C++ single quotes syntax - c++

Question

I am learning C++ and just started reading "Programming Principles and Practice" by Bjarne Stroustrup and he uses this code to illustrate a point:
#include "std_lib_facilities.h"
using namespace std;
int main() // C++ programs start by executing the function main
{
char c = 'x';
int i1 = c;
int i2 = 'x';
char c2 = i1;
cout << c << ' << i1 << ' << c2 << '\n';
return 0;
}
I am familiar in general with the difference between double and single quotes in the C++ world, but would someone kindly explain the construction and purpose of the section ' << i1 << '
Thanks

Answer 1

cout << c << ' << i1 << ' << c2 << '\n';
appears to be a typo in the book. I see it in Programming Principles and Practice Using C++ (Second Edition) Second printing. I do not see it listed in the errata.
According to the book, the intended output is
x 120 x
But what happens here is ' << i1 << ' attempts to compress the << i1 << to a multi-byte character and prints out an integer (most likely 540818464-> 0x203C3C20 -> ASCII values of ' ', '<', '<', ' ') because cout doesn't know wide characters. You'd need wcout for that. End result is output something like
x540818464x
and a warning or two from the compiler because while it's valid C++ code, it's almost certainly not what you want to be doing.
The line should most likely read
cout << c << ' ' << i1 << ' ' << c2 << '\n';
which will output the expected x 120 x
In other words, Linker3000, you are not crazy and not misunderstanding the example code.
Anyone know who I should contact to log errata or get a clarification on the off chance there is some top secret sneakiness going way over my head?

Answer 2

Before answering your question, here is a little background on what that is actually doing. Also note that there is a typo in the example, the string constant should have been double quoted:
cout << c << " << i1 << " << c2 << "\n";
In C++, operators can be overloaded so that they mean different things with different functions. In the case of cout, the << operator is overloaded as the "Insertion Operator". Think of it as taking the operand on the right, and inserting it (or sending it) into the operator on the left.
For example,
cout << "Hello World";
This takes the string "Hello World", and sends it to cout for processing.
So what beginners do not get is what something like this means:
cout << "Hello" << " World";
This is doing the same thing, but the operator precedence says to perform the injections from left to right. To make this work, the cout object returns itself as a function return value. Why is this important? Because the above statement is actually two separate operator evaluations:
(cout << "Hello") << " World";
This first injects "Hello" to cout, which outputs it, then continues to evaluate the next inject operator. Because cout returns itself, after the (cout << "Hello") is executed you have the following still to be evaluated:
cout << " World";
This expression injects " World" into the cout object, which then outputs " World", with the net effect being that you see "Hello World" just like the first time.
So in your example, what is it doing?
cout << c << " << i1 << " << c2 << "\n";
This is evaluated left to right as follows:
((((cout << c) << " << i1 << ") << c2) << "\n"); => Outputs value of c
((((cout ) << " << i1 << ") << c2) << "\n"); => Outputs string " << i1 << "
((( cout ) << c2) << "\n"); => Outputs value of c2
(( cout ) << "\n"); => Outputs newline character
( cout ); => No more output
Expression completes and returns the cout object as the expression value.
Assuming c='x' and c2='x', the final output from this expression is the following character string output on a single line:
x << i1 << x
For beginners, all those insertion operators << look a little strange. It is because you are dealing with objects. You could build the string up as a complete formatted object before injecting it into cout, and while that make the cout expression look simpler, we do not do that in C++ because it makes your code more complex and error prone. Note also, there is nothing special about the cout object. If you wanted to output to the standard error stream, you would use cerr instead. If you wanted to output to a file, your would instantiate a stream object that outputs to the desired file. That rest of the code in your example would be the same.
In C, the same thing would be done procedurally using a format string:
printf("%d << i1 << %d\n", i1, c2);
This is allowed in C++ too, because C++ is a superset of C. Many C++ programmers still use this output method, but that is because those programmers learned C first, and may not have fully embraced the object oriented nature of C++
Note that you may also have seen the << operator in the context of mathematical expressions like:
A = A << 8;
In this case, the << operator is the bitwise rotate operation. It has nothing to do with output to cout. It will rotate the bits in A to the left by eight bits.