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Below code gives error on compilation in a C compiler
++(-i);
error: lvalue required as increment operand
It means that -i returns rvalue.
while code
++(+i);
don't give any error . Why so? this link says that +i don't result in lvalue.
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Why is it not possible to set a function returning an address while it is possible to set a function that returns a reference. Short answer: You return a pointer by-value and anything returned by-value is (by definition) not an lvalue.
To return a value from a function, you must include a return statement, followed by the value to be returned, before the function's end statement. If you do not include a return statement or if you do not specify a value after the keyword return, the value returned by the function is unpredictable.
The result of a function is called its return value and the data type of the return value is called the return type. Every function declaration and definition must specify a return type, whether or not it actually returns a value.
In C++, a function which is defined as having a return type of void , or is a constructor or destructor, must not return a value. If a function is defined as having a return type other than void , it should return a value.
It is a glitch in your compiler. In C language all lvalues in expressions are converted to rvalues even before any operators are applied, with the exception of operands of sizeof, &, ++, -- and left-hand sides of . and assignment (see 6.3.2/2)
In other words, in C language +i must produce an rvalue not because unary + supposedly produces an rvalue, but rather because i is converted to rvalue before that unary + even gets a chance to do its thing.
For example, for an int i variable that holds value 42, expression +i is fully equivalent to expression +42. The lvalueness of i is lost and it gets turned into 42 before the semantics of unary + comes into play.
Needless to say, in this case there's no chance the result of unary + can be an lvalue.
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