I write a non-template simple string with the technique
that giving the same string a same storage with a count.
so i must give the necessary copy operation when the string will be
modified, I differ the two kinds of operator[] s.
// const object to use as const
const char& operator[] const
{
//directly get the ref
}
char& operator[]
{
// first copy. then give the new copy 's ref
}
but you see, when I just want to get the value ,not to modify it, a
unnecessary copy ocurrs.
Is there some methods to avoid this ?
Thanks. 3 1619
"belief" <su*********@163.com> schrieb im Newsbeitrag news:d5**********@news.yaako.com... I write a non-template simple string with the technique that giving the same string a same storage with a count. so i must give the necessary copy operation when the string will be modified, I differ the two kinds of operator[] s. // const object to use as const const char& operator[] const { //directly get the ref } char& operator[] { // first copy. then give the new copy 's ref } but you see, when I just want to get the value ,not to modify it, a unnecessary copy ocurrs. Is there some methods to avoid this ? Thanks.
Don't return a reference. Create your own class, which behaves like a reference, knows which string it refers to, and knows when a copy is required. Something like
class char_ref
{
public:
char_ref(string& base, int offset);
char_ref& operator=(char); // assigns to location refered to,
// copies if neccessary.
operator char() const; // get content of location
...
};
Good luck
Heinz
Hi,Heinz.
It's a good solution ! by giving a intermediate layer .
Maybe char_ref will give me more work to implement it's behavior
like inner type char, i.e. operator= operator+= opeartor*= operator/= etc.
As the solution , the necessary copy operation will be separated to two
pieces.
some ones in the String(e,g, insert, replace) ,and others in the char_ref
like operators just talking above, isn't it?
I think i get it .
Thanks a ton.
wisdo (belief).
//--------------------------------------------------------------------------
----------------------------------------------------------------------------
-
"Heinz Ozwirk" <ho**********@arcor.de> ????
news:42***********************@newsread4.arcor-online.net...
"belief" <su*********@163.com> schrieb im Newsbeitrag
news:d5**********@news.yaako.com...
Don't return a reference. Create your own class, which behaves like a
reference, knows which string it refers to, and knows when a copy is
required. Something like
class char_ref
{
public:
char_ref(string& base, int offset);
char_ref& operator=(char); // assigns to location refered to,
// copies if neccessary.
operator char() const; // get content of location
...
};
Good luck
Heinz
belief schreef: I write a non-template simple string with the technique that giving the same string a same storage with a count.
so i must give the necessary copy operation when the string will be modified, I differ the two kinds of operator[] s.
// const object to use as const const char& operator[] const { //directly get the ref }
It's probably more efficient to return a plain char. What's the point
in having a reference? That also prevents the following bug:
SimpleString a = "foo";
SimpleString const& b = a;
char const& f = b[0];
a[0]='x';
assert(f=='f');
char& operator[] { // first copy. then give the new copy 's ref } but you see, when I just want to get the value, not to modify it, a unnecessary copy ocurrs.
References are dumb (like pointers), use a smart ref class.
Regards,
Michiel Salters This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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