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]

"Black Ice" <sm*******@yahoo.com> wrote...

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

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]

"Black Ice" <sm*******@yahoo.com> wrote...

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

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