Make a function that Replace elements between two arrays [closed] - c

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Need to make a function that can replace first n elements from array A with last n elements of array B.
arguments in function should be like (arrayA, arrayB, n)
Works like:
int A[ ] = {1, 2, 3, 4, 5};
int B[ ] = {6, 7, 8, 9, 10};
function name(A[ ], B[ ], 2)
Output:
int A[ ] = {9, 10, 3, 4, 5};
int B[ ] = {6, 7, 8, 1, 2};

Answer 1

void *mymemcpy(void *a, const void *b, size_t size)
{
unsigned char *ac = a;
const unsigned char *bc = b;
while(size--) *ac++ = *bc++;
return a;
}
void replace(void *a, const void *b, size_t nelem, size_t elemsize, size_t size_b)
{
const char *bc = b;
mymemcpy(a, b + (size_b - nelem) * elemsize, nelem * elemsize);
}
int main(void)
{
int a[] = {1,2,3,4,5,6};
int b[] = {1,2,3,4,5,6,7,8,9,0};
replace(a,b,3,sizeof(a[0]), sizeof(b)/sizeof(b[0]));
for(size_t index = 0; index < sizeof(a)/sizeof(a[0]); index++)
{
printf("%2d ", a[index]);
}
printf("\n");
}
I would not use it directly as your teacher probably will not believe you.

Answer 2

Here you are.
#include <stdio.h>
void swap_n( int *a, int *b, size_t n )
{
while ( n-- )
{
int tmp = *a;
*a++ = *b;
*b++ = tmp;
}
}
int main(void)
{
int a[] = { 1, 2, 3, 4, 5 };
int b[] = { 6, 7, 8, 9, 10 };
const size_t N1 = sizeof( a ) / sizeof( *a );
const size_t N2 = sizeof( b ) / sizeof( *b );
for ( size_t i = 0; i < N1; i++ )
{
printf( "%2d ", a[i] );
}
putchar( '\n' );
for ( size_t i = 0; i < N2; i++ )
{
printf( "%2d ", b[i] );
}
putchar( '\n' );
size_t n = 2;
swap_n( a, b + N2 - n, n );
for ( size_t i = 0; i < N1; i++ )
{
printf( "%2d ", a[i] );
}
putchar( '\n' );
for ( size_t i = 0; i < N2; i++ )
{
printf( "%2d ", b[i] );
}
putchar( '\n' );
return 0;
}
The program output is
1 2 3 4 5
6 7 8 9 10
9 10 3 4 5
6 7 8 1 2
Pay attention to that the argument n of the function swap_n shall be not greater than the number of elements in the array a and in the array b.
Using the function swap_n you can exchange any parts of the size n of two arrays. For example to exchange two elements third and fourth of the array a with second and third elements of the array b you could write
size_t n = 2;
swap_n( a + 2, b + 1, n );
In this case the program output will be
1 2 3 4 5
6 7 8 9 10
1 2 7 8 5
6 3 4 9 10