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NNTP-Posting-Date: Tue, 28 Aug 2012 22:47:36 -0500
Message-ID: <503D90D7.E67@mindspring.com>
Date: Tue, 28 Aug 2012 23:47:35 -0400
From: pete <pfiland@mindspring.com>
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Subject: Re: #define BYTES *8
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Xref: csiph.com comp.lang.c:25351

JohnF wrote:
> 
> Anything wrong with that, i.e., with  #define BYTES *8
> to multiply by 8? It looked a little weird to me, but
> the more obvious #define BYTES(x) ((x)*8) isn't what
> I wanted to write. This was to express units of measurement,
> e.g., int bits = so_many BYTES; rather than
> int bits=BYTES(so_many); . I just wanted to read and write
> it the first way, and my test program
>  #define BYTES *8
>  #include <stdio.h>
>  int main ( int argc, char *argv[] ) {
>   int bytes=(argc<2?1:atoi(argv[1])),
>       bits = bytes BYTES;
>   printf("%d bytes = %d bits\n",bytes,bits);
>   return(0); }
> works fine (and compiles with no -pedantic warnings).
> But that #define BYTES *8 still looks a little funky to me.
> I realize 2+2 BYTES must be written (2+2)BYTES, but is there
> any other kind of "gotcha"?

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

#define BITS(X)  ((X) * CHAR_BIT)

int 
main (int argc, char *argv[]) 
{
    unsigned bytes = argc < 2 ? 1 : atoi(argv[1]);
    unsigned bits  = BITS(bytes);

    printf("%u bytes = %u bits\n", bytes, bits);

    return 0; 
}

-- 
pete