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

It is not an option of cout but an array index of "\n"

The array index [a==N] evaluates to [0] or [1], and indexes the character array represented by "\n" which contains a newline and a nul character.

However passing nul to the iostream will have undefined results, and it would be better to pass a string:

cout << &("\n"[a==N]) ;

However, the code in either case is not particularly advisable and serves no particular purpose other than to obfuscate; do not regard it as an example of good practice. The following is preferable in most instances:

cout << (a != N ? "\n" : "") ;

or just:

if( a != N ) cout << `\n` ;

Answer 2

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.

Answer 3

It's probably intended as a bizarre way of writing

if ( a != N ) {
    cout<<"\n";
}

The [] operator selects an element from an array. The string "\n" is actually an array of two characters: a new line '\n' and a string terminator '\0'. So cout<<"\n"[a==N] will print either a '\n' character or a '\0' character.

The trouble is that you're not allowed to send a '\0' character to an I/O stream in text mode. The author of that code might have noticed that nothing seemed to happen, so he assumed that cout<<'\0' is a safe way to do nothing.

In C and C++, that is a very poor assumption because of the notion of undefined behavior. If the program does something that is not covered by the specification of the standard or the particular platform, anything can happen. A fairly likely outcome in this case is that the stream will stop working entirely — no more output to cout will appear at all.

In summary, the effect is,

"Print a newline if a is not equal to N. Otherwise, I don't know. Crash or something."

… and the moral is, don't write things so cryptically.

Answer 4

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

This is actually not a cout option, what is happening is that "\n" is a string literal. A string literal has the type array of n const char, the [] is simply an index into an array of characters which in this case contains:

\n\0

note \0 is appended to all string literals.

The == operator results in either true or false, so the index will be:

• 0 if false, if a does not equal N resulting in \n
• 1 if true, if a equals N resulting in \0

This is rather cryptic and could have been replaced with a simple if.

For reference the C++14 standard(Lightness confirmed the draft matches the actual standard) with the closest draft being N3936 in section 2.14.5 String literals [lex.string] says (emphasis mine):

string literal has type “array of n const char”, where n is the size of the string as defined below, and has static storage duration (3.7).

and:

After any necessary concatenation, in translation phase 7 (2.2), ’\0’ is appended to every string literal so that programs that scan a string can find its end.

section 4.5 [conv.prom] says:

A prvalue of type bool can be converted to a prvalue of type int, with false becoming zero and true becoming one.

Writing a null character to a text stream

The claim was made that writing a null character(\0) to a text stream is undefined behavior.

As far as I can tell this is a reasonable conclusion, cout is defined in terms of C stream, as we can see from 27.4.2 [narrow.stream.objects] which says:

The object cout controls output to a stream buffer associated with the object stdout, declared in <cstdio> (27.9.2).

and the C11 draft standard in section 7.21.2 Streams says:

[...]Data read in from a text stream will necessarily compare equal to the data that were earlier written out to that stream only if: the data consist only of printing characters and the control characters horizontal tab and new-line;

and printing characters are covered in 7.4 Character handling <ctype.h>:

[...]the term control character refers to a member of a locale-specific set of characters that are not printing characters.199) All letters and digits are printing characters.

with footnote 199 saying:

In an implementation that uses the seven-bit US ASCII character set, the printing characters are those whose values lie from 0x20 (space) through 0x7E (tilde); the control characters are those whose values lie from 0 (NUL) through 0x1F (US), and the character 0x7F (DEL).

and finally we can see that the result of sending a null character is not specified and we can see this is undefined behavior from section 4 Conformance which says:

[...]Undefined behavior is otherwise indicated in this International Standard by the words ‘‘undefined behavior’’ or by the omission of any explicit definition of behavior.[...]

We can also look to the C99 rationale which says:

The set of characters required to be preserved in text stream I/O are those needed for writing C programs; the intent is that the Standard should permit a C translator to be written in a maximally portable fashion. Control characters such as backspace are not required for this purpose, so their handling in text streams is not mandated.

Answer 5

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

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

In C++ operator Precedence table, operator [] binds tighter than operator <<, so your code is equivalent to:

cout << ("\n"[a==N]);  // or cout.operator <<("\n"[a==N]);

Or in other words, operator [] does nothing directly with cout. It is used only for indexing of string literal "\n"

For example for(int i = 0; i < 3; ++i) std::cout << "abcdef"[i] << std::endl; will print characters a, b and c on consecutive lines on the screen.

---

Because string literals in C++ are always terminated with null character('\0', L'\0', char16_t(), etc), a string literal "\n" is a const char[2] holding the characters '\n' and '\0'

In memory layout this looks like:

+--------+--------+
|  '\n'  |  '\0'  |
+--------+--------+
0        1          <-- Offset
false    true       <-- Result of condition (a == n)
a != n   a == n     <-- Case

So if a == N is true (promoted to 1), expression "\n"[a == N] results in '\0' and '\n' if result is false.

It is functionally similar (not same) to:

char anonymous[] = "\n";
int index;
if (a == N) index = 1;
else index = 0;
cout << anonymous[index];

valueof "\n"[a==N] is '\n' or '\0'

typeof "\n"[a==N] is const char

---

If the intention is to print nothing (Which may be different from printing '\0' depending on platform and purpose), prefer the following line of code:

if(a != N) cout << '\n';

Even if your intention is to write either '\0' or '\n' on the stream, prefer a readable code for example:

cout << (a == N ? '\0' : '\n');