Is the return statement considered to be an expression statement in C?

Question

I heard that if an expression is followed by a semicolon, then it is considered to be an expression statement.

Source: http://farside.ph.utexas.edu/teaching/329/lectures/node11.html

int x = 7;
x = 8;
x++;
x—-;
x = x << 1;

These are all expression statements.

But is this an expression statement too?

return 5;

And if not, then please throughly explain why.

And I would also appreciate it if you could tell whether the return satetement can be considered an expression statement in other languages as well.

Answer 1

A return statement and an expression statement are two different things.

Section 6.8.3 of the C standard gives the syntax for an expression statement:

expression-statement:

• expression _opt ;

While section 6.8.6 gives the syntax of a return statement:

jump-statement:

• goto identifier ;
• continue;
• break;
• return expression _opt ;

Also, this is not an expression statement (in fact not a statement at all):

int x = 7;

But a declaration .

Answer 2

This is basically answered by Expression Versus Statement . The key question is: Does a return evaluate to a value (eg could you do x = return 5; ?). Clearly it does not, so it is a statement, not an expression. Expression statements are just expressions used as statements; if it's not an expression, it can't be an expression statement, so return does not form an expression statement.

Answer 3

No. Expression statements are (optional) expressions followed by ; .

return 5 isn't an expression. That's because it doesn't evaluate to a value (you can't assign return 5 to anything) and because it's specifically defined as a jump statement , which is a type of statement distinct from expression statements .