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        riddles >> cs >> C++ problem
        
(Message started by: curt_cobain on Sep 28^th, 2008, 1:06pm)

Title: C++ problem
Post by curt_cobain on Sep 28^th, 2008, 1:06pm

Following code will give error
switch(i)
{
int x=10;
case 1:cout<<"A";
}

Following code will node give error
switch(i)
{
int x;
case 1:cout<<"A";
}
Why?????//

Title: Re: C++ problem
Post by towr on Sep 28^th, 2008, 1:16pm

If I had to guess, I would say that declaring variables outside of cases is allowed (but probably a bad habit), whereas changing them isn't.

Title: Re: C++ problem
Post by 0.999... on Sep 28^th, 2008, 2:13pm

Also odd is the fact that I can easily undermine the system employed by simply doing this (which compiles in my GCC compiler):
#include <iostream>

using namespace std;

int main()
{
int i = 6;
switch(i) {
int x;
case 6:
cout << "Hello world!" << endl;
default: x = 1;
}
return 0;
}

Title: Re: C++ problem
Post by towr on Sep 29^th, 2008, 12:36am

How is that undermining the system exactly? I can't see any reason why that shouldn't compile; and it's basically an example of Curt's second example which didn't give an error (although it'd be clearer if he had used "not" instead of "node" ;)).

Title: Re: C++ problem
Post by 0.999... on Sep 29^th, 2008, 4:32pm

Disregard that. This works:
switch(i) {
int x;
x = 2;
case 6:
x = 1;
cout << "Hello world!" << endl;
}
So, it appears that you can't declare and initialize a new variable in the same statement, but you can do anything else with it.

Also interesting is that the following prints "Hello, world":
switch(i) {
i = 2;
case 6:
cout << "Hello world!" << endl;
}
Which means that the variable given to the switch is evaluated only once. I declared "i" as volatile to receive the same result.

Title: Re: C++ problem
Post by towr on Sep 30^th, 2008, 1:13am

on 09/29/08 at 16:32:05, 0.999... wrote:

Also interesting is that the following prints "Hello, world":
switch(i) {
i = 2;
case 6:
cout << "Hello world!" << endl;
}
Which means that the variable given to the switch is evaluated only once. I declared "i" as volatile to receive the same result.

When the program encounters the switch, it immediately jump to the relevant case label, so "i = 2;" is never executed.

Title: Re: C++ problem
Post by SMQ on Sep 30^th, 2008, 5:07am

on 09/30/08 at 01:13:36, towr wrote:

When the program encounters the switch, it immediately jump to the relevant case label, so "i = 2;" is never executed.

Which is also why switch(foo) { int bar = 1; ... } is illegal: by the rules of the language the variable bar must be initialized, but it's in a location where the initialization code will never be executed -- a contradiction.

--SMQ