Thread: Dynamically allocated array

I had a few questions regarding the way the program was posted by the op

ptr = (char*)malloc(2);

Now this allocates 2 bytes in memory but ptr can hold the address of only the first byte as its a char pointer.

ptr holds the address of one byte, but you still have room to add 2 bytes.

Code:

ptr = (char*)malloc(2); // Note that the cast to char* is not needed
ptr[0] = 'a';
ptr[1] = 'b';
ptr[2] = 'c'; // BAD, you only have room for 2 chars -- not 3! 

or you can do...

ptr = (char*)malloc(2);
*ptr = 'a';
ptr++;
*ptr = 'b';

The memory you get back from malloc() is completely undefined, never mind that the memory OUTSIDE this is not to be touched - it may not give you an error to read [or even write] it, but it doesn't belong to your program, so you shouldn't access it.

If you are using Microsoft compilers, you may find that the memory allcoated is filled with different patterns for the actual memory you allocated (1 byte) and the memory before/after - which may either be padding [because the system requires for example 8 byte alignment of any memory allocated by malloc] or meta-data used by malloc itself to know where the next allocation is or some such - or free memory that hasn't been used ever.

Even if the compiler is not filling the memory in any way, you may find that you get some "random" content when you allocate memory, because the memory contains whatever it was used to hold before the call to malloc (what it held when free was called).

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Mats