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Will the following give a compilation error? delete cout; delete cin; The answer is : No. It is a flaw in the implementation of stream classes from the Standard library. They have the following conversion function to void* type, which means, all stream objects can be implicitly converted to void* : operator void * ( ) const; This is very useful in general as it lets us write very idiomatic loop , say, when reading input from files. But at the same time, it lets user to write delete stream . As I said, you can delete any stream object. So all of these are allowed: delete ss; //declare std::stringstream ss; delete iss; //declare std::istringstream iss; delete oss; //declare std::ostringstream oss; Only that they'll give a warning, saying (see at ideone ): warning: deleting ‘void*’ is undefined which you can easily avoid just by casting, say, to char* . But the program has still issue, and most likely will crash when running it. -- So my question is, has this issue been addressed and fixed, in C++11? The following article provides one fix for this problem: The Safe Bool Idiom -- Edit: From @Xeo's comment on @Alf's answer: Is the safe-bool idiom obsolete in C++11?

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