
}
}
}
}
int main(void)
{
int tab[] = { 9, 5, 2, 3, 8, 4, 16, 20, 24 };
const size_t N = sizeof( tab ) / sizeof( *tab );
for ( size_t i = 0; i < N; i++ )
{
printf( "%d ", tab[i] );
}
putchar( '\n' );
ft_sort_int_tab( tab, N, ascending );
for ( size_t i = 0; i < N; i++ )
{
printf( "%d ", tab[i] );
}
putchar( '\n' );
ft_sort_int_tab( tab, N, descending );
for ( size_t i = 0; i < N; i++ )
{
printf( "%d ", tab[i] );
}
putchar( '\n' );
return 0;
}
The program output is
9 5 2 3 8 4 16 20 24
2 3 4 5 8 9 16 20 24
24 20 16 9 8 5 4 3 2
Your code swaps the two elements if needed, but then moves on to the next position in the array. In the beginning of your example, the 9 and 5 get swapped, so 5 ends up in the first position, but 5 never gets compared to other elements.
There are lots of sorting algorithms out there, and you should read about them. The one that your code seems most similar to is called “bubble sort.” To get there, modify your code do that it makes multiple passes over the array until the array is completely sorted. If you made a second pass, for example, the 5 and 2 would get swapped, and you’d be closer to a sorted result.
Bubble sort is not the fastest way to sort, but it’s easy to understand. Other ways include quicksort, mergesort, and heapsort. Wikipedia can explain each of those, and they’re worth looking into if your need better performance.

duplicate same following elements in array

I should build a function that gets an array and it's size and return a pointer
to new array (i need do create new array by using malloc and realloc) that find the identical numbers and duplicates them in the row
for example the array:{1,8,8,70,2,2,2,5,5,2} and size 10 suppose to return a pointer to this array
{1,8,8,8,8,70,2,2,2,2,2,2,5,5,5,5,2}. Any clue what's wrong with my code??
int * duplicateArray(int* arr, int n)
{
int g = 1;
int i,j=0;
int *p = (int*)(calloc)(n, sizeof(int));
assert(p);
for (i = 0; i < n-1; i++)
{
if (arr[i] == arr[i + 1])
{
p= (int*)(realloc)(p, n+g * sizeof(int));
n=n+g;
assert(p);
p[j] = arr[i];
j++;
p[j] = arr[i+1];
}
else
p[j] = arr[i];
j++;
}
return p;
}
The approach used by you when the memory is constantly reallocated is inefficient.
Also the function should return not only the pointer to the dynamically allocated array but also the number of elements in the allocated array. Otherwise the user of the function will not know how many elements are in the allocated array. To use a sentinel value for an integer dynamically allocated array is not a good idea.
I suggest to split the task into two separate tasks that will correspond to two separate functions..
The first function will count the number of repeated elements in a given array.
The second function will create dynamically an array with the specified size based on the returned value of the first function and copy elements of the source array to the newly created array.
Here is a demonstrative program
#include <stdio.h>
#include <stdlib.h>
size_t countRepeated( const int a[], size_t n )
{
size_t repeated = 0;
for ( size_t i = 0; i != n; )
{
size_t m = 1;
while ( ++i != n && a[i] == a[i-1] ) ++m;
if ( m != 1 ) repeated += m;
}
return repeated;
}
int * copyWithDuplication( const int a[], size_t n, size_t m )
{
int *result = m == 0 ? NULL : calloc( m, sizeof( int ) );
if ( result )
{
for ( size_t i = 0, j = 0; j != m && i != n; )
{
result[j++] = a[i++];
size_t k = 1;
while ( j != m && i != n && a[i] == a[i-1] )
{
result[j++] = a[i++];
++k;
}
if ( k != 1 )
{
while ( j != m && k-- ) result[j++] = a[i-1];
}
}
}
return result;
}
int main(void)
{
int a[] = { 1, 8, 8, 70, 2, 2, 2, 5, 5, 2 };
const size_t N = sizeof( a ) / sizeof( *a );
for ( size_t i = 0; i < N; i++ )
{
printf( "%d ", a[i] );
}
putchar( '\n' );
size_t m = N + countRepeated( a, N );
int *b = copyWithDuplication( a, N, m );
if ( b )
{
for ( size_t i = 0; i < m; i++ )
{
printf( "%d ", b[i] );
}
putchar( '\n' );
}
free( b );
return 0;
}
The program output is
1 8 8 70 2 2 2 5 5 2
1 8 8 8 8 70 2 2 2 2 2 2 5 5 5 5 2
And here is another more interesting demonstrative program.
#include <stdio.h>
#include <stdlib.h>
size_t countRepeated( const int a[], size_t n )
{
size_t repeated = 0;
for ( size_t i = 0; i != n; )
{
size_t m = 1;
while ( ++i != n && a[i] == a[i-1] ) ++m;
if ( m != 1 ) repeated += m;
}
return repeated;
}
int * copyWithDuplication( const int a[], size_t n, size_t m )
{
int *result = m == 0 ? NULL : calloc( m, sizeof( int ) );
if ( result )
{
for ( size_t i = 0, j = 0; j != m && i != n; )
{
result[j++] = a[i++];
size_t k = 1;
while ( j != m && i != n && a[i] == a[i-1] )
{
result[j++] = a[i++];
++k;
}
if ( k != 1 )
{
while ( j != m && k-- ) result[j++] = a[i-1];
}
}
}
return result;
}
int main(void)
{
int a[] = { 1, 8, 8, 70, 2, 2, 2, 5, 5, 2 };
const size_t N = sizeof( a ) / sizeof( *a );
for ( size_t i = 0; i < N; i++ )
{
printf( "%d ", a[i] );
}
putchar( '\n' );
size_t m = N + countRepeated( a, N );
for ( size_t i = 0; i < m; i++ )
{
int *b = copyWithDuplication( a, N, i + 1 );
if ( b )
{
for ( size_t j = 0; j < i + 1; j++ )
{
printf( "%d ", b[j] );
}
putchar( '\n' );
}
free( b );
}
return 0;
}
Its output is
1 8 8 70 2 2 2 5 5 2
1
1 8
1 8 8
1 8 8 8
1 8 8 8 8
1 8 8 8 8 70
1 8 8 8 8 70 2
1 8 8 8 8 70 2 2
1 8 8 8 8 70 2 2 2
1 8 8 8 8 70 2 2 2 2
1 8 8 8 8 70 2 2 2 2 2
1 8 8 8 8 70 2 2 2 2 2 2
1 8 8 8 8 70 2 2 2 2 2 2 5
1 8 8 8 8 70 2 2 2 2 2 2 5 5
1 8 8 8 8 70 2 2 2 2 2 2 5 5 5
1 8 8 8 8 70 2 2 2 2 2 2 5 5 5 5
1 8 8 8 8 70 2 2 2 2 2 2 5 5 5 5 2
Any clue what's wrong with my code??
If the parameter n is 1, then your program will allocate an array for 1 element of type int, but will write nothing to it. It won't copy anything from the input buffer.
You are accessing both the input arr and the output array p out of bounds, which causes undefined behavior. The loop
for (i = 0; i < n-1; i++)
will not count from 0 to the function parameter n minus 2, because n is being incremented inside the loop. This causes your loop to have more iterations than it is supposed to, which causes both the input and the output array to be accessed out of bounds.
Also, there doesn't seem to be much point in your variable g, as it never changes and always has the value 1.
The function prototype
int * duplicateArray(int* arr, int n);
does not seem meaningful, as the calling function has no way of knowing the size of the returned array. If you use the function prototype
void duplicateArray ( const int *p_input_array, int num_input, int **pp_output_array, int *p_num_output );
instead, the function duplicateArray can write the address of the new array to *pp_output_array and the number of elements in the array to *p_num_output. That way, the calling function will effectively be able to receive two "return values" instead of only one.
Here is my implementation of the function duplicateArray and also of the calling function:
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
void duplicateArray( const int *p_input_array, int num_input, int **pp_output_array, int *p_num_output )
{
int i = 0, j = 0;
// In the most extreme case, the output array must be 2 times larger than
// the input buffer, so we allocate double the size of the input buffer.
int *p_output_array = (int*)malloc( num_input * 2 * sizeof(int) );
assert( p_output_array != NULL );
while ( i < num_input )
{
int num_repetitions;
int k = p_input_array[i++];
//count the number of repetitions
for ( num_repetitions = 0; i < num_input && p_input_array[i] == k; num_repetitions++, i++ );
if ( num_repetitions == 0 )
{
p_output_array[j++] = k;
}
else
{
for ( int l = 0; l < num_repetitions + 1; l++ )
{
p_output_array[j++] = k;
p_output_array[j++] = k;
}
}
}
//shrink the array to the actually needed size
p_output_array = (int*)realloc( p_output_array, j * sizeof(int) );
assert( p_output_array != NULL );
*pp_output_array = p_output_array;
*p_num_output = j;
}
int main()
{
int arr[] = { 1, 8, 8, 70, 2, 2, 2, 5, 5, 2 };
int *p;
int num;
duplicateArray( arr, sizeof(arr)/sizeof(*arr), &p, &num );
for ( int i = 0; i < num; i++ ) {
printf( "%d\n", p[i] );
}
free( p );
}

Quicksort skipping one element

So today i tried to implement a quicksort. It's almost working but somehow skips one element.
example : 5 2 8 2 3 4 1 5 7 -5 -1 -9 2 4 5 7 6 1 4
output : -5 -1 -9 1 2 2 3 2 1 4 4 4 5 5 5 6 7 7 8
In this case it skips -9. Here is code of the function.
test = quicksort_asc(0,size,tab,size) // this is function call
int quicksort_asc(int l, int r, int tab[], int tabSize) //l is first index, r is last index, tabSize is tabsize-1 generally
{
int i,buffer,lim,pivot,flag=0;
if(l == r)
return 0;
lim = l-1;
pivot = tab[r-1];
printf("pivot:%d\n",pivot);
for (i = l; i <= r-1; ++i)
{
if(tab[i] < pivot) {
lim++;
buffer = tab[lim];
tab[lim] = tab[i];
tab[i] = buffer;
flag = 1;
}
}
if(flag == 0)
return 0;
buffer = tab[lim+1];
tab[lim+1] = pivot;
tab[r-1] = buffer;
quicksort_asc(l,lim+1,tab,lim+1);//left side
quicksort_asc(lim+1,tabSize,tab,tabSize);//right side
}
This is my array length count code. 100 is maximum size, 0 is a stop value.
Count is size.
int count=0;
for (int i = 0; i < 100; i++)
{
test = scanf("%d",&vec[i]);
if(vec[i] == 0) break;
count++;
}
return count;
It seems nobody hurries to help you.:)
For starters the last parameter is redundant.
int quicksort_asc(int l, int r, int tab[], int tabSize);
^^^^^^^^^^^
All you need is the pointer to the first element of the array and starting and ending indices.
Also instead of the type int for the indices it is better to use the type size_t. However as you are using the expression statement
lim = l-1;
then o'k let the indices will have the type int though you could use another approach without this expression statement.
So the function should be declared like
void quicksort_asc( int tab[], int l, int r );
The variable flag is redundant. When it is equal to 0 it means that all elements before the pivot value are greater than or equal to it. But nevertheless you have to swap the pivot value with the first element that is greater than or equal to the pivot.
This loop
for (i = l; i <= r-1; ++i)
has one iteration redundant. It should be set like
for (i = l; i < r-1; ++i)
This call
quicksort_asc(lim+1,tabSize,tab,tabSize);
^^^^^ ^^^^^^^
shall be substituted for this call
quicksort_asc( lim + 2, r, tab );
^^^^^^^ ^^
because the pivot value is not included in this sub-array.
Here is a demonstrative program.
#include <stdio.h>
void quicksort_asc( int tab[], int l, int r )
{
if ( l + 1 < r )
{
int lim = l - 1;
int pivot = tab[r - 1];
for ( int i = l; i < r - 1; ++i )
{
if ( tab[i] < pivot )
{
lim++;
int tmp = tab[lim];
tab[lim] = tab[i];
tab[i] = tmp;
}
}
tab[r - 1] = tab[lim + 1];
tab[lim + 1] = pivot;
quicksort_asc( tab, l, lim + 1 );
quicksort_asc( tab, lim + 2, r );
}
}
int main(void)
{
int a[] = { 5, 2, 8, 2, 3, 4, 1, 5, 7, -5, -1, -9, 2, 4, 5, 7, 6, 1, 4 };
const int N = ( int )( sizeof( a ) / sizeof( *a ) );
for ( int i = 0; i < N; i++ )
{
printf( "%d ", a[i] );
}
putchar( '\n' );
quicksort_asc( a, 0, N );
for ( int i = 0; i < N; i++ )
{
printf( "%d ", a[i] );
}
putchar( '\n' );
return 0;
}
Its output is
5 2 8 2 3 4 1 5 7 -5 -1 -9 2 4 5 7 6 1 4
-9 -5 -1 1 1 2 2 2 3 4 4 4 5 5 5 6 7 7 8

Moving elements in 2d array for c

So I have 2d array filled with random numbers .For example:
#define d 4
int main(void)
{
int a[d][d];
int primary[d], secondary[d];
size_t i, j;
srand(time(NULL));
/* fill array with random numbers */
for (i = 0; i < d; i++)
for (j = 0; j < d; j++)
a[i][j] = rand() % 100;
/* save diagonals */
for (i = 0; i < d; i++)
{
primary[i] = a[i][i];
secondary[i] = a[d - (i + 1)][i];
How to mirror horizontally diagonals?
For example:
1 0 0 2
0 3 4 0
0 5 6 0
7 0 0 8
8 0 0 7
0 6 5 0
0 4 3 0
2 0 0 1
Task is to print main matrix and then print matrix with mirrored diagonals however i got no ideas how cycle for this should look like.
I thought about cycle for rotating matrix for 180 degrees however I will loose the positions of elements that are not included to the diagonals.
Or can i save diagonal and then reverse it somehow.
Here is code for matrix and diagonal what should i do now.
Hope for your help.
One of approaches is the following
#include <stdio.h>
#define N 4
int main(void)
{
int a[N][N] =
{
{ 1, 0, 0, 2 },
{ 0, 3, 4, 0 },
{ 0, 5, 6, 0 },
{ 7, 0, 0, 8 }
};
for ( size_t i = 0; i < N; i++ )
{
for ( size_t j = 0; j < N; j++ ) printf( "%d ", a[i][j] );
printf( "\n" );
}
printf( "\n" );
for ( size_t i = 0; i < N * N / 2; i++ )
{
int tmp = a[i / N][i % N];
a[i / N][i % N] = a[(N * N - i - 1) / N][(N * N - i - 1) % N];
a[(N * N - i - 1) / N][(N * N - i - 1) % N] = tmp;
}
for ( size_t i = 0; i < N; i++ )
{
for ( size_t j = 0; j < N; j++ ) printf( "%d ", a[i][j] );
printf( "\n" );
}
printf( "\n" );
return 0;
}
The program output is
1 0 0 2
0 3 4 0
0 5 6 0
7 0 0 8
8 0 0 7
0 6 5 0
0 4 3 0
2 0 0 1
The same can be written using pointers. For example
#include <stdio.h>
#define N 4
int main(void)
{
int a[N][N] =
{
{ 1, 0, 0, 2 },
{ 0, 3, 4, 0 },
{ 0, 5, 6, 0 },
{ 7, 0, 0, 8 }
};
for ( size_t i = 0; i < N; i++ )
{
for ( size_t j = 0; j < N; j++ ) printf( "%d ", a[i][j] );
printf( "\n" );
}
printf( "\n" );
for ( int *first = ( int * )a, *last = ( int * )a + N * N;
first < last;
++first, --last )
{
int tmp = first[0];
first[0] = last[-1];
last[-1] = tmp;
}
for ( size_t i = 0; i < N; i++ )
{
for ( size_t j = 0; j < N; j++ ) printf( "%d ", a[i][j] );
printf( "\n" );
}
printf( "\n" );
return 0;
}

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