运算符<<：std :: cout << i <<（i << 1）;

Question

I use the stream operator << and the bit shifting operator << in one line. I am a bit confused, why does code A) not produce the same output than code B)?

A)

int i = 4;  
std::cout << i << " " << (i << 1) << std::endl;   //4 8

B)

myint m = 4;
std::cout << m << " " << (m << 1) << std::endl;   //8 8

class myint:

class myint {
    int i;
public:
    myint(int ii) {
        i = ii;
    }
    inline myint operator <<(int n){
        i = i << n;
        return *this;
    }
    inline operator int(){
        return i;
    }
};

thanks in advance
Oops

Answer 1

Your second example is undefined behavior.

You have defined the << operator on your myint class as if it were actually <<= . When you execute i << 1 , the value in i is not modified, but when you execute m << 1 , the value in m is modified.

In C++, it is undefined behavior to both read and write (or write more than once) to a variable without an intervening sequence point, which function calls and operators are not, with respect to their arguments. It is nondeterministic whether the code

std::cout << m << " " << (m << 1) << std::endl;  

will output the first m before or after m is updated by m << 1 . In fact, your code may do something totally bizarre, or crash. Undefined behavior can lead to literally anything, so avoid it.

One of the proper ways to define the << operator for myint is:

myint operator<< (int n) const
{
   return myint(this->i << n);
}

(the this-> is not strictly necessary, just my style when I overload operators)

Answer 2

Because int << X returns a new int. myint << X modifies the current myint. Your myint << operator should be fixed to do the former.

The reason you're getting 8 for the first is that apparently m << 1 is called first in your implementation. The implementation is free to do them in any order.

Answer 3

Your << operator is in fact a <<= operator. If you replace the line with

std::cout << i << " " << (i <<= 1) << std::endl;   //8 8

you should get 8 8.

Answer 4

since m is a myInt your second example could be rewritten as:

std::cout << m << " " << (m.operator<<( 1)) << std::endl; 

The order of evaluation for the subexpressions m and (m.operator<<( 1)) is unspecified, so there's no saying which " m " you'll get for the 1'st expression that m is used in (which is a simple m expression). So you might get a result of "4 8" or you might get "8 8".

Note that the statement doesn't result in undefined behavior because there are sequence points (at least one function call) between when m is modified and when it's 'read'. But the order of evaluation of the subexpressions is unspecified, so while the compiler has to produce a result (it can't crash - at least not legitimately), there's no saying which of the two possible results it should produce.

So the statement is about as useful as one that has undefined behavior, which is to say it's not very useful.

Answer 5

Well (m << 1) is evaluated before m and therefore m holds 8 already, as in your operator<< you overwrite your own value.

This is wrong behaviour on your side, the operator<< should be const and not change your object.

Answer 6

Because the << operator of myint modifies its lhs. So after evaluating m << 1 , m will actually have the value 8 (while i << 1 only returns 8, but does not make i equal to 8). Since it is not specified whether m<<1 executes before cout << m (because it's not specified in which order the arguments of a function or operator are evaluated), it is not specified whether the output will be 8 8 or 4 8 .

Answer 7

The C++ language does not define the order of evaluation of operators. It only defines their associativity.

Since your results depend on when the operator<< function is evaluated within the expression, the results are undefined.

Algebraic operator $ functions should always be const and return a new object:

inline myint operator <<(int n) const { // ensure that "this" doesn't change
    return i << n; // implicit conversion: call myint::myint(int)
}