CODEWITHSUNDEEP
cpointersnullallocationrealloc
KeyC0de
KeyC0de

Reputation: 5257

Why NULL ing pointer before using it in realloc function?

If i do the following with malloc

HashTableEntry *util = malloc(sizeof(HashTableEntry) * T->capacity);

all works well. However if i do the following with realloc

HashTableEntry *util = realloc(util, sizeof(HashTableEntry) * T->capacity);

realloc returns NULL. I also get a warning by gcc, saying that i haven't initialized util before using it.

But, the following works perfectly well, like malloc

HashTableEntry *util = NULL;
util = realloc(util, sizeof(HashTableEntry) * T->capacity);

My question is why?

What i believe is happening is that, in all cases, i am creating a pointer and assigning to it the starting address of a block of heap memory i am allocating. Am i wrong somewhere?

Upvotes: 0

Views: 1206

Answers (5)

user3629249
user3629249

Reputation: 16540

Some items to note:

When calling realloc() do NOT assign directly into the target pointer. Rather assign to a temporary pointer, then check (!=NULL) before assigning to the target pointer.

Then, if/when realloc() fails, the original pointer will not be lost. Losing such a pointer results in a memory leak.

Note: the first parameter of realloc() is a 'source' pointer. It must be pointing to something meaningful, like NULL or a known place in the heap.

Upvotes: 0

Vic
Vic

Reputation: 301

  1. If the first argument is a null pointer, realloc() behaves like malloc() for the specified size.

  2. If the first arg is an uninitialized pointer (bad pointer), realloc() won't work properly, so it returns NULL.

  3. If the first arg is a valid address, realloc() deallocates the old object and returns a pointer to a new object or it expands the memory of the old object (it's up to the OS to decide).

Upvotes: 3

Persixty
Persixty

Reputation: 8579

In the second case util is uninitialised when passed into realloc(). So the behaviour of realloc() is undefined.

HashTableEntry *util = realloc(util, sizeof(HashTableEntry) * T->capacity);
                             //^^^^ This value is undefined.

In the third case you initialise it to NULL in which case realloc() behaves like malloc().

Upvotes: 1

David Grayson
David Grayson

Reputation: 87386

In general, functions look at the values of their arguments and use the values to do something, so it's important to pass a specific value instead of something random or unspecified. When you write HashTableEntry *util = realloc(util, sizeof(HashTableEntry) * T->capacity);, you haven't told the compiler what value to put in the util variable when it calls the realloc function, so you'll probably get undefined behavior, which is bad.

Upvotes: 3

Vlad from Moscow
Vlad from Moscow

Reputation: 310920

As it is seen from the function call

HashTableEntry *util = realloc(util, sizeof(HashTableEntry) * T->capacity);

the function realloc uses argument util. If it has an indeterminate value then the function has undefined behavior.

Also before using any function you should read its description and the purpose of each its parameter.

From the C Standard (7.22.3.5 The realloc function)

2 The realloc function deallocates the old object pointed to by ptr and returns a pointer to a new object that has the size specified by size....

and

3 If ptr is a null pointer, the realloc function behaves like the malloc function for the specified size.

Upvotes: 1

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