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From: johnhlee@CS.Cornell.EDU (John H. Lee)
Newsgroups: comp.sys.amiga.programmer
Subject: Re:   4 bytes to a long?
Message-ID: <1991Jun11.214051.16596@cs.cornell.edu>
Date: 11 Jun 91 21:40:51 GMT
References: <55908@nigel.ee.udel.edu> <mykes.3486@amiga0.SF-Bay.ORG>
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In article <mykes.3486@amiga0.SF-Bay.ORG> mykes@amiga0.SF-Bay.ORG (Mike Schwartz) writes:
>In article <55908@nigel.ee.udel.edu> jleonard@pica.army.mil writes:
>>mrimages@beach.gal.utexas.edu in <451.28517f01@beach.gal.utexas.edu> write:
>>		Is there a fast, simple way to convert 4 bytes into a long?
>>	
>>I use :
>>
>>char ch[4];
>>long l;
>>
>>memcpy( (char *)&l, ch, sizeof(long));
>
>If portability isn't a concern:
>
>	ULONG	l;
>	char	*ch = &l;
>
>ch[0] is MSB of l, ch[3] is LSB.  This is dependant on byte order of ULONG.

A little clarification is in order.  When you obtain the 4 bytes in the
first place (from for instance, fgetc()), place in the sizeof(long)-byte
buffer pointed to by "ch".  You can then access the long through "l" without
copying.  However, if the 4 bytes are say, in the middle of some other
buffer, you must still do a copy of the sizeof(long) bytes to ensure they
end up properly long-word aligned.

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	John Lee		Internet: johnhlee@cs.cornell.edu
The above opinions are those of the user, and not of this machine.