06-29-2008
Thanks jim ..
i tried this , it is showing access vilolation error at the memcpy line . I tried with debugging . I can able to find the error only at memcpy line . Can you please help me to correct this.
It is compiling , but the error is comming at runtime
Thanks,
Arun
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MEMORY(2) System Calls Manual MEMORY(2)
NAME
memccpy, memchr, memcmp, memcpy, memmove, memset - memory operations
SYNOPSIS
#include <u.h>
#include <libc.h>
void* memccpy(void *s1, void *s2, int c, long n)
void* memchr(void *s, int c, long n)
int memcmp(void *s1, void *s2, long n)
void* memcpy(void *s1, void *s2, long n)
void* memmove(void *s1, void *s2, long n)
void* memset(void *s, int c, long n)
DESCRIPTION
These functions operate efficiently on memory areas (arrays of bytes bounded by a count, not terminated by a zero byte). They do not check
for the overflow of any receiving memory area.
Memccpy copies bytes from memory area s2 into s1, stopping after the first occurrence of byte c has been copied, or after n bytes have been
copied, whichever comes first. It returns a pointer to the byte after the copy of c in s1, or zero if c was not found in the first n bytes
of s2.
Memchr returns a pointer to the first occurrence of byte c in the first n bytes of memory area s, or zero if c does not occur.
Memcmp compares its arguments, looking at the first n bytes only, and returns an integer less than, equal to, or greater than 0, according
as s1 is lexicographically less than, equal to, or greater than s2. The comparison is bytewise unsigned.
Memcpy copies n bytes from memory area s2 to s1. It returns s1.
Memmove works like memcpy, except that it is guaranteed to work if s1 and s2 overlap.
Memset sets the first n bytes in memory area s to the value of byte c. It returns s.
SOURCE
All these routines have portable C implementations in /sys/src/libc/port. Most also have machine-dependent assembly language implementa-
tions in /sys/src/libc/$objtype.
SEE ALSO
strcat(2)
BUGS
ANSI C does not require memcpy to handle overlapping source and destination; on Plan 9, it does, so memmove and memcpy behave identically.
If memcpy and memmove are handed a negative count, they abort.
MEMORY(2)