Does C++ compiler treat all postfix operator overloading as same (postfix version of - and --)?

Question

#include <cstdio> #include <iostream> using namespace std; class Int32 { int num; public: Int32(int num = 0) : num(num) {} ~Int32() {} int value() { return num; } Int32 & operator - (int x) { cout << "Postfix of -" << endl; return *this; } Int32 & operator -- (int x) { cout << "Postfix of --" << endl; return *this; } }; int main() { Int32 x(100); x--; x-; // [Error] expected primary-expression before ';' token x.operator-(0); return 0; }

From the above code I overloaded postfix increment and postfix unary minus. I know postfix unary minus doesn't make sense, but I wonder why I have compilation error for x- and don't have any issue with x-- and x.operator-(0) operations.

I compiled this code in DevC++ and I got following error.

[Error] expected primary-expression before ';' token

What is wrong with x- ?

x- doesn't exist. where you use this operator? there will be only unary minus. — Sanjeev
– Sanjeev, Commented Aug 8, 2020 at 16:48
@Sanjeev: His point is that the compiler lets him declare a function that is a postfix unary minus operator, and he can even call it, but he can't use the operator syntax. — Nicol Bolas
– Nicol Bolas, Commented Aug 8, 2020 at 16:49
Parsing is done independently of overload resolution. x-; fails because the language grammar doesn't allow it. — Barmar
– Barmar, Commented Aug 8, 2020 at 16:50
You can define the function because you can call it using the x.operator-() syntax. — Barmar
– Barmar, Commented Aug 8, 2020 at 16:51
That's not a "unary postfix - operator", it's a "binary infix - operator". — Eljay
– Eljay, Commented Aug 8, 2020 at 17:17

JeJo · Accepted Answer · 2020-08-08 16:51:05Z

2

What is wrong with x- ?

Nothing wrong with it; This by language design. You will see the same error with

1 - ;

meaning, the operator - expect an argument to work with like you did it in the next line

x.operator-(0);

answered Aug 8, 2020 at 16:51

JeJo

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2 Comments

vinoth kumar Over a year ago

x--; also expecting one argument and implicitly ZERO has passed. I am expecting same for x-;. As @Barmer said, parsing is done independently of overload resolution. Does not mean it I can use any operator syntax to call Int32 & operator - (int x)?

Artyer Over a year ago

@vinothkumar That is for x - 0 (or any Int32{} - int{})

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