Why do we even need the “delete[]” operator?

Question

This is a question that\'s been nagging me for some time. I always thought that C++ should have been designed so that the delete operator (without brackets) wor

Answer 1

I'm a bit confused by Aaron's answer and frankly admit I don't completely understand why and where delete[] is needed.

I did some experiments with his sample code (after fixing a few typos). Here are my results. Typos:~Base needed a function body Base *b was declared twice

struct Base { virtual ~Base(){ }>; };
struct Derived : Base { };
int main(){
Base* b = new Derived;
delete b; // this is good

Base b = new Derived[2];
delete[] b; // bad! undefined behavior
}

Compilation and execution

david@Godel:g++ -o atest atest.cpp 
david@Godel: ./atest 
david@Godel: # No error message

Modified program with delete[] removed

struct Base { virtual ~Base(){}; };
struct Derived : Base { };

int main(){
    Base* b = new Derived;
    delete b; // this is good

    b = new Derived[2];
    delete b; // bad! undefined behavior
}

Compilation and execution

david@Godel:g++ -o atest atest.cpp 
david@Godel: ./atest 
atest(30746) malloc: *** error for object 0x1099008c8: pointer being freed was n
ot allocated
*** set a breakpoint in malloc_error_break to debug
Abort trap: 6

Of course, I don't know if delete[] b is actually working in the first example; I only know it does not give a compiler error message.