Why does this C program compile without an error? - c

Question

I'm a beginner in C, and I was playing with C. I typed a C code like this:
#include <stdio.h>
int main()
{
printf("hello world\n");
\
return 0;
}
Even though I used \ knowingly, the C compiler doesn't throw any error. What is this symbol used for in the C language?
Edit:
Even this works:
"\n";

Answer 1

The sequence backslash-newline is removed from the code in a very early phase (phase 2) of the translation process. It used to be how you created long string literals before there was string concatenation, and is how you still extend macros over multiple lines.
See §5.1.1.2 Translation Phases of the C99 standard:
The precedence among the syntax rules of translation is specified by the following
phases.5)
Physical source file multibyte characters are mapped, in an implementation defined
manner, to the source character set (introducing new-line characters for
end-of-line indicators) if necessary. Trigraph sequences are replaced by
corresponding single-character internal representations.
Each instance of a backslash character (\) immediately followed by a new-line
character is deleted, splicing physical source lines to form logical source lines.
Only the last backslash on any physical source line shall be eligible for being part
of such a splice. A source file that is not empty shall end in a new-line character,
which shall not be immediately preceded by a backslash character before any such
splicing takes place.
The source file is decomposed into preprocessing tokens6) and sequences of
white-space characters (including comments). A source file shall not end in a
partial preprocessing token or in a partial comment. Each comment is replaced by
one space character. New-line characters are retained. Whether each nonempty
sequence of white-space characters other than new-line is retained or replaced by
one space character is implementation-defined.
Preprocessing directives are executed, macro invocations are expanded, and
_Pragma unary operator expressions are executed. If a character sequence that
matches the syntax of a universal character name is produced by token
concatenation (6.10.3.3), the behavior is undefined. A #include preprocessing
directive causes the named header or source file to be processed from phase 1
through phase 4, recursively. All preprocessing directives are then deleted.
Each source character set member and escape sequence in character constants and
string literals is converted to the corresponding member of the execution character
set; if there is no corresponding member, it is converted to an implementation defined
member other than the null (wide) character.7)
Adjacent string literal tokens are concatenated.
White-space characters separating tokens are no longer significant. Each
preprocessing token is converted into a token. The resulting tokens are
syntactically and semantically analyzed and translated as a translation unit.
All external object and function references are resolved. Library components are
linked to satisfy external references to functions and objects not defined in the
current translation. All such translator output is collected into a program image
which contains information needed for execution in its execution environment.
5) Implementations shall behave as if these separate phases occur, even though many are typically folded together in practice.
6) As described in 6.4, the process of dividing a source file’s characters into preprocessing tokens is
context-dependent. For example, see the handling of < within a #include preprocessing directive.
7) An implementation need not convert all non-corresponding source characters to the same execution
character.
If you had a blank or any other character after your stray backslash, you would have a compilation error. We can tell that you don't have anything after it because you don't have a compilation error.
The other part of your question, about:
"\n";
is quite different. It is a simple expression that has no side-effects and therefore no effect on the program. The optimizer will completely discard it. When you write:
i = 1;
you have an expression with a value that is discarded; it is evaluated for its side-effect of modifying i.
Sometimes, you'll find code like:
*ptr++;
The compiler will warn you that the result of the expression is discarded; the expression can be simplified to:
ptr++;
and will achieve the same effect in the program.

Answer 2

The \, when immediately followed by a newline, is consumed by preprocessing and causes the next "physical" line to be joined to the current logical line. This is very important for writing long preprocessing directives, which have to be all on one logical line:
#define SHORT very log macro \
consisting of lots and \
lots of preprocessor \
tokens
If you remove the backslash-newline sequences, it is no longer correct. Some other languages from the Unix culture have a similar backslash line continuation syntax: the POSIX shell language derived from the Bourne shell, and also makefiles.
$ this is \
one shell command
About "\n";, that is a primary expression used to form an expression-statement. In C, expressions can be used as statements, and this is exploited all the time. Your printf call, for instance, is an expression statement. printf("hello world\n") is a postfix expression which calls a function, obtaining a return value. Because you used this expression as a statement, the return value is thrown away. The return value of printf
indicates how many characters were printed, or whether it was successful at all, so by throwing it away, your program makes itself oblivious to whether the printf call actually worked.
Since the value of an expression-statement is discarded, if such a statement also has no side effects, it is a useless statement which does nothing (like your "\n"). But such useless expression statements are not erroneous. If you add warning options to your compiler command line you might get a warning such as "statement with no effect" or something like that.

Answer 3

The backslash \ get interpreted by the C preprocessor. It protect its following character (the new line character on your case).

Answer 4

The backslash is simply escaping the next character. In this case, probably a line end (CR) character. Perfectly reasonable.

Answer 5

The backslash plus what is following it is an escape sequence; "\n" together is the newline character (prints a newline). Another important one is "\t", for tab.