what is function call as Lvalue

[srikumar]

Hi,
In C++, we can use function calls on L value (left side of assignment operator) where the function returns non-const reference object.
It looks something like f()=10;

But i didnt understand the usage of the above statement, how the value 10 is given to function f() and where does return value of f() gone?

Please help me here!

[helios]

You're not assigning 10 to the function, just like when you do a=f(), you're not assigning the function to a. You're assigning 10 to [the data that's pointed to by the reference that's returned by the function].

int a=-1;

int &f(){ return a; }

int main(){
    std::cout <<a<<std::endl;
    f()=1;
    std::cout <<a<<std::endl;
}

As for the return value of f() (a reference to data), just like all data (a reference to data and a pointer to data are both also data) that's returned by a function, it is destroyed at the end of the statement.

[srikumar]

helios: Thanks for your reply.
What is the practical advantage of this feature?
As i understand from your explanation, We are executing a function f() and assigning the R value(1) to the return object of the function f(). In addition, you are telling that the object will be destroyed after the functions returns as its L-Value. I didnt understand the advantage or practical usecase to use the above statement.

Please clarify
Thanks in advance
-sri

[jsmith]

A common usage is to allow

v[ 4 ] = 42;

if v is a std::vector<int>.

[helios]

assigning the R value(1) to the return object of the function f(). In addition, you are telling that the object will be destroyed after the functions returns as its L-Value.

No, you misunderstood. It's not the object whose value has changed that will be destroyed, it's the reference that pointed to that object.
Consider this:

int a;
{
    int &b=a;
    b=1;
}

At the end of the block, it's b that will go out of scope, not a. a will continue to exist with its modified value.