Why " throw" will display a warning?

Question

#include <iostream>
#include <sstream>

using std::string;
using std::cout;
using std::endl;
using std::ostringstream;

class illegalParameterValue {

private:
    string message;
public:
    illegalParameterValue() : message("Illegal parameter value") {}
    explicit illegalParameterValue(const char* theMessage) {message = theMessage;}
    void outputMessage() {cout << message << endl;}

};

int main() {

    ostringstream s;
    s << "index = " << 1 << " size = " << 2;
    throw illegalParameterValue(s.str().c_str());

    return 0;
}

I just use some code like this, but the throw will remind some warnings which called

Clang-Tidy: Throwing an exception whose type 'illegalParameterValue' is not derived from 'std::exception'

How can I solve this problem?

The message is literally telling you what to do. What is confusing you? — Mad Physicist, Mar 27 '21 at 05:34
If I do not derive from std::exception, will it go wrong ? @Mad Physicist — ZCdream, Mar 27 '21 at 05:40
If I don’t derive from the exception, I haven’t found any problems, but I’m afraid there will be hidden problems, so I want to know why it is.@Mad Physicist — ZCdream, Mar 27 '21 at 05:47
There is nothing wrong by throwing something else than a `std::exception`. If you have some reason to throw another type, just ignore this warning. — prapin, Mar 27 '21 at 06:38
You have to decide who is in control of your program design, you or the compiler. The code is legal and its meaning is well defined. The compiler is giving you style advice. — Pete Becker, Mar 27 '21 at 14:22

score 2 · Accepted Answer · answered Mar 27 '21 at 06:20

consider the following code:

try {
  // do something that might throw an exception
}
catch (const std::exception &ex) {
  // handle the exception
}

If you derive your illegalParameterValue class from std::exception, then the catch clause above will catch it (along with other kinds of exceptions derived from std::exception). As currently written, it will not. Your users will have to add another catch clause:

try {
  // do something that might throw an exception
}
catch (const std::exception &ex) {
  // handle std exceptions
}
catch (const illegalParameterValue &ip) {
  // handle illegal parameter exeception
}

Maybe that's what you want, maybe not. But there's a lot of code out there like the first case.

Why " throw" will display a warning?

1 Answers1