variable shadow

[Black Ice]

If any compiler will give warning when meed variable show.
Following is the example code:
void foo()
{
int total;
if(...)
{
int total;
...
}
}


--


Regards,
Black Ice

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[Victor Bazarov]

Re: variable shadow

"Black Ice" <smartnoho@yaho o.com> wrote...[color=blue]
> If any compiler will give warning when meed variable show.[/color]

I am not sure I understand that sentence. 'Meed' is a noun. Used
with another noun ('variable') it becomes an attribute. "Meed
variable" is "earned reward variable"?... I don't understand.
[color=blue]
> Following is the example code:
> void foo()
> {
> int total;
> if(...)
> {
> int total;
> ...
> }
> }[/color]

No, in most cases no warning will be given. Name hiding (that's
the C++ term) is a legal C++ construct and used often.

Victor

[Victor Bazarov]

Re: variable shadow

"Black Ice" <smartnoho@yaho o.com> wrote...[color=blue]
> If any compiler will give warning when meed variable show.[/color]

I am not sure I understand that sentence. 'Meed' is a noun. Used
with another noun ('variable') it becomes an attribute. "Meed
variable" is "earned reward variable"?... I don't understand.
[color=blue]
> Following is the example code:
> void foo()
> {
> int total;
> if(...)
> {
> int total;
> ...
> }
> }[/color]

No, in most cases no warning will be given. Name hiding (that's
the C++ term) is a legal C++ construct and used often.

Victor