在C中：为什么在函数之外存在堆栈分配结构？

Question

my function:

struct hostent * gethost(char * hostname){
    if(/*some condition under which I want 
         to change the mode of my program to not take a host*/){
       return null
    }
    else{
        struct hostent * host = gethostbyname(hostname);
        return host;
    }
}

in main:

struct hostent * host = gethost(argv[2]);

(ignore any minor errors in the code, I'm spewing from memory)

this works fine. and Valgrind doesn't tell me I'm losing memory, despite the fact I'm not freeing.

Why? I thought stuff allocated on the stack disappears with the function call returning? or is it because I return the pointer? is this dangerous in any way?

Answer 1

host is not allocated on the stack, only a pointer to it is on the stack. The pointer gets copied when the function returns, so there is nothing wrong with the code.

Note that gethostbyname does not actually dynamically allocate memory. It always returns a pointer to the same statically allocated block of memory, which is why valgrind doesn't report a leak. Be careful, though, because that means you have to copy the hostent returned by your function if you want to save the value for later because further calls to gethost will overwrite it.

Answer 2

It's fine and does leak because the returned pointer doesn't point to data on stack or heap, but some static variable.

http://linux.die.net/man/3/gethostbyname :

The functions gethostbyname() and gethostbyaddr() may return pointers to static data, which may be overwritten by later calls . Copying the struct hostent does not suffice, since it contains pointers; a deep copy is required.

Answer 3

from the manual :

RETURN VALUE
       The gethostbyname() and gethostbyaddr() functions  return  the  hostent
       structure  or a NULL pointer if an error occurs.  On error, the h_errno
       variable holds an error number.  When non-NULL, the  return  value  may
       point at static data, ...

Some memory is reserved at the compile time (ie. inside the binary the code) for the structure, the function returns a pointer to this memory.

Answer 4

Well the memory is not leaked until all references to it is lost, in your example the pointer is returned so there is still a reference to it.

However imho it's a bad design decision on most cases to rely on another part of the code to release dynamic memory. If the function needs to return a struct for example, the caller should do it and pass a pointer to the struct.