compare string and "" string literal

[lovecreatesbea...]

Is it correct and safe to compare a string object with "", a pair of
quotation marks quoted empty string?If the string object: s = ""; does
s contain a single '\'? Is it better to use std::string::size or
std::string::empty to deal with this condition? Thank you.

string s = "";
if (s == "");
if (s.size == 0);
if (s.empty());

 

[lovecreatesbea...]

Is it correct and safe to compare a string object with "", a pair of
quotation marks quoted empty string?If the string object: s = ""; does
s contain a single '\'? Is it better to use std::string::size or
std::string::empty to deal with this condition? Thank you.

string s = "";
if (s == "");
if (s.size == 0);
if (s.empty());

(sorry for the mistakes in my previous post)

Is it correct and safe to compare a string object with "", a pair of
quotation marks quoted empty string? If the string object: s = ""; does
s contain a single '\0'? Is it better to use std::string::size or
std::string::empty to deal with this condition?

string s = "";
if (s == "");
if (s.size == 0);
if (s.empty());

 

[Michael DOUBEZ]

(e-mail address removed) a ¨¦crit :

Is it correct and safe to compare a string object with "", a pair of
quotation marks quoted empty string?

Yes, the compiler creates a temporary string initialized with "" to
perform the comparison.

If the string object: s = ""; does
s contain a single '\0'?

This is implementation depedant, the only requirement is that the
c_str() method returns a POD char* which is 0 terminated.

Is it better to use std::string::size or
std::string::empty to deal with this condition?

Unless you have a specific reason to do otherwise, the
std::string::empty is the clearer.

string s = "";
if (s == "");

Costs a intermediary object and a call to a comparison function unless
there is some kind of optimisation (?).

if (s.size == 0);

It is to note that std::string::size complexity may be linear in the
container's size; but it should work fine with main string implementations.

if (s.empty());

It may be equivalent to a.size() == 0 but it is expected to run in
amortized constant time; it may be faster if that does count.


Michael

 

[Roland Pibinger]

(e-mail address removed) a ¨¦crit :

Yes, the compiler creates a temporary string initialized with "" to
perform the comparison.

The compiler does not create a temporary string. Some implementations
internally create a temporary string but that is a different question.

Best wishes,
Roland Pibinger

 

[Michael DOUBEZ]

Roland Pibinger a écrit :

The compiler does not create a temporary string. Some implementations
internally create a temporary string but that is a different question.

Which implementation are you talking about, the compiler's or the
library's ?

The standard defines only:
bool std::string:perator==(const string& c1, const string& c2);

So encoutering:
s == "blah"

The compiler matched "blah" as a string through the relevant constructor:
std::string::string( const char* str );

And uses an intermediary object. Now, dependending on the implementation
of operator== and the constructor, the actual instanciation of a
std::string may be avoided by the compiler. Is it what you mean ?

Regards,
Michael

 

[Roland Pibinger]

Roland Pibinger a écrit :

Which implementation are you talking about, the compiler's or the
library's ?

the library

The standard defines only:
bool std::string:perator==(const string& c1, const string& c2);

In 21.2 the C++ Standard defines the following function templates:

template<class charT, class traits, class Allocator>
bool operator==(const basic_string<charT,traits,Allocator>& lhs,
const basic_string<charT,traits,Allocator>& rhs);

template<class charT, class traits, class Allocator>
bool operator==(const charT* lhs,
const basic_string<charT,traits,Allocator>& rhs);

template<class charT, class traits, class Allocator>
bool operator==(const basic_string<charT,traits,Allocator>& lhs,
const charT* rhs);

Implementations of the latter two templates need not create a
temporary string object.

Best regards,
Roland Pibinger

 

[Michael DOUBEZ]

Roland Pibinger a écrit :

the library


In 21.2 the C++ Standard defines the following function templates:

template<class charT, class traits, class Allocator>
bool operator==(const basic_string<charT,traits,Allocator>& lhs,
const basic_string<charT,traits,Allocator>& rhs);

template<class charT, class traits, class Allocator>
bool operator==(const charT* lhs,
const basic_string<charT,traits,Allocator>& rhs);

template<class charT, class traits, class Allocator>
bool operator==(const basic_string<charT,traits,Allocator>& lhs,
const charT* rhs);

Implementations of the latter two templates need not create a
temporary string object.

You are right, I had never seen it.
But then, what about section 21.3.8.2:

template<class charT, class traits, class Allocator>
bool operator==(const basic_string<charT,traits,Allocator>& lhs,
const basic_string<charT,traits,Allocator>& rhs);

1 Returns:lhs.compare(rhs) == 0.

template<class charT, class traits, class Allocator>
bool operator==(const charT* lhs,
const basic_string<charT,traits,Allocator>& rhs);

2 Returns:basic_string<charT,traits,Allocator>(lhs) == rhs.

template<class charT, class traits, class Allocator>
bool operator==(const basic_string<charT,traits,Allocator>& lhs,
const charT* rhs);

3 Returns:lhs == basic_string<charT,traits,Allocator>(rhs).


Michael

 

[Roland Pibinger]

But then, what about section 21.3.8.2:

template<class charT, class traits, class Allocator>
bool operator==(const basic_string<charT,traits,Allocator>& lhs,
const charT* rhs);

3 Returns:lhs == basic_string<charT,traits,Allocator>(rhs).

I'd call that a contradiction in terms.

 

[Ron Natalie]

Is it correct and safe to compare a string object with "", a pair of
quotation marks quoted empty string? If the string object: s = ""; does
s contain a single '\0'? Is it better to use std::string::size or
std::string::empty to deal with this condition?

You must realize that once you have a std::string, the null char has
no special meaning.

However, the std::string member functions that accept parameters of
char* (and no explicit length) effectively do what "strlen" does to
compute the length. Hence "" in the context of these parameters
is zero sized NOT an array of one char of '\0'.

 

[Bo Persson]

I'd call that a contradiction in terms.

Sort of, but not really.

The "Returns:" element of the description prescribes the value returned by
the function/operator, not how the value has to be computed.

An implementor can note that basic_string::compare() is overloaded for const
charT*, and use

lhs.compare(rhs) == 0

instead.


Bo Persson

 

[Roland Pibinger]

Sort of, but not really.

The "Returns:" element of the description prescribes the value returned by
the function/operator, not how the value has to be computed.

Seems to be a plausible interpretation since the C++ Standard (AFAIK)
never prescribes a certain library implementation.

Best regards,
Roland Pibinger