int main () {
char *str = new char[3];
//delete[] str; // OK
delete str; // OK too
return 0;
}
I know I need to call delete[] for new []. However, this "delete str" passes compilation. Is there memory leaking in "delete str" in this case?
Yes there is a sure shot memory leak in case of plain delete.
Matching combination must always be:-
new;
delete;
new[];
delete[];
Memory allocated with new[] should be deleted with delete[] - otherwise you leak memory.
i.e.
int main () {
char *str = new char[3];
delete[] str; // OK
char *str2 = new char;
delete str2; // OK
char *str3 = new char[3];
delete str3; // Not OK
char *str4 = new char;
delete[] str4; // Not OK
return 0;
}
For this example there may not be much difference on most of the systems but it is UB. Deleting buffers using delete which were allocated with new invokes UB.
But if char is replaced with some class, delete[] is guaranteed to call destructors of all objects, while delete may call destructor of obj[0] only which may cause resource leak or some other unwanted issue.
Conclusion: This is UB, so never do this.
