C / C++ / MFC

29-Jun-05 3:28

how do you find the number of bytes in a multibyte string?

_mbslen and _mbstrlen return the number of characters in the string, according to: http://msdn.microsoft.com/library/default.asp?url=/library/en-us/vccore98/html/_crt_strlen.2c_.wcslen.2c_._mbslen.2c_._mbstrlen.asp[^]

With _MBCS defined, size of TCHAR is 1 byte, so you cant even multiply _tcslen with this.

RChin29-Jun-05 4:40

RChin

29-Jun-05 4:40

Flace wrote:
With _MBCS defined, size of TCHAR is 1 byte, so you cant even multiply _tcslen with this

I'm confused by this statement. But the number of occupied bytes in your string will be :

<br />
int nBytes = (_tcslen(pszString) * sizeof(TCHAR));<br />

The sizeof() will return the number of bytes occupied per character. This will obviously vary depending if you are working in single byte or double byte character mode.

I Dream of Absolute Zero

Jose Lamas Rios29-Jun-05 4:41

Jose Lamas Rios

29-Jun-05 4:41

If you know you have a multibyte string, you can simply use strlen() to obtain the number of bytes.

void SomeFunction(LPSTR pString)
{
	int nChars = _mbslen(pString);
	int nBytes = strlen(pString);
	[...]
}

If you are using the TCHAR macros you can do it like this:

void SomFunction(LPTSTR pString)
{
	int nChars = _tcsclen(pString);
	int nBytes = _tcslen(pString) * sizeof(TCHAR);
}

Note _tcslen is different than _tcsclen.

If _UNICODE is not defined, _tcslen resolves to strlen, and sizeof(TCHAR)==1.
If _UNICODE is defined, _tcslen resolves to wcslen, and sizeof(TCHAR)==2.

So, _tcslen(pString) * sizeof(TCHAR) will give you the number of bytes in any case.

Hope that helps,

--
jlr
http://jlamas.blogspot.com/[^]

Chintoo72329-Jun-05 4:54

29-Jun-05 4:54

Jose Lamas Rios wrote:
If you know you have a multibyte string, you can simply use strlen() to obtain the number of bytes.

can you not have a NULL in the middle of a multi-byte string?

Jose Lamas Rios wrote:
If _UNICODE is not defined, _tcslen resolves to strlen, and sizeof(TCHAR)==1.
If _UNICODE is defined, _tcslen resolves to wcslen, and sizeof(TCHAR)==2.
So, _tcslen(pString) * sizeof(TCHAR) will give you the number of bytes in any case.

If _UNICODE is not defined, and _MBCS is defined, _tcslen resolves to _mbslen. And you know, _MBCS is defined by default in the project settings. So, _tcslen(pString) * sizeof(TCHAR) will not always give you the number of bytes all the time...

Jose Lamas Rios29-Jun-05 5:04

Jose Lamas Rios

29-Jun-05 5:04

Flace wrote:
can you not have a NULL in the middle of a multi-byte string?

No, you can't. That's why I say strlen() works fine in this case.

Flace wrote:
If _UNICODE is not defined, and _MBCS is defined, _tcslen resolves to _mbslen

Not really. If _UNICODE is not defined, and _MBCS is defined, _tcsclen does resolve to _mbslen, but _tcslen still resolves to strlen. You can check it here:

http://msdn.microsoft.com/library/default.asp?url=/library/en-us/vclib/html/_crt_strlen.2c_.wcslen.2c_._mbslen.2c_._mbstrlen.asp[^]

--
jlr
http://jlamas.blogspot.com/[^]

Andrew Kirillov29-Jun-05 4:58

Andrew Kirillov

29-Jun-05 4:58

Two above posts are correct, except only the one thing. As MSDN says :
"these functions returns the number of characters in string, excluding the terminal NULL."

So, I suppose, the correct answer on the question will be
(_tcslen(str) + 1) * sizeof(TCHAR)

Andrew

any limitations to VTable ???

Jose Lamas Rios29-Jun-05 5:07

Jose Lamas Rios

29-Jun-05 5:07

You are right, of course. I stand corrected.

--
jlr
http://jlamas.blogspot.com/[^]

Abhi Lahare29-Jun-05 3:24

Abhi Lahare

29-Jun-05 3:24

Hi all,

Can I make a variable argrument function like foofunc(char*, const char *,...); as a virtual function in C++ class.?.

Thanks in Advance
Abhi Lahare

C^hris L^osinger29-Jun-05 3:33

29-Jun-05 3:33

yes , you can make a virtual variable argrument function.

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Chintoo72329-Jun-05 3:41

29-Jun-05 3:41

I dont have a compiler to test this right now, but c++ allow the following two to be present simultaneously? My guess is No. How about the first one in the base class and the second on in the derived class? Myy guess again is no, but this time I am less confident. Can someone clarify with reasons?

foofunc(char*, const char *,...);
foofunc(char*, const char *, int);

Chintoo72329-Jun-05 3:42

29-Jun-05 3:42

insert a 'does' there - read it as "...,but does c++ allow..."

C^hris L^osinger29-Jun-05 3:48

29-Jun-05 3:48

Flace wrote:
My guess is No

why?

this works exactly as expected:

class foo
{
public:
   virtual void doo(int i, ...)
   {
      TRACE("foodoo\n");
   }
} ;

class voo : public foo
{
public:
   virtual void doo(int i, int x)
   {
      TRACE("voodoo\n");
   }
};

...

   foo f;
   f.doo(0, 0);

   voo v;
   v.doo(0, 0);

foodoo
voodoo

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Chintoo72329-Jun-05 3:55

29-Jun-05 3:55

I see. How about this:
foo *f = new doo;
f->doo();

Chintoo72329-Jun-05 3:56

29-Jun-05 3:56

eeek...I meant foo *f= new voo;

C^hris L^osinger29-Jun-05 4:03

29-Jun-05 4:03

foo *f = new voo;
f->doo(3,3);

gives

voodoo

as you would expect

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Chintoo72329-Jun-05 4:18

29-Jun-05 4:18

The first one is no surprise as the compiler will pick the method of class on which you are calling the function. The second one is interesting, but it too is not a surprise to me as you have defined doo as virtual in the base class. Can you do a couple more tests:

1. Do not define doo as virtual in foo, and try this:

<br />
foo *f = new voo;<br />
f->doo(3,3); <br />

2. what if both the prototypes are in the same class, which one does the compiler pick:

<br />
class foo {<br />
   void doo(int i, ...) {<br />
       TRACE ("int i, ...");<br />
   }<br />
   void doo(int i, int j) {<br />
       TRACE("int i, int j");<br />
   }   <br />
}<br />
foo f;<br />
f.doo(3, 3);<br />

C^hris L^osinger29-Jun-05 4:23

29-Jun-05 4:23

1. foodoo
2. int i, int j

why do you think this would behave any differently than a function that doesn't have a vararg parameter ?

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Chintoo72329-Jun-05 4:34