如果从 lambda 表达式生成的类没有默认构造函数，那么如何创建生成的 class 类型的 object

Question

I am learning C++ using the resource listed here . In particular, i read about lambda expressions in C++ Primer by Lippman. There i came to know that lambdas are function objects. Moreover, classes generated from lambda expressions do not have a default constructor . So take for example:

auto wc = find_if(words.begin(), words.end(),
            [sz](const string &a){
                    return s.size() >= sz;
            };

It is written that the above lambda expression will generate a class that will look something like:

class SizeComp {
    SizeComp(size_t n): sz(n) { } // parameter for each captured
variable
    // call operator with the same return type, parameters, and body as the lambda
    bool operator()(const string &s) const
        { return s.size() >= sz; }
private:
    size_t sz; // a data member for each variable captured by value
};

So what happens is that an unnamed object of this compiler generated class is created and is passed as the third argument to the std::find_if shown above. I can understand this.

Now it is my thinking/understanding that since this compiler generated class has no default ctor, so the third argument that is passed to the std::find_if must not be created using a default ctor since that would fail. So, internally the object that is passed should be created using the parameterized ctor something like:

auto wc = find_if(words.begin(), words.end(), SizeComp(sz)); //note the third argument uses the parameterized ctor

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My first question is that since there is no default ctor in case of an empty capture list in C++11 , so how will an object of that compiler generated class be created. I mean in case of an empty capture list, there will be no data members inside the compiler generated class so there will be no parameterized ctor. Which means that this compiler generated class has neither default nor parameterized ctor. Then how can an object of this class be generated. For example, the situation shown below won't work when the capture list is empty:

auto wc = find_if(words.begin(), words.end(),
            [](const string &a){           //NOTE THE CAPTURE LIST IS EMPTY
                    return s.size() >= 5;
            };

Now, how will the third argument to std::find_if be created:

auto wc = find_if(words.begin(), words.end(), CompGeneratedClass()); //this can't work because there is no default ctor

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I also know that for a normal/ordinary user-defined class if we provide a user-defined ctor, then the compiler will not generate a default ctor for that class. So, it seems to me that since in case of an empty capture list lambda expression there is no user-defined ctor, the compiler should generate a default ctor for this compiler generated class. But the C++ standard says that there won't be a default ctor. So my second question is that why does this compiler generated class has different behavior than an user-defined class.

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Also, note that this explanation is not only limited to C++ Primer by Lippman. I've seen the same explanation in CppCon presentations as well.

Answer 1

The problem here is that of frame of mind. While the Primer is explaining the behavior in ways that one could easily understand, it is not to be taken literally. The normative text describes behavior, not ways to achieve it. A compiler doesn't actually need to create a class the same way we do.

For instance, a vendor can always sneak in some secret tag into the constructor's parameter list. Imagine this is what it generates:

struct __secret_tag_at_src_line{};

struct __lambda_at_src_line{
   __lambda_at_src_line(__secret_tag_at_src_line, /*Other arguments*/)
   // ...
};

// ...

auto wc = find_if(words.begin(), words.end(), __lambda_at_src_line(__secret_tag_at_src_line{}, ...)};

We can't name that tag or even know of its existence, so we can't create the lambda on another line. And the class will never have a default constructor. That's a way to do it. But even then, a compiler doesn't have to do that.

The implementation is not bound by the same rules as us; the standard gives great leeway to implementations, so long as the program is translated correctly. A compiler could just bless a byte and say "this is a lambda", no need to call a constructor. Why?; Because it accomplishes what the standard says, the observable behavior is intact (think about it? can the initialization of a capture-less lambda be observed by a C++ program?)

Sometimes implementations need to "cheat"; heck, it's been known that implementations will call private constructors to get the job done . But it's not really cheating, because the same rules do not apply.