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Re: [PATCH] Call memcpy in generic mempcpy

On 02/07/2013 08:54 PM, OndÅej BÃlka wrote:
On Wed, Feb 06, 2013 at 03:19:37PM -0800, Roland McGrath wrote:
On 02/06/13 12:53, Ond$,1 yej B,Amlka wrote:
unsigned long int dstp = (long int) dstpp;
+  memcpy (dstpp, srcpp, len);
+  return (void *) (dstp + len);

There's no need to use an integer type, or to do any casts. Something like this, perhaps?

   char *dstp = dstpp;
   memcpy (destpp, srcpp, len);
   return dstp + len;

It's GNU code and can assume GNU C extensions. So there isn't even a need for a cast. And, it's more efficient to use the return value.

I also renamed dstpp to dest as that distinction is no longer needed.

2013-02-06 OndÅej BÃlka <>

* string/mempcpy.c: Implement by calling memcpy.

diff --git a/string/mempcpy.c b/string/mempcpy.c
index ba4d1c6..e1aad75 100644
--- a/string/mempcpy.c
+++ b/string/mempcpy.c
@@ -20,48 +20,17 @@
     <>.  */

  #include <string.h>
-#include <memcopy.h>
-#include <pagecopy.h>

  #undef mempcpy
  #undef __mempcpy

  void *
-__mempcpy (dstpp, srcpp, len)
-     void *dstpp;
-     const void *srcpp;
+__mempcpy (dest, src, len)
+     void *dest;
+     const void *src;
       size_t len;

Shouldn't this be: __mempcpy (void dest, const void *src, size_t len) ? We're converting the code to get rid of the old prototypes.

 Andreas Jaeger aj@{,} Twitter/Identica: jaegerandi
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