In the C programming language, it is possible to omit a code block in the case of a single statement, for example:
if(1) exit();
Now, does this only apply to conditionals ? Why is this not valid in the case of functions:
void f(int a) exit();
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4Because the standard says so?Georg Fritzsche– Georg Fritzsche2011-07-07 14:40:52 +00:00Commented Jul 7, 2011 at 14:40
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1never tried it, but I suppose that, if it is not allowed, it is just a matter of syntax, maybe legacy: once you would have written void f(a) int a; { ... }, so the { } were needed to mark the end of argument type declaration block ... maybe...ShinTakezou– ShinTakezou2011-07-07 14:42:50 +00:00Commented Jul 7, 2011 at 14:42
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thanks, I knew of the old function syntax, but this also means compilers are able to correctly parse this then..overscore– overscore2011-07-07 14:48:21 +00:00Commented Jul 7, 2011 at 14:48
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compilers are able to correctly parse that (at least, gcc can).ShinTakezou– ShinTakezou2011-07-07 14:59:29 +00:00Commented Jul 7, 2011 at 14:59
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See this related question, asking the same question about C#: stackoverflow.com/questions/6016654/…Eric Lippert– Eric Lippert2011-07-07 16:29:13 +00:00Commented Jul 7, 2011 at 16:29
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3 Answers
It's a feature of C syntax. In BNF, a function definition is something like
FUNC_DEF ::= TYPE IDENTIFIER "(" PARAM_LIST ")" BLOCK
while a statement is
STATEMENT ::= (EXPRESSION | DECLARATION | CONTROL | ) ";" | BLOCK
BLOCK ::= "{" STATEMENT* "}"
(simplifying to allow intermixed declarations and statements, which C++ allows but C doesn't), and an if statement is
CONDITIONAL ::= "if" "(" EXPRESSION ")" STATEMENT
omitting the else part for now.
The reason for this is that otherwise, you could write the function
void no_op() {}
as
void no_op();
but the latter syntax is already in use to denote a declaration.
6 Comments
Fedor Skrynnikov
I think it's intrasting question from the C parser point of view. Because I dont see any contradiction in the second statment for parser. But any way +1
overscore
Thanks ! This is exactly the kind of answer I was looking for, and you remind me why I like CS !
Michas
The
int no_op() {} as no_op() looks more confusing.
Sebastian Mach
Though you could forbid the
void no_op(); syntax for definitions, e.g. FUNC_DEF ::= TYPE IDENTIFIER "(" PARAM_LIST ")" (BLOCK|STATEMENT). But that would be really confusing to newbies and even apprentices.
Fred Foo
@phresnel: I updated the
STATEMENT rule to more accurately reflect C syntax; ; by itself is a valid statement, so your suggestion would not work. |
The syntax of a conditional statement is this:
if(expression) statementA compound statement is a statement.
A
compound statementis defined as{ zero or more statements }The syntax of a function definition is this
function_declaration compound_statementSo, by definition a function body must be a compound statement and have
{}QED :)
Comments
There is a very old dialect of C, the K&R C. In this dialect the function declaration may look like this:
fun_a(a,b)
char a;
float b;
{
fun_b(b,a);
}
I think it would be too hard to parse it without { and }.
1 Comment
Fred Foo
This is probably the historical reason, +1, but there's also a reason why the syntax hasn't changed in ANSI C.