Absence of compilation error when using parametrized constructor

问题

Today at work I came across a behavior in C++ which I don't understand. I have produced the following example code to illustrate my problem:

#include <string>
#include <iostream>

class MyException
{
    public:
        MyException(std::string s1) {std::cout << "MyException constructor, s1: " << s1 << std::endl;}
};

int main(){
    const char * text = "exception text";
    std::cout << "Creating MyException object using std::string(const char *)." << std::endl;
    MyException my_ex(std::string(text));
    std::cout << "MyException object created." << std::endl;
    //throw my_ex;

    std::string string_text("exception text");
    std::cout << "Creating MyException object using std::string." << std::endl;
    MyException my_ex2(string_text);
    std::cout << "MyException object created." << std::endl;
    // throw my_ex2;

    return 0;
}

This code snippet compiles without any errors and produces the following output:

 $ g++ main.cpp
 $ ./a.out
Creating MyException object using std::string(const char *).
MyException object created.
Creating MyException object using std::string.
MyException constructor, s1: exception text
MyException object created.

Note that for my_ex the constructor I have defined was not called. Next, if I want to actually throw this variable:

throw my_ex;

I get a compilation error:

 $ g++ main.cpp
/tmp/ccpWitl8.o: In function `main':
main.cpp:(.text+0x55): undefined reference to `my_ex(std::string)'
collect2: error: ld returned 1 exit status

If I add braces around the conversion, like this:

const char * text = "exception text";
std::cout << "Creating MyException object using std::string(const char *)." << std::endl;
MyException my_ex((std::string(text)));
std::cout << "MyException object created." << std::endl;
throw my_ex;

Then it works as I would have expected:

 $ g++ main.cpp
 $ ./a.out
Creating MyException object using std::string(const char *).
MyException constructor, s1: exception text
MyException object created.
terminate called after throwing an instance of 'MyException'
Aborted (core dumped)

I have the following questions:

Why does my first example compile? How come I don't get a compilation error?
Why doesn't the code compile, when I try to throw my_ex;?
Why do the braces resolve the problem?

回答1:

According to most vexing parse, MyException my_ex(std::string(text)); is a function declaration; the function is named my_ex, taking a parameter named text with type std::string, returns MyException. It's not an object definition at all, then no constructor will be called.

Note the error message undefined reference to 'my_ex(std::string)' for throw my_ex; (you're trying to throw a function pointer in fact), which means that can't find the definition of the function my_ex.

To fix it you can add additional parentheses (as you has shown) or use braces which supported from C++11:

MyException my_ex1((std::string(text)));
MyException my_ex2{std::string(text)};
MyException my_ex3{std::string{text}};

回答2:

The answer is to use {} (braced-init) as much as possible. Sometimes, though, it might be missed inadvertently. Luckily, compilers (like clang, under no extra warning flags) can hint:

warning: parentheses were disambiguated as a function declaration [-Wvexing-parse]
    MyException my_ex(std::string(text));
                     ^~~~~~~~~~~~~~~~~~~
test.cpp:13:23: note: add a pair of parentheses to declare a variable
    MyException my_ex(std::string(text));
                      ^
                      (                )
1 warning generated.

which will immediately point you the issue.

来源：https://stackoverflow.com/questions/40767681/absence-of-compilation-error-when-using-parametrized-constructor

标签

c++

most-vexing-parse