Question

Consider the following code (and assume that it is embedded in a com- plete and correct program and then run): cout << "Enter a line of input:\n"; char next; do { cin.get(next); cout << next; } while ( (! isdigit(next)) && (next != '\n') ); cout << "

Short answer

Answer: The output will be "I' ".

Step-by-step solution

Understand the Code

Let's first understand the code provided. - It first outputs "Enter a line of input:" to prompt the user for input. - It initializes a char variable 'next'. - A do-while loop starts, which performs the following steps: - It takes the first character of the input using cin.get(next). - It prints the character. - It checks if the character is a digit or a newline character ('\n'). If the character is not a digit and not a newline character, the loop continues with the next character. - The loop stops when a digit or a newline character is encountered. - After the loop, it prints "<END OF OUTPUT".

Apply Code on Given Input

Now, we will apply the code to the given input "I' 11 see you at \(10: 3 \theta\) AM.". We'll go through the input characters one by one until the loop stops: 1. 'I': Not a digit or newline. Print 'I'. 2. ''': Not a digit or newline. Print '''. 3. ' ': Not a digit or newline. Print ' '. 4. '1': It's a digit, so the loop stops. After the loop stops, the code prints "<END OF OUTPUT". So the next line of output will be "I' <END OF OUTPUT".