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Home/ Questions/Q 839411
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T05:27:34+00:00 2026-05-15T05:27:34+00:00

I tried following code : #include<iostream> #include<string> using namespace std; string f1(string s) {

  • 0

I tried following code :

#include<iostream> 
#include<string>
using namespace std;

string f1(string s)
{
   return s="f1 called";
}

void f2(string *s)
{
   cout<<*s<<endl;
}

int main()
{
   string str;
   f2(&f1(str));
}

But this code doesn’t compile.
What I think is : f1 returns by value so it creates temporary, of which I am taking address and passing to f2.
Now Please explain me where I am thinking wrong?

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  1. Editorial Team
    Editorial Team
    2026-05-15T05:27:35+00:00Added an answer on May 15, 2026 at 5:27 am

    The unary & takes an lvalue (or a function name). Function f1() doesn’t return an lvalue, it returns an rvalue (for a function that returns something, unless it returns a reference, its return value is an rvalue), so the unary & can’t be applied to it.

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