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; ```

Transcript text: 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

Solution Steps

Step 1: Define the function myfun

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.

Step 2: Return the pointer p

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

Step 3: Call myfun in main

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.

Step 4: Analyze the scope of mychar

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

The code has undefined behavior because it returns a pointer to a local variable that goes out of scope after the function returns. Accessing this pointer in main results in undefined behavior, and the output is unpredictable.

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