Why does this C++ program cause a memory leak?

Question

Consider the following piece of code:

char* str1 = new char [30];    
char* str2 = new char [40];   

strcpy(str1, "Memory leak");    
str2 = str1;     

delete [] str2;     
delete [] str1; 

Why does the above program cause a memory leak? How do I avoid this?

Answer 1

The above doesn't just lead to a memory leak; it causes undefined behavior, which is much, much worse.

The problem is in the last three lines:

str2 = str1; 
delete [] str2; 
delete [] str1; 

If we ignore this first line, then the last two lines correctly reclaim all the memory allocated in this function. However, this first line sets str2 to point to the same buffer as str1. Since str2 is the only pointer in the program to the dynamic memory it references, this line leaks the memory for that buffer. Worse, when you then execute the next two lines to clean up the two pointers, you delete the same block of memory twice, once through str2 and once through str1. This leads to undefined behavior and often causes crashes. Particularly malicious users can actually use this to execute arbitrary code in your program, so be careful not to do this!

But there's one higher-level issue to take into account here. The whole problem with this setup is that you have to do all your own memory management. If you opt to use std::string instead of raw C-style strings, then you can write the code like this:

string str1 = "Memory leak"; // Actually, it doesn't. :-)
string str2;

str2 = str1; // Okay, make str2 a copy of str1

// All memory reclaimed when this function or block ends

Now, there's no need to explicitly manage the memory, and you don't have to worry about buffer overruns or double-frees. You're saved by the magic of object memory allocation.