Pointer Division possible? [closed] - c

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How can i divide and multiplicate 2 variables in C with Pointers?
I tried this:
int multiplicate(int *x,int *y)
{
*x**y;
}
int divide(int *x,int *y)
{
*x/*y;
}

Answer 1

You are missing the return statement:
int multiplicate(int* x, int* y)
{
return (*x) * (*y);
}
int divide(int *x,int *y)
{
return (*x) / (*y);
}

Answer 2

Use *x/(*y) instead. Else it's interpreted as a multiline comment. And you forgot return

Answer 3

Since the indirection operator has higher precedence that multiplication operator, parenthesis are not needed.
Simply adding some whitespace will do:
int multiplicate(int *x,int *y)
{
return *x * *y;
}
int divide(int *x,int *y)
{
return *x / *y;
}

Answer 4

Pointer Division possible?
You do not try to devide pointers here, but as using the dereferencing *-operator, the values the pointers are pointing to are use.
However, as all examples given so far fully miss essential error checking, I propose the following approaches:
int multiply(int * pr, int * px, int * py)
{
int result = 0; /* Be optimistic. */
if (NULL == pr || NULL == px || NULL == py)
{
errno = EINVAL;
result = -1;
}
else
{
*pr = *px * *py;
}
return result;
}
int devide(int * pr, int * px, int * py)
{
int result = 0; /* Be optimistic. */
if (NULL == pr || NULL == px || NULL == py)
{
errno = EINVAL;
result = -1;
}
elseif (0 = *py) /* Division by zero is not defined. */
{
errno = EDOM;
result = -1;
}
else
{
*pr = *px / *py;
}
return result;
}
Call those functions like this:
#include <stdio.h>
#include <errno.h>
/* The two functions above go here. */
int main(int argc, char ** argv)
{
int result = EXIT_SUCCESS;
if (2 > argc)
{
fprintf("Missing or invalid arguments.\n");
printf("Usage: %s x y\n", argv[0]);
result = EXIT_FAILURE;
}
if (EXIT_SUCCESS == result)
{
int x = atoi(argv[1]); /* Better use strtol here. */
int y = atoi(argv[2]); /* Better use strtol here. */
printf("Got x = %d and y = %d.\n", x, y);
{
int r;
if (-1 == multiply(&r, &x, &y)
{
perror("multiply() failed");
}
else
{
printf("%d * %d = %d\n", x, y, r);
}
}
{
int r;
if (-1 == multiply(&r, &x, &y)
{
perror("devide() failed");
}
else
{
printf("%d / %d = %d\n", x, y, r);
}
}
}
return result;
}
Finally a more sane approach would be to use pointers only where needed.
A possible approch modifying the above examples would look like this:
int multiply(int * pr, int x, int y)
{
int result = 0; /* Be optimistic. */
if (NULL == pr)
{
errno = EINVAL;
result = -1;
}
else
{
*pr = x * y;
}
return result;
}
int devide(int * pr, int x, int y)
{
int result = 0; /* Be optimistic. */
if (NULL == pr)
{
errno = EINVAL;
result = -1;
}
elseif (0 = y) /* Division by zero is not defined. */
{
errno = EDOM;
result = -1;
}
else
{
*pr = x / y;
}
return result;
}
Call those functions like this:
#include <stdio.h>
#include <errno.h>
/* The two functions above go here. */
int main(int argc, char ** argv)
{
int result = EXIT_SUCCESS;
if (2 > argc)
{
fprintf("Missing or invalid arguments.\n");
printf("Usage: %s x y\n", argv[0]);
result = EXIT_FAILURE;
}
if (EXIT_SUCCESS == result)
{
int x = atoi(argv[1]); /* Better use strtol here. */
int y = atoi(argv[2]); /* Better use strtol here. */
printf("Got x = %d and y = %d.\n", x, y);
{
int r;
if (-1 == multiply(&r, x, y)
{
perror("multiply() failed");
}
else
{
printf("%d * %d = %d\n", x, y, r);
}
}
{
int r;
if (-1 == multiply(&r, x, y)
{
perror("devide() failed");
}
else
{
printf("%d / %d = %d\n", x, y, r);
}
}
}
return result;
}