How to convert const char *' to 'CHAR [32]'

[chadsxe]

I have to assing and std::string to a D3DXFONT_DESC font value and I can't figure the conversion.

        D3DXFONT_DESC lf;
        std::string name = "blah";
        lf.FaceName = name.c_str();

This produces

error C2440: '=' : cannot convert from 'const char *' to 'CHAR [32]'

I have boost::lexical_cast avaiable to me but I am not sure what the type is. Or is there a simpler way to do this.

Regards

Chad

[matsp]

You probably don't WANT to do that, what you do want to do is assign an array lt.facename with the content of the string.c_str(). If that is what you want to do, then you should use strcpy() to copy the c_str() to facename.

--
Mats

[chadsxe]

You probably don't WANT to do that, what you do want to do is assign an array lt.facename with the content of the string.c_str(). If that is what you want to do, then you should use strcpy() to copy the c_str() to facename.

--
Mats

Hmm not sure if I am grasping what you are suggesting.

	char str1[] = strcpy(str1, name.c_str());

Obviously this peice of code does not compile. Any suggestions?

Thanks

Chad

[tabstop]

You have to define how many characters are in the str1 array. And you cannot assign to an array (you can initialize it, yes, but strcpy is not initialization).

[Angus]

Try this

strcpy(lf.FaceName, name.c_str());

Or if you want to be optimal about it

strncpy(lf.FaceName, name.c_str(), name.length() + 1);

And if you're like me you'll put

assert(name.length() < 32);

before the str*cpy()

[matsp]

Hmm not sure if I am grasping what you are suggesting.

	char str1[] = strcpy(str1, name.c_str());

Obviously this peice of code does not compile. Any suggestions?

Thanks

Chad

strcpy is a function that copies the second string argument into the first one - it does return the string you have as input, but the syntax of the line you describe isn't meaningull.

Note that in C (and C++) you can not assign a values to an array (other than as a constant initializer) using = - you have to assign each element as a single item - or call a function that does it for you, in this case.

I'm 99% sure that what you want in this case is

strcpy(lf.FaceName, name.c_str());

--
Mats

[chadsxe]

To all poster.

Thanks for the responses.

Both the strcpy and strncpy method work, but is it not possible to use the _s version of these functions?

Also, what is the difference between strcpy and strcpy. Is it just a length parameter?

Regards

Cjad

[chadsxe]

Correct me if I am wrong but this is how you would use the _s versions correct?

strncpy_s(lf.FaceName, 32, name.c_str(), name.length() + 1);
strcpy_s(lf.FaceName, 32, name.c_str());

Now to figure out which one to use.

Regards

Chad

[laserlight]

Also, what is the difference between strcpy and strcpy. Is it just a length parameter?

I think you meant "strcpy and strncpy", in which case the answer is yes. When it comes to avoiding buffer overflow, strncpy is generally safer than strcpy.

Now to figure out which one to use.

For better or for worse, the _s versions are non-standard.

[chadsxe]

I think you meant "strcpy and strncpy", in which case the answer is yes. When it comes to avoiding buffer overflow, strncpy is generally safer than strcpy.


For better or for worse, the _s versions are non-standard.

Yeah I realize this now. There Microsoft platform depenedent. For educational sakes am I correct in my assumption that since lf.FaceName is a CHAR[32] then my destination size parameter of the _s functions would have to be 32?

Regards

Chad

[Elysia]

Yes, but sizeof(lf.FaceName) would work better.
Though the *_s versions are non-standard, they do work better in finding bugs and making your application a tad safer. Plus they are generally easy to write wrappers around.
So it's up to you if you want to use them.
And as for why the first example didn't work... the API function has a buffer that it wants you to fill in with the name of the font. It does not want a pointer to the string (it's an array).
Also, you should never really cast away const, unless absolutely necessary (very few valid reasons there).