Questions: What is the output of the following code snippet? ``` char* myfun() char mychar = 'A'; char* p; p = mychar; return p; int main() cout << *(myfun())<< endl; return 0; ```

Solution Steps

The function myfun is defined to return a pointer to a character (char*). Inside the function, a local character variable mychar is initialized with the value 'A'. A pointer p is then assigned the address of mychar.

The function myfun returns the pointer p, which points to the local variable mychar.

In the main function, myfun is called, and the returned pointer is dereferenced to print the value it points to. The cout statement attempts to print the character pointed to by the returned pointer.

The variable mychar is a local variable within the myfun function. Once the function returns, the local variable mychar goes out of scope, and the pointer p becomes a dangling pointer.

Final Answer