c - 指针的编译问题

Question

我刚刚被介绍给 C 中的指针，并且一直在玩弄它们以尝试更熟悉它们。

有人可以向我解释为什么下面的代码无法编译吗？

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.