Path: utzoo!utgpu!news-server.csri.toronto.edu!cs.utexas.edu!uunet!pdn!palan!ckctpa!crash
From: crash@ckctpa.UUCP (Frank J. Edwards)
Newsgroups: comp.sys.amiga.programmer
Subject: Re:  4 bytes to a long?
Message-ID: <1991Jun14.161152.1421@ckctpa.UUCP>
Date: 14 Jun 91 16:11:52 GMT
References: <55908@nigel.ee.udel.edu> <19488689.ARN0eac@swinjm.UUCP>
Organization: Edwards & Edwards Consulting
Lines: 35

In article <19488689.ARN0eac@swinjm.UUCP> forgeas@swinjm.UUCP (Jean-Michel Forgeas) 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));
>
>To convert four bytes into a long:
>    l = *((long*) ch)
>This will generate only one instruction:
>    move.l _ch,_l

Yes, but watch out for alignment restrictions.  To take care of
that, put the "long l" before the "char ch".  On stack-oriented
machines, *usually* the long will be allocated on the stack first
and then the char array, so the array must be properly aligned.

A better method, though, is to use a union:

	union {
		char ch[4];
		long l;
	} uvar;

Now, put the individual bytes into uvar.ch[0] through uvar.ch[3] and
then read'em back as uvar.l

>Jean-Michel Forgeas
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