Thread: switches and classes... so very very confused :( plz clarify

can someone plz clarify to me in simple terms, why the brackets are needed.

Otherwise a would not go out of scope until the end of the switch, it would sort of be possible to reach code using a without constructing it.

Similarly:

Code:

    if (x) goto skip;
    std::string s;

skip:
    s = "Hello";

If this was allowed, we might end up storing "Hello" in a string object that was never constructed (if x is true).

the problem with this is that i wasn't able to input my "first input" in the first place.. it just skipped the 'cin' and went to the next one..

operator >> leaves newline (and any other unread characters) in the stream, so the first getline has something to read (in effect, if the input stream was empty, you'd get the same effect by pressing Enter at that place).
Solution is to clear to stream (should be in a FAQ).

Because in case 2: you might have written:

Code:

case 2:
cout << a.first << endl;

Which will probably crash when slct equals 2 because as it says, it skips initialisation of a. When you add the extra brackets you can't write the above in case 2, so it knows there can never be a problem.

Originally Posted by MarkZWEERS

that's pretty annoying: I think a temporary should be allowed if it is not used out side its case...

It is, as long as you make it local scope - the problem is that in a

Code:

switch(xx)
{
...
}

the whole switch statement is one large scope - so if you declare a variable in the middle, it's the same as if you declared a variable on the other side of a goto as described above - it will not be constructed because you don't run the constructor code, and unconstructed objects aren't a particularly good idea.

--
Mats

While we're on the subject, let's discuss some other things.
First, if you don't know, I'd suggest you actually read up on what classes are.
Second, input should be a local array, not global.
Void main is undefined. Don't use it.
You allocate with new, but you never free with delete. Free in the destructor.
And you should probably indent the code in main once more so it's +1 level than the main function header itself.

No, the a_class a; will not be skipped. Once a case is matched in a switch, control jumps to that point and keeps on going until the end of the switch or it encounters a break (or a return, or an exception is thrown, etc).

EDIT:
Consider a typical use of a switch:

Code:

switch (x)
{
case 1:
    foo1();
    foo2();
    break;
case 2:
case 3:
    bar();
    break;
default:
    baz();
}

The above is functionally equivalent to:

Code:

if (x == 1)
{
    foo1();
    foo2();
}
else if (x == 2 || x == 3)
{
    bar();
}
else
{
    baz();
}

Incidentally, we have a tutorial section that includes a tutorial on this very topic, if I remember correctly.

I understand there is something in the buffer which is why it skips the first cin.. where in my code did i leave something in the buffer? and isn't cin supped to empty buffer before expecting an input??

You left whitespace in the buffer with the use of formatted input (operator>>). If the use of formatted input emptied the buffer, then you would find that reading a string of words would only leave you with one word, since the rest would be gone as soon as you had read the first one. As such, you are responsible for ignoring what's left in the buffer, e.g., with the use of cin.ignore().

The switch limitation doesn't apply only to classes. You can't declare any variable in a way that it is possible for code execution to skip the declaration to a place where the variable is technically in scope (this can happen with goto and switch). If you want to declare variables within a case, you have to enclose it in {} to make sure that it's scope ends where the case ends.

where in my code did i leave something in the buffer?

Code:

cin >> slct;

This leaves (at least) the newline that came when you pressed Enter left to be read. That's the way >> works.

Therefore, after using >> and before using getline, you must empty the stream.

Code:

cin.ignore();

This would clear one unread character.

Code:

cin.ignore(numeric_limits<streamsize>::max(), '\n');

This would clear everything up to and including a newline.

Originally Posted by MegaManZZ

thanks Elysia for you concern, but i'm already aware of all those things you have mentioned. I just was looking for the answer to a particular question.

Alright, good to know!

Code:

If (a > 0)
     a++;
     a--;
else
     cout << "do nothing" << endl;

in this case the

Code:

a--

would be skipped in the same way the

Code:

a_class a

; did; As i understand it.
If thats the case why does the

Code:

break"

statement not get skipped?

Incidentally, I'm not sure if a-- is skipped here, but I can't be sure. It might depend on how the compiler generates code.

this:

Code:

If (a > 0)
     a++;
     a--;
else
     cout << "do nothing" << endl;

should be formatted as:

Code:

If (a > 0)
     a++;
a--;
else
     cout << "do nothing" << endl;

Since you now have an "else without an if", the compiler will complain about that, and there will be no code generated. "Syntax Error"

--
Mats

Incidentally, I'm not sure if a-- is skipped here, but I can't be sure. It might depend on how the compiler generates code.

I think it will just result in a syntax error.

I suppose that's possible too. Never tried, from what I can remember.

Originally Posted by Elysia

I suppose that's possible too. Never tried, from what I can remember.

It's not just possible, it's certain. If you have "if" without braces, then the next statement is part of the "true" branch. To have an else-statement, that MUST be the next statement after the true-branch - any further statements will be taken as "no else". If the compiler then finds an else somewhere floating about in the code it will say something along the lines of "unexpected else".

--
Mats