| Topic: |
DEVELOP > c-Plus-Plus |
| User: |
"jimjim" |
| Date: |
25 Mar 2006 07:55:02 AM |
| Object: |
lvalue (s) and rvalue (s) |
Hello,
void func( const string& ISIN) {
cout << ISIN << endl;
}
int main() {
char bufIsin[31];
char *charIsin = new char[31];
strcpy(bufIsin, "FIRST");
strcpy(charIsin, "SECOND");
func(bufIsin);
func(charIsin);
}
1. The above compiles fine. In both instances, a char * is passed to the
func( ). On the other hand, func( ) recieves a string &. What is the exact
argument passing process? (i.e is the string(char *) ctr invoked? is this
called type promotion?)
2. If the func( )'s signature is void func(string& ISIN) the code does not
compile. Can you explain to me the meaning of the error, please?
~/MyTests $CC Passing_CharBuf_Char\*_to_func_accepting_string.cc
"Passing_CharBuf_Char*_to_func_accepting_string.cc", line 21: Error: Formal
argument ISIN of type std::basic_string<char,
std::char_traits<char>, std::allocator<char>>& in call to
func(std::basic_string<char, std::char_traits<char>,
std::allocator<char>>&) requires an lvalue.
"Passing_CharBuf_Char*_to_func_accepting_string.cc", line 22: Error: Formal
argument ISIN of type std::basic_string<char, std::char_traits<char>,
std::allocator<char>>& in call to func(std::basic_string<char,
std::char_traits<char>, std::allocator<char>>&) requires an lvalue.
2 Error(s) detected.
TIA
jimjim
.
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| User: "Heinz Ozwirk" |
|
| Title: Re: lvalue (s) and rvalue (s) |
25 Mar 2006 09:32:44 AM |
|
|
"jimjim" <netuser@blueyonder.co.uk> schrieb im Newsbeitrag =
news:WYbVf.41838$wl.15595@text.news.blueyonder.co.uk...
Hello,
=20
void func( const string& ISIN) {
cout << ISIN << endl;
}
=20
int main() {
char bufIsin[31];
char *charIsin =3D new char[31];
strcpy(bufIsin, "FIRST");
strcpy(charIsin, "SECOND");
func(bufIsin);
func(charIsin);
}
=20
1. The above compiles fine. In both instances, a char * is passed to =
the=20
func( ). On the other hand, func( ) recieves a string &. What is the =
exact=20
argument passing process? (i.e is the string(char *) ctr invoked? is =
this=20
called type promotion?)
The type of func's parameter is "const string&", not just "string&". The =
compiler will generate a temporary string object, initialize it with the =
contents of the character array and pass a reference to the temporary =
object to the function.
2. If the func( )'s signature is void func(string& ISIN) the code does =
not=20
compile. Can you explain to me the meaning of the error, please?
A reference to a non-const object cannot be bound to a temporary.
HTH
Heinz
.
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|
| User: "jimjim" |
|
| Title: Re: lvalue (s) and rvalue (s) |
26 Mar 2006 09:56:27 AM |
|
|
The type of func's parameter is "const string&", not just "string&". The
compiler will generate a temporary string
object, initialize it with the contents of the character array and pass a
reference to the temporary object to the function.
By the way, is the temporary object created on the stack?
.
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| User: "jimjim" |
|
| Title: Re: lvalue (s) and rvalue (s) |
26 Mar 2006 08:28:29 AM |
|
|
void func( const string& ISIN) {
cout << ISIN << endl;
}
int main() {
char bufIsin[31];
char *charIsin = new char[31];
strcpy(bufIsin, "FIRST");
strcpy(charIsin, "SECOND");
func(bufIsin);
func(charIsin);
}
2. If the func( )'s signature is void func(string& ISIN) the code does not
compile. Can you explain to me the meaning of the error, please?
A reference to a non-const object cannot be bound to a temporary.
Does the compiler require a const argument because changing the state of a
temporary object is practically without any essence?
What does "func(std::basic_string<char, std::char_traits<char>,
std::allocator<char>>&) requires an **lvalue**" mean?
TIA
.
|
|
|
| User: "Kai-Uwe Bux" |
|
| Title: Re: lvalue (s) and rvalue (s) |
26 Mar 2006 10:31:38 AM |
|
|
jimjim wrote:
void func( const string& ISIN) {
cout << ISIN << endl;
}
int main() {
char bufIsin[31];
char *charIsin = new char[31];
strcpy(bufIsin, "FIRST");
strcpy(charIsin, "SECOND");
func(bufIsin);
func(charIsin);
}
2. If the func( )'s signature is void func(string& ISIN) the code does
not
compile. Can you explain to me the meaning of the error, please?
A reference to a non-const object cannot be bound to a temporary.
Does the compiler require a const argument because changing the state of a
temporary object is practically without any essence?
No. The reasons for that provision are lost to history. In my opinion, it is
just a bug in the language. In my experience, it makes for much hassle
without tangible gain.
Modifying a temporary makes perfect sense: just think of a proxy class; it
undergoes some changes and at the end of its lifetime writes back the
changes to the object from which it was created. Or think of a temporary
stream object:
#include <fstream>
#include <string>
#include <iomanip>
int main ( void ) {
std::fstream ( "/dev/stdout" )
<< std::dec
<< std::string( "hello world!\n" );
}
Now, you may wonder that this compiles as the temporary is clearly modified.
The reason is that the standard does *not* prohibit calling non-const
members on temporaries. That is the reason for std::dec. In comparison, the
following does not compile:
#include <fstream>
#include <string>
#include <iomanip>
int main ( void ) {
std::fstream ( "/dev/stdout" )
<< std::string( "hello world!\n" );
}
If you can find a rational for this, please let me know.
Best
Kai-Uwe Bux
ps.: this is one of my pet peeves. So please excuse the ranting.
.
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| User: "Bo Persson" |
|
| Title: Re: lvalue (s) and rvalue (s) |
26 Mar 2006 11:22:03 AM |
|
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"Kai-Uwe Bux" <jkherciueh@gmx.net> skrev i meddelandet
news:e06fla$6m1$1@murdoch.acc.Virginia.EDU...
jimjim wrote:
Does the compiler require a const argument because changing the
state of a
temporary object is practically without any essence?
No. The reasons for that provision are lost to history. In my
opinion, it is
just a bug in the language. In my experience, it makes for much
hassle
without tangible gain.
But jimjim is right on the original motivation. The C promotion rules
for built in types is the real "bug".
void inc(float& x)
{ ++x; }
int i = 0;
inc(i);
or even
inc(1.0);
What would happen here, if it was allowed? Nothing!
The parameter would be converted to a float temporary, which is then
incremented.
Modifying a temporary makes perfect sense: just think of a proxy
class; it
undergoes some changes and at the end of its lifetime writes back
the
changes to the object from which it was created. Or think of a
temporary
stream object:
#include <fstream>
#include <string>
#include <iomanip>
int main ( void ) {
std::fstream ( "/dev/stdout" )
<< std::dec
<< std::string( "hello world!\n" );
}
On the other hand, *this* is a bug in the language. :-)
Bo Persson
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| User: "Kai-Uwe Bux" |
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| Title: Re: lvalue (s) and rvalue (s) |
26 Mar 2006 12:37:29 PM |
|
|
Bo Persson wrote:
"Kai-Uwe Bux" <jkherciueh@gmx.net> skrev i meddelandet
news:e06fla$6m1$1@murdoch.acc.Virginia.EDU...
jimjim wrote:
Does the compiler require a const argument because changing the
state of a
temporary object is practically without any essence?
No. The reasons for that provision are lost to history. In my
opinion, it is
just a bug in the language. In my experience, it makes for much
hassle
without tangible gain.
But jimjim is right on the original motivation. The C promotion rules
for built in types is the real "bug".
void inc(float& x)
{ ++x; }
int i = 0;
inc(i);
or even
inc(1.0);
What would happen here, if it was allowed? Nothing!
The parameter would be converted to a float temporary, which is then
incremented.
Ok, I agree that inc(i) doing nothing might be surprising. On the other
hand, I would not want it to do the expected thing either: increment i as a
float and then truncate the result (thus overall being just unpredictable
due to rounding).
However, I wonder: is it really the initialization of a non-const reference
from a temporary that causes the problem or the implicit conversion. But I
guess, we agree on that: as you pointed out, the bug lies within the
promotion rules.
[snip]
Best
Kai-Uwe Bux
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| User: "jimjim" |
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| Title: Re: lvalue (s) and rvalue (s) |
25 Mar 2006 08:00:45 AM |
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P.S: what happens if the char * is not null terminated and the string(const
char *) ctr is invoked? Does string dim = "hello"; ensure that the "hello"
is terminated?
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