在 C++ 中省略 return 语句

I just had some weird behavior from a version of g++ for Windows that I got with Strawberry Perl. It allowed me to omit a return statement.

I have a member function that returns a structure consisting of two pointers, called a boundTag:

struct boundTag Box::getBound(int side) {
    struct boundTag retBoundTag;
    retBoundTag.box = this;
    switch (side)
    {
        // set retBoundTag.bound based on value of "side"
    }
}

This function gave me some bad output, and I discovered that it had no return statement. I had meant to return retBoundTag but forgot to actually write the return statement. Once I added return retBoundTag; everything was fine.

But I had tested this function and gotten correct boundTag output from it. Even now, when I remove the return statement, g++ compiles it without warning. WTF? Does it guess to return retBoundTag?

解决方案

Omitting the return statement in a non-void function [Except main()] and using the returned value in your code invokes Undefined Behaviour.

ISO C++-98[Section 6.6.3/2]

A return statement with an expression can be used only in functions returning a value; the value of the expression is returned to the caller of the function. If required, the expression is implicitly converted to the return type of the function in which it appears. A return statement can involve the construction and copy of a temporary object (class.temporary). 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.

For example

int func()
{
    int a=10;
    //do something with 'a'
    //oops no return statement
}


int main()
{
     int p=func();
     //using p is dangerous now
     //return statement is optional here 
}

Generally g++ gives a warning: control reaches end of non-void function. Try compiling with -Wall option.