C++中的函数返回引用

Question

double& squaredX (double x)
{
   return x*x;
}

1. What would be the potential problem of this function?
2. What is the best way to rewrite the function?

Answer 1

1. What would be the potential problem of this function?

It doesn't compile, because you cannot bind a non-const reference to a temporary (which is the result of x*x , a temporary int , and the return type of double& is not even compatible) and even if you correct for that, you are returning a reference of a local variable which is incorrect, as the variable goes out of scope when the function returns.

2. What is the best way to rewrite the function?

The type of x*x is int , so it makes sense to return an int

int squaredX (int x)
{
   return x*x;
}

It's possible that the result of x*x is too large to fit into an int , in which case you may want to use a larger type for the calculation and return value. But that depends on the particulars of the situation.

Answer 2

One way that could be "better" is to use templates and universal forwarding references (C++11 or newer).

 template<typename T>
 T square(const T&& x)
 {
     return x*x;
 }

T could be double , int8_t , int , MyComplexMath::Quaternion ... anything that supports operator* . It also supports pass by reference in case the argument is expensive to copy.

Arguably, this might (rightly so) scare away people, so may not "best". It still doesn't address overflows.