Why do you specify the size when using malloc in C?

Asked 2009-08-06T19:47:36.587

Take the following code :

int *p = malloc(2 * sizeof *p);

p[0] = 10;  //Using the two spaces I
p[1] = 20;  //allocated with malloc before.

p[2] = 30;  //Using another space that I didn't allocate for. 

printf("%d", *(p+1)); //Correctly prints 20
printf("%d", *(p+2)); //Also, correctly prints 30
                      //although I didn't allocate space for it

With the line malloc(2 * sizeof *p) I am allocating space for two integers, right ? But if I add an int to the third position, I still gets allocated correctly and retrievable.

So my question is, why do you specify a size when you use malloc ?

c memory-management malloc

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2 Answers

When you use * (p+3), you're addressing out of bounds even with using 2*sizeof(* p), hence you're accessing an invalid memory block, perfect for seg faults.

You specify the size b/c otherwise, the function doesn't know how big of a block out of the heap memory to allocate to your program for that pointer.

answered 2009-08-06T19:51:12.603

Because malloc is allocating space on the heap which is part of the memory used by your program which is dynamically allocated. The underlying OS then gives your program the requested amount (or not if you end up with some error which implies you always should check return of malloc for error condition ) of virtual memory which it maps to physical memory (ie. the chips) using some clever magic involving complex things like paging we don't want to delve into unless we are writing an OS.

answered 2009-08-06T20:02:00.797

Why do you specify the size when using malloc in C?

2 Answers

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