省略 C++ 中的 return 语句

发布于 2024-09-12 08:54:20 字数 582 浏览 10 评论 0原文

我刚刚使用 Strawberry Perl 获得的 Windows 版 g++ 出现了一些奇怪的行为。它允许我省略返回语句。

我有一个成员函数，它返回一个由两个指针组成的结构，称为boundTag：

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

这个函数给了我一些错误的输出，我发现它没有 return 语句。我本想返回 retBoundTag ，但忘记实际编写返回语句。一旦我添加了 return retBoundTag; 一切都很好。

但我已经测试了这个函数并从中获得了正确的 boundTag 输出。即使现在，当我删除 return 语句时，g++ 也会在没有警告的情况下编译它。搞什么？它是否猜测返回retBoundTag？

原文

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?

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无声无音无过去 2024-09-19 08:54:20

省略 non-void 函数中的 return 语句 [除 main()] 并在代码中使用返回值会调用未定义行为。

ISO C++-98[第 6.6.3/2 节]

可以使用带表达式的 return 语句
仅在返回值的函数中；表达式的值为
返回给函数的调用者。如果需要，表达式
隐式转换为函数的返回类型
出现。 return 语句可以涉及构造和复制
临时对象（class.temporary）。从a的末端流出
函数相当于没有值的返回；这导致
返回值函数中的未定义行为。

例如

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 
}

，通常 g++ 会给出警告：控制到达非 void 函数的末尾。尝试使用 -Wall 选项进行编译。

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.

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