[Webkit-unassigned] [Bug 14124] [CURL] Support data URLs

[bugzilla-daemon at webkit.org]

http://bugs.webkit.org/show_bug.cgi?id=14124





------- Comment #5 from andersca at apple.com  2007-10-26 16:30 PDT -------
(In reply to comment #4)
> (In reply to comment #3)
> > (From update of attachment 16876 [edit] [edit])
> > >     unsigned len = in.size();
> > >     const char* data = in.data();
> > >
> > 
> > Can these just be folded into the new base64Decode call now?
> 
> Like this?
> 
>     return base64Decode(in.data(), in.size(), out);
> 
> Sure.

Good.

> 
> > 
> > >+    return base64Decode(data, len, out);
> > >+}
> > >+
> > >+bool base64Decode(const char* data, unsigned len, Vector<char>& out)
> > >+{
> > >+    out.clear();
> > >+    if (len == 0)
> > >+        return true;
> > >+
> > >+    // If the input string is pathologically large, just return nothing.
> > >+    if (len > UINT_MAX)
> > >+        return false;
> > >+
> > 
> > Why is this change needed?
> 
> These checks were in the original base64Decode(). The first is a base case and
> seems to still be necessary, but the second is probably redundant.
>

Are they still in the method that takes a vector? If so, can they be removed
from there?

> > 
> > >     while (len && data[len-1] == '=')
> > >         --len;
> > > 
> > 
> > >+static void parseDataUrl(ResourceHandle* job)
> > 
> > Please name the variable "handle" instead of "job".
> 
> OK.
> 
> > 
> > >+{
> > >+    DeprecatedString data = job->request().url().url().latin1();
> > 
> > Why do you need to call latin1() here? Also, we recommend against using
> > DeprecatedString in newly written code.
> 
> Its features were closest to the original code (eg. DeprecatedString::mid()).
> KURL's KURL::decode_string() also expects and returns DeprecatedString, which
> we call for percent-encoded data.
>

String::substring is the equivalent of DeprecatedString::mid. 


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