How could I select a certain number of elements from an array by giving a start and ending index number to create a new array?
For example, if my original array was {1,2,3,4,5,6}, and I say x=0 and y=2 for their index values, I would have a new array that is {1,2,3}.
Thank you.
If your compiler supports variable length arrays then you can do this the following way
#include <stdio.h>
#include <string.h>
int main(void)
{
int a[] = { 1, 2, 3, 4, 5, 6 };
size_t n1 = 0, n2 = 2;
int b[n2 - n1 + 1];
memcpy( b, a + n1, ( n2 - n1 + 1 ) * sizeof( int ) );
size_t n = sizeof( b ) / sizeof( *b );
for ( size_t i = 0; i < n; i++ )
{
printf( "%d ", b[i] );
}
putchar( '\n' );
return 0;
}
The program output is
1 2 3
Otherwise the new array should be allocated dynamically as for example
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main(void)
{
int a[] = { 1, 2, 3, 4, 5, 6 };
size_t n1 = 0, n2 = 2;
size_t n = n2 - n1 + 1;
int *b = malloc( n * sizeof( *b ) );
memcpy( b, a + n1, n * sizeof( int ) );
for ( size_t i = 0; i < n; i++ )
{
printf( "%d ", b[i] );
}
putchar( '\n' );
free( b );
return 0;
}
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void print_arr(int* arr, int len) {
for (int i = 0; i < len; i ++) {
printf("%d", arr[i]);
if (i != len - 1) {
printf(" ");
} else {
printf("\n");
}
}
}
int main() {
int arr1[] = {1, 2, 3, 4, 5, 6};
int start, end;
printf("input the beginning and the end: ");
scanf("%d %d", &start, &end);
int len = end - start + 1;
// we use malloc to dynamically allocate an array of the correct size
int* arr2 = (int*)malloc(len * sizeof(int));
// we use memcpy to copy the values
memcpy(arr2, arr1 + start, len * sizeof(int));
print_arr(arr1, 6);
print_arr(arr2, len);
// we have to free the memory
free(arr2);
return 0;
}
