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Topic: C++ problem (Read 592 times) |
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curt_cobain
Newbie
Posts: 33
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C++ problem
« on: Sep 28th, 2008, 1:06pm » |
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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 ??//
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towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
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Posts: 13730
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Re: C++ problem
« Reply #1 on: Sep 28th, 2008, 1:16pm » |
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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.
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0.999...
Full Member
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Posts: 156
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Re: C++ problem
« Reply #2 on: Sep 28th, 2008, 2:13pm » |
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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;
}
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| « Last Edit: Sep 28th, 2008, 2:13pm by 0.999... » |
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towr
wu::riddles Moderator
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Some people are average, some are just mean.
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Re: C++ problem
« Reply #3 on: Sep 29th, 2008, 12:36am » |
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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"  ).
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0.999...
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Re: C++ problem
« Reply #4 on: Sep 29th, 2008, 4:32pm » |
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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.
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towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
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Re: C++ problem
« Reply #5 on: Sep 30th, 2008, 1:13am » |
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on Sep 29th, 2008, 4:32pm, 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. |
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When the program encounters the switch, it immediately jump to the relevant case label, so "i = 2;" is never executed.
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SMQ
wu::riddles Moderator
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Re: C++ problem
« Reply #6 on: Sep 30th, 2008, 5:07am » |
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on Sep 30th, 2008, 1:13am, towr wrote:| When the program encounters the switch, it immediately jump to the relevant case label, so "i = 2;" is never executed. |
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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
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--SMQ
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