02-14-2008
*I understood from the beginning. I said they are not truly contiguous in memory.
Why? they are 20 chars long. 20 % 8 == 4 There are 4 bytes between them. Very probably.
The first position in memory after str1 (str1["21"] which does not really exist) is very likely NOT str2[0].
Why only very likely NOT true?
How things are positioned in memory is implementation dependent. So I cannot guarantee where str1 and str2 live in memory because you may be running a non-standard compiler or running in a special unix environment.
In the environments I do know they will not be back-to-front in memory. There will be a gap. Why don't you just modify my code and see?
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LEARN ABOUT SUSE
streqvcmp
streqvcmp(3) Programmer's Manual streqvcmp(3)
NAME
streqvcmp - compare two strings with an equivalence mapping
SYNOPSIS
#include <your-opts.h>
cc [...] -o outfile infile.c -lopts [...]
int streqvcmp(char const* str1, char const* str2);
DESCRIPTION
Using a character mapping, two strings are compared for "equivalence". Each input character is mapped to a comparison character and the
mapped-to characters are compared for the two NUL terminated input strings. This function name is mapped to option_streqvcmp so as to not
conflict with the POSIX name space.
str1 first string
str2 second string
RETURN VALUE
the difference between two differing characters
ERRORS
none checked. Caller responsible for seg faults.
SEE ALSO
The info documentation for the -lopts library.
ao_string_tokenize(3), configFileLoad(3), optionFileLoad(3), optionFindNextValue(3), optionFindValue(3), optionFree(3), optionGetValue(3),
optionLoadLine(3), optionNextValue(3), optionOnlyUsage(3), optionProcess(3), optionRestore(3), optionSaveFile(3), optionSaveState(3),
optionUnloadNested(3), optionVersion(3), pathfind(3), strequate(3), streqvmap(3), strneqvcmp(3), strtransform(3),
2010-07-05 streqvcmp(3)