#include <iostream>
using namespace std;
void abc (char &x1, char &y1)
{
char x;
x=y1;
cout<<endl<<x<<x1<<y1;
{
char x='T';
x++;
(x1)++;
cout<<endl<<x<<x1<<y1;
}
cout<<endl<<x<<x1<<y1;
}
main()
{
char x='A', y='B';
abc(x,y);
cout<<endl<<x<<y;
}
//output:
//ab pb ub bb
small program in C++
Author
can someone tell me how can the following program work/compile?
those inner brackets in the function make no sense to me.
Generally it helps if you say what the exact problem is. For instance, by copy and pasting the error messages you received after trying to compiling. Or if the program compiled successfully, what happened and what did you expect to happen when you ran the program?
Anyway, before 'main()' add 'int'. Main requires int as a return type in c++. If you need more help, give a better description of your problem.
Anyway, before 'main()' add 'int'. Main requires int as a return type in c++. If you need more help, give a better description of your problem.
Author
i am sorry, i was in a hurry...
nothing is wrong with the program per se, but i _do not understand_ the inner brackets that are located within the function.
what is the purpose of those brackets and why does it compile?
nothing is wrong with the program per se, but i _do not understand_ the inner brackets that are located within the function.
what is the purpose of those brackets and why does it compile?
They create an inner or nested scope. They're perfectly legal, that's why it compiles.
You can nest brackets as much as you want; each pair delimits a new scope for identifiers. In this case, for example, the 'x' in the innermost scope hides the 'x' in the outer scope, and the innermost scope's x is destroyed when execution leaves the scope.
They're not very common, although they're sometimes used to force destruction of automatic variables prior the end of the function, which occasionally has some use.
You can nest brackets as much as you want; each pair delimits a new scope for identifiers. In this case, for example, the 'x' in the innermost scope hides the 'x' in the outer scope, and the innermost scope's x is destroyed when execution leaves the scope.
They're not very common, although they're sometimes used to force destruction of automatic variables prior the end of the function, which occasionally has some use.
quick lesson!
Now until you learn about scope, we'll stop right here [smile]
{ cout << "This is called a code block." << endl; cout << "You can have many of these in a function." << endl;}Now until you learn about scope, we'll stop right here [smile]
The brackets inside of the function create a new scope which has a char x variable that is set to the value 'T'. This x variable is only valid inside the brackets and is different than the x variable that is declared outside of the brackets.
Also, your output seems to be wrong. Running this in VC 2005 I get
BAB
UBB
BBB
BB
which is what I would expect when looking at the code.
Also, your output seems to be wrong. Running this in VC 2005 I get
BAB
UBB
BBB
BB
which is what I would expect when looking at the code.
This topic is closed to new replies.
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