Why C++ operation priority is different when dealing with objects or primitive types ++a*b/b-- - a

Question

I confused why the two identical expressions give different results? The only difference is only that in the expression () I use custom primitive wrapper types.

I'm providing code for clarity: struct Int { int value;

Int operator*(Int b) const{
    return Int{value*b.value};
}

Int operator/(Int b) const{
    return Int{value/b.value};
}

Int operator-(Int b) const{
    return Int{value - b.value};
}

const Int& operator++() {
    ++value;
    return *this;
}

Int operator--(int) {
    Int copy{value};
    value--;
    return copy;
    }
};

ostream& operator<<(ostream& out, Int x) {
    out << x.value;
}

int main(){
    int a = 11; Int A{11};
    int b = 4; Int B{4};
    cout << ++a / b * b-- - a << endl; // 0
    cout << ++A / B * B-- - A << endl; // 4
}

Also Clang compiler shows a warning -Wunsequenced but only for the expression with primitive types.

Ok, but why are we getting warning only with primitive types? Does it means that with classes the order of operations is defined? — Ivan Ehreshi, Dec 02 '15 at 14:36
Both are undefined, but clang can only detect unsequenced modification for the primitive types - it doesn't "know" that your operators modify the objects. — molbdnilo, Dec 02 '15 at 14:37

score 2 · Accepted Answer · answered Dec 02 '15 at 14:40

2

All expressions like ++a / b * b-- - a have undefined behavior because variables are read from and written to without a sequence point in between.

answered Dec 02 '15 at 14:40

Paul Evans

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Why C++ operation priority is different when dealing with objects or primitive types ++a*b/b-- - a

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