NetNews Offline 2/NetNews Offline Volume 2.iso/news/comp/lang/c-part2/13517

Internet Message Format | 1996-08-05 | 1.6 KB

Path: mail2news.demon.co.uk!genesis.demon.co.uk
From: Lawrence Kirby <[e-mail]>
Newsgroups: comp.lang.c
Subject: Re: convert 4 8-bit bytes in a foat
Date: Mon, 08 Apr 96 12:21:00 GMT
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Message-ID: <[e-mail]>
References: <4k5pr7$[e-mail]>
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In article <4k5pr7$[e-mail]>
           [e-mail] "LEONG TUCK WAI" writes:

>Hi,
>
>I have 4 8-bit bytes. If I line them up in order, they actually represent
>the 32-bit float data type version of a decimal point number. So, how do
>I get the compiler to understand that, and print out the float number
>instead of giving me 4 integers?

    float f;
    unsigned char a[sizeof f];

    /* Write your bytes to a by whatever means */

    memcpy(&f, a, sizeof f);

    printf("%f\n", f);

Another possibility is:

    float f;
    unsigned char *ptr = (unsigned char *)&f;

You can then use ptr to write the bytes into f directly. Using unions, or
char instead of unsigned char will typically work but is more likely to
trip up on the letter of the standard. If you are reading from a binary
file you could write:

    fread(&f, sizeof f, 1, fp)

That does assume that byte order and so on in the file are correct since
you don't get a chance to alter them prior to loading into f.

--
-----------------------------------------
Lawrence Kirby | [e-mail]
Wilts, England | [e-mail]
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