[英]c++ : returning reference to temporary
I have made this c++ code : 我做了这个C ++代码:
std::string const & Operand::toString() const
{
std::ostringstream convert;
convert << this->value;
return convert.str();
}
The compiler tells me : returning reference to temporary
编译器告诉我: returning reference to temporary
Am I forced to put convert.str()
in my Operand
class ? 我是否必须将convert.str()
放入我的Operand
类中?
EDIT : It's for a school exercise, I can't change the prototype 编辑:这是一次学校练习,我无法更改原型
convert.str();
this returns an std::string
object which will be destroyed after Operand::toString()
returns. 这将返回一个std::string
对象,该对象将在Operand::toString()
返回之后销毁。 Thus this is temporary variable with lifetime limited to scope of this function. 因此,这是临时变量,其生存期限于此功能的范围。 You should return just the string
itself, by value: 您应该只按值返回string
本身:
std::string Operand::toString() const
{
std::ostringstream convert;
convert << this->value;
return convert.str();
}
or: 要么:
const std::string Operand::toString() const
{
std::ostringstream convert;
convert << this->value;
return convert.str();
}
Just change the function to return a std::string
instead of a std::string const &
. 只需更改函数以返回std::string
而不是std::string const &
。 You want to return by value here. 您想在此处按值返回。
Just remove the &
and const 只需删除&
和const
ie 即
std::string Operand::toString() const
{
std::ostringstream convert;
convert << this->value;
return convert.str();
}
If you can't change the prototype - it must return reference instead the string itself - you have to keep the string somewhere and return reference to it. 如果您不能更改原型-它必须返回引用而不是字符串本身-您必须将字符串保留在某个地方并返回对其的引用。 Maybe create a static string, assign convert.str()
into it and return reference to this static string. 也许创建一个静态字符串,在其中分配convert.str()
并返回对该静态字符串的引用。
std::string const & Operand::toString() const
{
static std::string s;
std::ostringstream convert;
convert << this->value;
s = convert.str();
return s;
}
Another solution is recommended in comment by David Schwartz (at different answer), but it depends if you can add that member string to Operand
. David Schwartz在评论中建议使用另一种解决方案(答案不同),但这取决于您是否可以将该成员字符串添加到Operand
。
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