Should a function have only one return statement?

Answer 1

My preference would be for single exit unless it really complicates things. I have found that in some cases, multiple exist points can mask other more significant design problems:

public void DoStuff(Foo foo)
{
    if (foo == null) return;
}

On seeing this code, I would immediately ask:

• Is 'foo' ever null?
• If so, how many clients of 'DoStuff' ever call the function with a null 'foo'?

Depending on the answers to these questions it might be that

1. the check is pointless as it never is true (ie. it should be an assertion)
2. the check is very rarely true and so it may be better to change those specific caller functions as they should probably take some other action anyway.

In both of the above cases the code can probably be reworked with an assertion to ensure that 'foo' is never null and the relevant callers changed.

There are two other reasons (specific I think to C++ code) where multiple exists can actually have a negative affect. They are code size, and compiler optimizations.

A non-POD C++ object in scope at the exit of a function will have its destructor called. Where there are several return statements, it may be the case that there are different objects in scope and so the list of destructors to call will be different. The compiler therefore needs to generate code for each return statement:

void foo (int i, int j) {
  A a;
  if (i > 0) {
     B b;
     return ;   // Call dtor for 'b' followed by 'a'
  }
  if (i == j) {
     C c;
     B b;
     return ;   // Call dtor for 'b', 'c' and then 'a'
  }
  return 'a'    // Call dtor for 'a'
}

If code size is an issue - then this may be something worth avoiding.

The other issue relates to "Named Return Value OptimiZation" (aka Copy Elision, ISO C++ '03 12.8/15). C++ allows an implementation to skip calling the copy constructor if it can:

A foo () {
  A a1;
  // do something
  return a1;
}

void bar () {
  A a2 ( foo() );
}

Just taking the code as is, the object 'a1' is constructed in 'foo' and then its copy construct will be called to construct 'a2'. However, copy elision allows the compiler to construct 'a1' in the same place on the stack as 'a2'. There is therefore no need to "copy" the object when the function returns.

Multiple exit points complicates the work of the compiler in trying to detect this, and at least for a relatively recent version of VC++ the optimization did not take place where the function body had multiple returns. See Named Return Value Optimization in Visual C++ 2005 for more details.