Final answer:
In the C++ code provided, both 'r' and 'x' will be printed as '20' due to the use of the pre-increment operator before 'x'. The given typo 'e end;' seems to be incorrect syntax and likely meant to be 'endl' for ending the output line.
Step-by-step explanation:
The question pertains to what will be printed after executing a certain block of C++ code. The code in question performs a pre-increment operation on the variable x, which initializes at 19, and then assigns this new value to another variable r. After that, it prints the values of r and x to the standard output. With the pre-increment operator (++x), the value of x is incremented by 1 before it is assigned to r. Therefore, the value of both x and r will be 20 when they are printed out.
Due to a typo in the original question, 'e end;' is not recognized C++ syntax. Presumably, this was meant to be 'endl', which is used in C++ to end the line of output. In summary, the output will be '2020' followed by a newline if the typo is corrected.