Teach Yourself Borland C++ in 14 Days - portal

82 Day 3
As mentioned in the preceding section, memory in a C++ program is allocated dynamically
using operator new. You free memory using the delete operator. Unless you have previously
programmed in C, you might not appreciate the simplicity of new and delete. In C programs,
you use malloc(), calloc(), realloc(), and free() to dynamically allocate memory.
Windows really complicates things by offering a whole raft of local and global memoryallocation
functions. Although this is not exactly difficult, it can be confusing to say the least.
C++ removes those headaches through the use of new and delete.
A new World Order
NOTE
You’ve already seen new in action, so let’s review. As discussed earlier, you can allocate
memory locally (from the stack) or dynamically (from the heap). The following code snippet
shows examples of allocating two character arrays. One is allocated from the stack (local
allocation), and the other is allocated from the heap (dynamic allocation):
char buff[80];
char* bigBuff = new char[4096];
In the first case the buffer size is insignificant, so it doesn’t really matter whether the stack
or the heap is used. In the second case a large char array is needed, so it makes sense to allocate
it from the heap rather than from the stack. This preserves stack space. In the case of arrays
(remember, a string is just an array of type char), the dynamic and local flavors can be used
interchangeably. That is, they use the same syntax:
strcpy(buff, “Ricky Rat”);
strcpy(bigBuff, “A very long string that goes on and on...”);
// later on...
strcpy(bigBuff, buff);
Remember that the name of an array when used by itself points to the first memory location
of the array. A pointer also points to the first memory location of the array, so that is why the
two forms can be used interchangeably.
If the new operator fails to allocate the requested memory, it returns
NULL. In theory, you should check the pointer after calling new to ensure
that it contains a non-zero value:
char* buff = new char[1024];
if (buff) strcpy(buff, “Buteo Regalis”);
else ReportError(); // something went wrong
In reality, if the new operator fails in a 32-bit Windows program, the
entire system is in trouble, and neither your program nor any other will
be running for long.