what does cout << “\n”[a==N]; do?

Question

In the following example:

cout<<\"\\n\"[a==N];

I have no clue about what the [] option does in cout, but it

Answer 1

Each of the following lines will generate exactly the same output:

cout << "\n"[a==N];     // Never do this.
cout << (a==N)["\n"];   // Or this.
cout << *((a==N)+"\n"); // Or this.
cout << *("\n"+(a==N)); // Or this.

As the other answers have specified, this has nothing to do with std::cout. It instead is a consequence of

• How the primitive (non-overloaded) subscripting operator is implemented in C and C++.
  In both languages, if array is a C-style array of primitives, array[42] is syntactic sugar for *(array+42). Even worse, there's no difference between array+42 and 42+array. This leads to interesting obfuscation: Use 42[array] instead of array[42] if your goal is to utterly obfuscate your code. It goes without saying that writing 42[array] is a terrible idea if your goal is to write understandable, maintainable code.

• How booleans are transformed to integers.
  Given an expression of the form a[b], either a or b must be a pointer expression and the other; the other must be an integer expression. Given the expression "\n"[a==N], the "\n" represents the pointer part of that expression and the a==N represents the integer part of the expression. Here, a==N is a boolean expression that evaluates to false or true. The integer promotion rules specify that false becomes 0 and true becomes 1 on promotion to an integer.

• How string literals degrade into pointers.
  When a pointer is needed, arrays in C and C++ readily degrade into a pointer that points to the first element of the array.

• How string literals are implemented.
  Every C-style string literal is appended with the null character '\0'. This means the internal representation of your "\n" is the array {'\n', '\0'}.

Given the above, suppose a==N evaluates to false. In this case, the behavior is well-defined across all systems: You'll get a newline. If, on the other hand, a==N evaluates to true, the behavior is highly system dependent. Based on comments to answers to the question, Windows will not like that. On Unix-like systems where std::cout is piped to the terminal window, the behavior is rather benign. Nothing happens.

Just because you can write code like that doesn't mean you should. Never write code like that.