What are the minimum requirements to get a job in?

[Cameron Simpson]

On 14Dec2012 16:57, Dennis Lee Bieber <wlfraed at ix.netcom.com> wrote:
| On Fri, 14 Dec 2012 12:42:27 +0100, Christian Heimes
| <christian at python.org> declaimed the following in
| gmane.comp.python.general:
| > 
| > To be fair, memcpy() is a pretty simple function. It can be implemented
| > in just about two lines of C code plus five lines of boiler plate. It
| > shows, if you have very basic understanding about memory layout and
| > pointer arithmetics.
| > 
| > You have to translate something like
| > 
| > def memcpy(dest, src, n):
| >     for i in xrange(n):
| >         dest[i] = src[i]
| > 
| > into C code. Here is a rather nonperformance solution. It copies just
| > one byte per cycle. A better implementation could copy 4 or 8 bytes at
| > once and handle the tail in a switch statement.
| > 
| > void *memcpy(void *dest, const void *src, size_t n)
| > {
| >     char *destptr = (char*)dest;
| >     char *srcptr = (char*)src;
| >     while (n--) {
| >        *destptr++ = *srcptr++;
| >     }
| >     return dest;
| > }
| > 
| > destptr and srcptr are the memory addresses of a byte (char). *destptr
| > is the value at a specific memory location. The code in the loop copies
| > the value at address srcptr to the location at destptr and then moves
| > both pointers one step to the right (++). That's all. ;)
| 
| 	That is ignoring the possibility of overlapping source/destination
| ranges (in which one may need to copy from the end rather than the
| start).

If you're going to be picky, memcpy() is not required to allow for that.
That allows a high speed implementation. memmove() exists to cover the
more general case.
-- 
Cameron Simpson <cs at zip.com.au>

NOTWORK: n. A network when it is acting flaky. Origin (?) IBM.
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