Compilation issue with Pointers [closed] - c

Question

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Closed 10 years ago.
I've just been introduced to pointers in C, and have been playing around with them to try and get more familiar with them.
Can someone explain to my why the code below doesn't compile?
int *high = (int *)malloc(sizeof(int)),*low = (int *)malloc(sizeof(int));
*high = 100;
*low = 0;

Answer 1

If you have #included stdlib.h it should compile.
To ensure that your code is correct, replace it with this:
#include <stdlib.h>
int *high = malloc(sizeof(int));
int *low = malloc(sizeof(int));
if(high == NULL || low == NULL)
{
// no memory, error handling here
}
*high = 100;
*low = 0;
...
free(high);
free(low);

Answer 2

Wrapping the code like shown below whould make it compile and work as expected:
#include <stdlib.h>
int main()
{
int *high = malloc(sizeof(int)),*low = malloc(sizeof(int));
*high = 100;
*low = 0;
return 0;
}
Anyway, the code lacks error checking on system calls. Always check the return value of system calls, In the case of malloc() a value of NULL would be returned if memory could not have been allocated as requested. In such case the line
*high = 100;
would most likely cause the program to crash, as dereferencing *NULL provokes undefined behavior.
In C it is not necessary to cast functions returning a value of void*. Even more it is not recommended to do so, as doing so might hide errors, for example if protoyping for the function in question is missing, the error/warning which would be given by the compiler without casting is surpressed if using casting.