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

Because you’re deleting str1 (the memory it points to) twice and you don’t delete the memory allocated under where str2 was first pointing to.

EDIT:
I’m not sure what you’re trying to achieve.

char* str1 = new char [30];
// str1 = 0x00c06810; (i.e.)
char* str2 = new char [40];
// str2 = 0x00d12340; (i.e.)
strcpy(str1, "Memory leak");

// delete [] str2; should be here

str2 = str1; 
// now str2 == str1, so str2 = 0x00c06810 and str1 = 0x00c06810
// deleting 0x00c06810
delete [] str2; 
// deleting 0x00c06810 once again
delete [] str1;
// 0x00d12340 not deleted - memory leak

If you want that assignment (str2 = str1) then delete str2 first.