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From:	mambuhl AT ripco DOT com (Martin Ambuhl)
Subject:	Re: [Q] Question about clock()
Sender:	usenet AT rci DOT ripco DOT com (Net News Admin)
Cc:	jim AT sun3 DOT gl DOT rhbnc DOT ac DOT uk (Jim Hu)
Organization:	Ripco Internet BBS Chicago
Date:	Tue, 4 Jul 1995 22:36:04 GMT
Lines:	37
To:	djgpp AT sun DOT soe DOT clarkson DOT edu
Dj-Gateway:	from newsgroup comp.os.msdos.djgpp

jim AT sun3 DOT gl DOT rhbnc DOT ac DOT uk (Jim Hu)
in <3tb4u3$5cl AT sun DOT rhbnc DOT ac DOT uk> asks:


>   I compiled the program below with both Borland C 4.0 and djgpp (v1.12) with
>no flag at all. The size generated by bcc is larger and run slower compared
>with one generated by djgpp. However, the values printed out by the excutable
>from djgpp is much larger and suggesting the excutable is slower. I was cheated
>by clock(). Any comments?

[code snipped]

>output of clock()

>  bcc    Start 0  End 4840
>  djgpp  Start 0  End 384517


>  Why the excutable generated by bcc gives small number but actually takes
>much (much) longer time than one generated by djgpp. Does the small values
>actually mean the value is wrapped around ?

The return value from clock() must be converted to seconds by division by
CLOCKS_PER_SEC

bcc:
#define CLOCKS_PER_SEC 18.2
    4840 / 18.2 = 265.93 sec

djgpp:
#define CLOCKS_PER_SEC 1000000
    384517 / 1000000 = 0.38 sec

                                                                  
--
* Martin Ambuhl       net: mambuhl AT ripco DOT com
* Chicago, IL (USA)        martin DOT ambuhl AT chessboard DOT com

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