CODEWITHSUNDEEP
c++
Devesh Agrawal
Devesh Agrawal

Reputation: 9212

How is this getting evaluated?

I am feeling very stupid to asking this question. but cant figure out the reason on my own.

int main()
{
    int target;
    int buffer =10;
    const int source = 15;
    target = (buffer+=source) = 20;
    cout << target+buffer;
    return 0;
}

target = (buffer+=source) = 20; will become target = (25) = 20.

But if I am giving same statement in my source file, it is giving l-value error.

How the value of target+buffer is printing 40.

Upvotes: 4

Views: 141

Answers (1)

masoud
masoud

Reputation: 56509

Some predefined operators, such as +=, require an operand to be an lvalue when applied to basic types [§13.5/7]

buffer+=source returns a lvalue reference to buffer. So you have not compile error.

your statement can be evaluate as:

buffer+=source;
buffer=20;
target=20;

But modifying buffer twice in a statement is undefined behavior and another compiler can evaluate something else as result. (Not sure in this case also!)

Upvotes: 5

Related Questions