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操作符不匹配*使用boost :: variant时

[英]No match for operator * when using boost::variant

I define my own variant type like so: 我定义自己的variant类型,如下所示:

typedef variant<myobj **, ... other types> VariantData;

One of my class methods gets this data type as a parameter and tries to do something like: 我的一种类方法将这种数据类型作为参数,并尝试执行以下操作:

void MyMethod(VariantData var){
    //method body
    if(some_cond){ // if true, then it implies that var is of type
                   // myobj **
        do_something(*var); // however, I'm unable to dereference it
    }
    // ... ther unnecessary stuff
}

As a result, when I compile my program, I get this error message: 结果,当我编译程序时,出现以下错误消息:

error: no match for 'operator*' (operand type is 'VariantData ....'

I do not know how to fix this error. 我不知道如何解决此错误。 PS. PS。 On the whole the code works well - if I comment out this part related to dereferencing, then everything runs smoothly. 总体而言,该代码运行良好-如果我注释掉与解引用有关的这一部分,那么一切都会顺利进行。

The error message is quite self-explanatory: you can't dereference boost::variant , it doesn't have such semantics. 错误消息是不言自明的:您不能取消引用boost::variant ,它没有这种语义。 You should first extract the value, ie the pointer, and then dereference it. 您应该首先提取值(即指针),然后取消引用。

To extract the value relying on a run-time logic, just call get(): 要提取依赖于运行时逻辑的值,只需调用get():

//method body
if(some_cond){ // if true, then it implies that var is of type myobj **
    do_something(*get<myobj **>(var));
}

Note, however, that if the run-time logic fails (eg. due to a bug), get() will throw bad_get exception. 但是请注意,如果运行时逻辑失败(例如,由于错误),则get()将引发bad_get异常。

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