Question

Possible Duplicate:
Is it safe to delete a void pointer?

Will the following code cause memory leak?

void *ptr = new long [10];
delete[] ptr; // note: ptr is a void*

[EDIT]
The code above will generate a warning message during compiling to specify it “undefined”.
I ask this cause I’m wondering how does C++ handle memory ranges when delete[] is called.
I should change my question to make it more specified.

Will the following code cause memory leak?

char *ptr = (char *)(new long [10]);
delete[] ptr; // note: ptr is a char*

Answer 1

No. Leaving delete[] out will cause a leak. BTW, it should be long* ptr. I don’t think the delete[] will even compile with a void* argument.