C ++是否允许函数的默认返回类型？

Question

In C the following horror is valid:

myFunc()
{
  return 42;  // return type defaults to int.
}

But, what about in C++? I can't find a reference to it either way...

My compiler (Codegear C++Builder 2007) currently accepts it without warning, but I've had comments that this is an error in C++.

Answer 1

It's ill-formed in C++. Meaning that it doesn't compile with a standard conforming compiler. Paragraph 7.1.5/4 in Annex C of the Standard explains the change "Banning implicit int".

Answer 2

Implicit return types are valid in C89, but a lot of compilers warn about it.

They are not valid in C++, nor in C99.

Answer 3

So, it's definitely 'ill formed' C++, but it seems many compilers accept it with a warning at best.

• Codegear C++Builder 2007: No error or warning at all
• G++: Requires -W -Wall to generate warning , or -pedantic to generate error (Piotr)
• MSVC 8: produces an error ( tfinniga )
• others...?

Please add to/correct this list!

Answer 4

这不是合法的C ++，但是一些编译器会以静默方式或使用诊断方式接受它。

Answer 5

As posted, it is ill-formed. MSVC 8 gives the following error:

error C4430: missing type specifier - int assumed. Note: C++ does not support default-int