C: stores n integers in the array in reverse order.

Stores n integers in the array in reverse order.

Solution: Program 1:

#include <stdio.h>

int inv (int x[], int n)

{

int temp, I, j, M = (n-1)/2;

for (i = 0; I <= m; i++)

{

j = n-1-I;

temp = X[i];

X[i] = X[j];

X[J] = temp;

}

Return

}

int main ()

{

int I, a[10] = {1,2,3,4,5,6,7,8,9,10};

printf ("The original array:\n");

for (i = 0; i <; i++)

{

printf ("%d", a[i]);

}

printf ("\ n");

INV (A, 10);

printf ("The array has been inverted:\n");//inverted: Reverse

for (i = 0; i <; i++)

{

printf ("%d", a[i]);

}

printf ("\ n");

return 0;

}

Program 2: Using pointer variables as arguments

#include <stdio.h>

int inv (int *x, int n)

{

int *p, temp, *i, *j, M = (n-1)/2;

Shape parameter group name x receives the address of the real parameter group first element a[0]

i = x; j = x + n-1; p = x + M;

for (; I <= p; i++, j--)

{

temp = *i;

*i = *j;

*j = temp;

}

Return

}

int main ()

{

int I, a[10] = {1,2,3,4,5,6,7,8,9,10};

printf ("The original array:\n");

for (i = 0; i <; i++)

{

printf ("%d", a[i]);

}

printf ("\ n");

INV (A, 10);

printf ("The array has been inverted:\n");//inverted: Reverse

for (i = 0; i <; i++)

{

printf ("%d", a[i]);

}

printf ("\ n");

return 0;

}

Results:

He original array:

1 2 3 4 5 6 7 8 9 10

The array has been inverted:

10 9 8 7 6 5 4 3 2 1

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C: stores n integers in the array in reverse order.