Is there any problem with throwing an object constructed on the stack in a try-block by non-const reference, catching it and modifying it, then throwing it by reference to another catch block?
Below is a short example of what I'm refering to.
struct EC {
EC(string msg) { what = msg; }
string where;
string what;
void app(string& t) { where += t; }
string get() {开发者_如何学Python return what; }
};
try {
try {
try {
EC error("Test");
throw error;
}
catch (EC& e) {
e.app("1");
throw e;
}
}
catch (EC& e) {
e.app("2");
throw e;
}
}
catch (EC& e) {
e.app("3");
cout << e.where << endl;
cout << e.get() << endl;
}
Is it possible that this could cause e.what to contain junk, but e.where to remain intact? For example:
e.where is "123" e.get() returns a lot of garbage data, until it happens to hit a null byte.There's no such thing as "throwing by reference". It is simply impossible. There's no syntax for that. Every time you try to "throw a reference", a copy of the referenced object is actually thrown. Needless to say, there are no attempts to throw by reference in your code.
It is possible to catch a previously thrown exception by reference (even by a non-const one) and modify the temporary exception object through it. It will work. In fact, you can re-throw the now-modified existing exception object instead of creating a new one. I.e. you can just do
throw;
instead of
throw e;
in your catch clauses and still get the correctly behaving code, i.e. the original object (with modifications) will continue its flight throgh the handler hierarchy.
However, your code is ill-formed at the
e.app("1");
call (and other calls to app
) since the parameter is non-const reference. Change the app
declaration to either
void app(const string& t) { where += t; } // <- either this
void app(string t) { where += t; } // <- or this
for it to compile.
Otherwise, you code should work fine. You are not supposed to get any garbage from get()
. If you do, it must be either a problem with your compiler or with your code that you don't show.
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