Simple question about \x escape sequence

[Christiano]

See the following statement:
std::cout << "\105ba" << std::endl;

In the ascii table 105 octal means 'E', so the result is: "Eba"

I want to do the same thing using hexadecimal escape sequence

std::cout << "\x45ba" << std::endl;

And the result is:

$ CC a.cpp
a.cpp:7:16: error: hex escape sequence out of range
std::cout << "\x45ba" << std::endl;
^~~~~~
1 error generated.

The book C++ Primer Fifth edition - Stanley Lippman - in the page 39 says:

""""" C++ Primer 5th, page 39 """""""""""""""""""""""""""""""""
"We can also write a generalized escape sequence , which is \x followed
by one or more hexadecimal digits or a \ followed be one, two, or three
octal digits. The value represents the numerical value of the character."
""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""

My question is:
How can I write the sequence \x45 + b + a ("Eba") without the compiler
understanding 'b' and 'a' as hexadecimal characters ?

In Octal this problem doesn't exist, example:
std::cout << "\10510" << std::endl;
will print "E10", that is, '1'+'0' will not be interpreted as octal
characters.

[Reinhardt Behm]

write it as "\x45" "ba" (not tested)

--
Reinhardt

[Christiano]

It works.
std::cout << "\x45""ba" << std::endl;