这段代码中指针运算的意图是什么？

Question

What does this line of code do, newnode -> item.string = (char *)newnode + sizeof(log_t); , in the below example?

int nodesize = sizeof(log_t) + strlen(data.string) + 1;
newnode = (log_t *)malloc(nodesize);
if (newnode == NULL) return -1;

// What is this line doing:
newnode -> item.string = (char *)newnode + sizeof(log_t);

strcpy(newnode -> item.string, data.string);

where newnode is the variable of type log_t ( log_t has a variable named item of type data_t ). data_t has a property char *string .

Is this code setting up the buffer size of string?

Answer 1

newnode -> item.string = (char *)newnode + sizeof(log_t);

The right side will take the pointer newnode , cast it to a character pointer then add the size of an object to it.

This gives a pointer n bytes beyond newnode , where n is the size of the log_t object.

It then places this pointer value into the string member of the item member of the object pointed to by newnode .

Without seeing the actual structures in use, it's a little hard to tell why this is being done but my best guess would be that it's to provide an efficient self-referential pointer.

In other words, the pointer within newnode will point to an actual part of the newnode itself, or part of a larger memory block that was allocated which contains a newnode object at the start of it. And, since you state that newnode is of the type log_t , it must be the latter case (a type cannot contain a copy of itself - it can contain a pointer to the type itself but not a actual copy).

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An example of where this may come in handy is an object allocation where small sizes are satisfied completely by the object itself but larger ones are handled differently, such as with an int -to- string map entry:

typedef struct {
    int  id;
    char *string;
} hdr_t;

typedef struct {
    hdr_t hdr;
    char  smallBuff[31];
} entry_t;

In the case where you want to populate an entry_t variable with a 500-character string, you would allocate the string separately then just set up string to point to it.

However, for a string of thirty characters or less, you could just create it in smallBuff then set string to point to that instead (no separate memory needed). That would be done with:

entry_t *entry = malloc (sizeof (*entry)); // should check for NULL.
entry->hdr.id = 7;
entry->hdr.string = (char*)entry + sizeof (hdr_t);
strcpy (entry->hdr.string, "small string");

The third line in that sample above is very similar to what you have in your code.

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Similarly (and probably more apropos to your case), you can allocate more memory than you need and use it:

typedef struct {
    int  id;
    char *string;
} entry_t;

char *str = "small string";
entry_t *entry = malloc (sizeof (*entry) + strlen (str) + 1); // with extra bytes.
entry->id = 7;
entry->string = (char*)entry + sizeof (entry_t);
strcpy (entry->string, str);

Answer 2

I'm assuming the memory for newnode was allocated with additional space for the string.

The result is the same, but I'd personally write that as newnode->item.string = (char*) &newnode[1];

That is, the storage space for string is immediately after the log_t object. This is sometimes done when a single chunk of memory has been allocated in advance, and objects and their members all point to memory in this chunk. It's been done in the past to cut down on the overhead of small memory allocations.

If log_t is 32 bytes, and the string is 16 bytes long (including the nul terminator!), you could malloc 48 bytes, point the string member to the 32nd byte of this memory allocation and copy the string there.

Answer 3

It's doing pointer arithmetic. In C, when you add a value to a pointer, it moves the pointer by that value * the size of the type pointed to by the pointer.

What this is doing is casting newnode to a char*, so that the pointer arithmetic is done assuming that newnode is a char*, thus the size of the data it points to is 1. It them adds the sizeof(log_t) which is the size of type log_t. Based on your description of log_t, it looks like it contains a single char*, so its size would be the size of a pointer, either 4 or 8 bytes, depending on the architecture.

So, this will set newnode.item.string to be sizeof(log_t) bytes after address that newnode contains.