Why does this C++ snippet compile (non-void function does not return a value) [duplicate]

I found this in one of my libraries this morning:

static tvec4 Min(const tvec4& a, const tvec4& b, tvec4& out)
{
    tvec3::Min(a,b,out);
    out.w = min(a.w,b.w);
}

I'd expect a compiler error because this method doesn't return anything, and the return type is not void.

The only two things that come to mind are

In the only place where this method is called, the return value isn't being used or stored. (This method was supposed to be void - the tvec4 return type is a copy-and-paste error)
a default constructed tvec4 is being created, which seems a bit unlike, oh, everything else in C++.

I haven't found the part of the C++ spec that addresses this. References (ha) are appreciated.

Update

In some circumstances, this generates an error in VS2012. I haven't narrowed down specifics, but it's interesting, nonetheless.

Solution 1:

This is undefined behavior from the C++11 draft standard section 6.6.3 The return statement paragraph 2 which says:

[...] Flowing off the end of a function is equivalent to a return with no value; this results in undefined behavior in a value-returning function. [...]

This means that the compiler is not obligated provide an error nor a warning usually because it can be difficult to diagnose in all cases. We can see this from the definition of undefined behavior in the draft standard in section 1.3.24 which says:

[...]Permissible undefined behavior ranges from ignoring the situation completely with unpredictable results, to behaving during translation or program execution in a documented manner characteristic of the environment (with or without the issuance of a diagnostic message), to terminating a translation or execution (with the issuance of a diagnostic message).[...]

Although in this case we can get both gcc and clang to generate a wanring using the -Wall flag, which gives me a warning similar to this:

warning: control reaches end of non-void function [-Wreturn-type]

We can turn this particular warning into an error using the -Werror=return-type flag. I also like to use -Wextra -Wconversion -pedantic for my own personal projects.

As ComicSansMS mentions in Visual Studio this code would generate C4716 which is an error by default, the message I see is:

error C4716: 'Min' : must return a value

and in the case where not all code paths would return a value then it would generate C4715, which is a warning.

Solution 2:

Maybe some elaboration on the why part of the question:

As it turns out, it is actually quite hard† for a C++ compiler to determine whether a function exits without a return value. In addition to the code paths that end in explicit return statements and the ones that fall off the end of the function, you also have to consider potential exception throws or longjmps in the function itself, as well as all of its callees.

While it is quite easy for a compiler to identify a function that looks like it might be missing a return, it is considerably harder to prove that it is missing a return. In order to lift compiler vendors of this burden, the standard does not require this to generate an error.

So compiler vendors are free to generate a warning if they are quite sure that a function is missing a return and the user is then free to ignore/mask that warning in those rare cases where the compiler was actually wrong.

†: In the general case, this is equivalent to the halting problem, so it is actually impossible for a machine to decide this reliably.

Solution 3:

Compile your code with -Wreturn-type option:

$ g++ -Wreturn-type source.cpp

This will give you warning. You can turn the warning into error if you use -Werror too:

$ g++ -Wreturn-type -Werror source.cpp

Note that this will turn all warnings into errors. So if you want error for specific warning, say -Wreturn-type, just type return-type without -W part as:

$ g++ -Werror=return-type source.cpp

In general you should always use -Wall option which includes most common warnings — this includes missing return statement also. Along with -Wall, you can use -Wextra also, which includes other warnings not included by -Wall.

Solution 4:

Maybe some additional elaboration on the why part of the question.

C++ was designed so that a very large body of pre-existing body of C code compiles with minimum amount of changes. Unfortunately, C itself was paying a similar duty to earliest pre-standard C which did not even have the void keyword and instead relied on a default return type of int. C functions usually did return values, and whenever code superficially similar to Algol/Pascal/Basic procedures was written without any return statements, the function was, under the hood, returning whichever garbage was left on the stack. Neither the caller nor the callee assigns the value of the garbage in a reliable way. If the garbage is then ignored by every caller, everything is fine and C++ inherits the moral obligation to compile such code.

(If the returned value is used by the caller, the code may behave non-deterministically, similar to processing of an uninitialized variable. Could the difference be reliably identified by a compiler, in a hypothetical successor language to C? This is hardly possible. The caller and the callee may be in different compilation units.)

The implicit int is just a part of the C legacy involved here. A "dispatcher" function might, depending on a parameter, return a variety of types from some code branches, and return no useful value from other code branches. Such a function would generally be declared to return a type long enough to hold any of the possible types and the caller might need to cast it or extract it from a union.

So the deepest cause is probably the C language creators' belief that procedures that do not return any value are just an unimportant special case of functions that do; this problem got aggravated by the lack of focus on type safety of function calls in the oldest C dialects.

While C++ did break compatibility with some of the worst aspects of C (example), the willingness to compile a return statement without a value (or the implicit value-less return at the end of a function) was not one of them.