Is there a difference between these two statements inside a function?
bool returnValue = true;
// Code that does something
return(returnValue);
No difference!!
People use parenthesis if there's a complex expression involved.
BTW return
is a statement not a function.
C++14 adds a fringe case where parentheses around a return value may alter the semantics. This code snippet shows two functions being declared. The only difference is parentheses around the return value.
int var1 = 42;
decltype(auto) func1() { return var1; } // return type is int, same as decltype(var1)
decltype(auto) func1() { return(var1); } // return type is int&, same as decltype((var1))
In the first func1
returns an int
and in the second one func1
returns an int&
. The difference in semantics is directly related to the surrounding parentheses.
The auto
specifier in its latest form was introduced in C++11. In the C++ Language Spec it is described as:
Specifies that the type of the variable that is being declared will be automatically deduced from its initializer. For functions, specifies that the return type is a trailing return type or will be deduced from its return statements (since C++14)
As well C++11 introduced the decltype
specifier which is described in the C++ Language Spec:
Inspects the declared type of an entity or queries the return type of an expression.
[snip]
If the argument is either the unparenthesised name of an object/function, or is a member access expression (object.member or pointer->member), then the decltype specifies the declared type of the entity specified by this expression.
If the argument is any other expression of type T, then
a) if the value category of expression is xvalue, then the decltype specifies T&&
b) if the value category of expression is lvalue, then the decltype specifies T&
c) otherwise, decltype specifies T
[snip]
Note that if the name of an object is parenthesised, it becomes an lvalue expression, thus decltype(arg) and decltype((arg)) are often different types.
In C++14 the ability to use decltype(auto)
was allowed for function return types. The original examples are where the semantic difference with parentheses comes into play. Revisiting the original examples:
int var1 = 42;
decltype(auto) func1() { return var1; } // return type is int, same as decltype(var1)
decltype(auto) func1() { return(var1); } // return type is int&, same as decltype((var1))
decltype(auto)
allows the trailing return type in the function to be deduced from the entity/expression on the return statement. In the first version return var1;
is effectively the same as returning the type decltype(var1)
(an int
return type by rule 1 above) and in the second case return (var1);
it's effectively the same as decltype((var1))
(an int &
return type by rule 2b).
The parentheses make the return type int&
instead of int
, thus a change in semantics. Moral of the story - "Not all parentheses on a return type are created equal"
They are identical. I see the parenthesis syntax quite often, and I always ask those who use it: why? And none can answer why they use it.
To bluntly sum it up, parenthesis around returning expressions are used by people who don't quite grasp the difference between function-like macros and functions, or who are confused about the operator precedence or order of evaluation rules in C. There is no coding style benefit from using parenthesis.
Thus
return value;
is more correct than
return (value)
because the latter suggests you don't quite know what you are doing :)
No, there are no difference in your code.
AFAIK, nothing is different.
In C++, expressions can have the form: expr
or (expr)
. So, the latter is an expression with more typing. For further reading about this, refer to a grammar (look for "expression").
I assume boo
is a typo, and you are asking whether there is a difference between
return expr;
and
return(expr);
The answer is no.