have a question about malloc

could you tell me what this code is doing in words because I get lost in b\w.

int *p;
if (p = (int *) malloc (sizeof(int)))==Null)
 { return (Null) }

Who is Participating?
int *p; - declares 'p' as a pointer to integer.

malloc(sizeof(int)) - alloccates block of memory of the size of an int, this returns a void* pointer to the memory if sucessful or NULL if error.

p=(int*) Cast the return value from malloc to void* to type int* (pointer to integer) and copy the value into 'p'

if(p  ==NULL) - If 'p' is NULL then there has been an error allocating the memory, return NULL from the function.
memset(p,0,sizeof(int)) - Otherwise set the new memory pointed to by 'p' to zeros for a total of size of an integer bytes.
it is allocating space for an integer and ...

if successful setting the bytes of the integer to 0 and returning a pointer to it

if unsuccessful it is returning null

PS 3 paragraph should read:
"from malloc __from__ void* to type int"
Cloud Class® Course: Ruby Fundamentals

This course will introduce you to Ruby, as well as teach you about classes, methods, variables, data structures, loops, enumerable methods, and finishing touches.

beat me to it robpitt *s*
>> if (p = (int *) malloc (sizeof(int)))==Null)
This code is trying to confuse all of us :)
When compiled, it will invite the compiler to issue an error message. When evaluating it will look like:


But we'll try to work from the inside out

this expression result in the size in bytes of an integer

  malloc (sizeof(int))
Calls malloc passing the size of an integer as parameter. malloc should respond by allocating memory with at least the requested size. If it fails, it returns NULL, if it succeeds it returns a pointer to the allocated storage. This pointer is of type void*

  (int *) malloc (sizeof(int))
Converts the type of the pointer returned by malloc to int*

Now the explanation can not continue, as there is a missing opening parenthesis, which might be important to the evaluation of the if-expression.
the cod edoes the following:
int *p;//defines a pointer p
if (p = (int *) malloc (sizeof(int)))==Null)//checks if the space asigned for the array is empty
{ return (Null) }//if it is,we return NULL
memset(p,0,sizeof(int))//places the reading pointer in the beginning of the array.

Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

All Courses

From novice to tech pro — start learning today.