strings

[alex]

what data type is a line like this:

"adslfkj adsfjk adf"

 

[JKop]

alex posted:

what data type is a line like this:

"adslfkj adsfjk adf"

It's an expression of type char* , which points to the first in an array of
characters.


-JKop

 

[Alf P. Steinbach]

what data type is a line like this:

"adslfkj adsfjk adf"

It is an array of n constant chars. Which in some but not all contexts
decays to a pointer to constant char. And in some but not all contexts,
as backward compatibility with C, decays to pointer to plain char.

 

[alex]

thanks

alex posted:


It's an expression of type char* , which points to the first in an array of
characters.


-JKop

 

[alex]

thanks alf

It is an array of n constant chars. Which in some but not all contexts
decays to a pointer to constant char. And in some but not all contexts,
as backward compatibility with C, decays to pointer to plain char.

--
A: Because it messes up the order in which people normally read text.
Q: Why is top-posting such a bad thing?
A: Top-posting.
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[Jorge Rivera]

what data type is a line like this:

"adslfkj adsfjk adf"

This is also considered a 'string literal'. As well pointed out, you
can use this as a const char* or a char[].

JLR

 

[Rolf Magnus]

alex posted:


It's an expression of type char* , which points to the first in an
array of characters.

No. It's of type array of const char with 19 elements.
You can see the difference if you look at sizeof(char*) vs.
sizeof("adslfkj adsfjk adf").

 

[JKop]

Rolf Magnus posted:

No. It's of type array of const char with 19 elements.
You can see the difference if you look at sizeof(char*) vs.
sizeof("adslfkj adsfjk adf").

Seeing as you are right, look at this.


void Chocolate(char monkey[19])
{
;
}

void Chocolate(char* monkey)
{
;
}


int main(void)
{
Chocolate("adslfkj adsfjk adf");

return 0;
}



If char* and char[19] are two different types, why they hell won't the
above compile! I wish C++ would make it's mind up and determine the type of:

"adslfkj adsfjk adf"



....How does sizeof() work? Does it generate a figure based upon the type of
the object passed to it?


-JKop

 

[Phlip]

void Chocolate(char monkey[19])
{
;
}

void Chocolate(char* monkey)
{
;
}


int main(void)
{
Chocolate("adslfkj adsfjk adf");

return 0;
}

If char* and char[19] are two different types, why they hell won't the
above compile! I wish C++ would make it's mind up and determine the type

of:

They won't compile because an array in a function prototype decays to a
pointer. Then either pointers or arrays, passed in, decay too. But the
compiler sees to ambiguous signatures at lookup time.

C++ is not strictly a typesafe language. But references are more typesafe
than copies, so...

void Chocolate( char (&monkey) [19] )
{
assert(19 == sizeof monkey);
}

 

[Rolf Magnus]

Rolf Magnus posted:

No. It's of type array of const char with 19 elements.
You can see the difference if you look at sizeof(char*) vs.
sizeof("adslfkj adsfjk adf").

Seeing as you are right, look at this.


void Chocolate(char monkey[19])
{
;
}

void Chocolate(char* monkey)
{
;
}


int main(void)
{
Chocolate("adslfkj adsfjk adf");

return 0;
}



If char* and char[19] are two different types, why they hell
won't the above compile!

This is a quirk of the C++ language. In the first function, 'monkey' is
_not_ of type char[19], but rather of type char*.

I wish C++ would make it's mind up and determine the type of:

"adslfkj adsfjk adf"

The type of "adslfkj adsfjk adf" is still const char[19].

...How does sizeof() work? Does it generate a figure based upon the
type of the object passed to it?

It returns the number of storage bytes that the object needs.