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From: Keith Thompson <kst-u@mib.org>
Newsgroups: comp.unix.programmer,comp.lang.c
Subject: Re: Odd compiler behaviour?
Date: Tue, 01 Mar 2016 08:51:46 -0800
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spud@potato.field writes:
> On Tue, 01 Mar 2016 08:36:27 -0800
> Keith Thompson <kst-u@mib.org> wrote:
>>spud@potato.field writes:
>>> The 1 << s actually returns 1 , not zero. The hard coded value
>>> returns zero.  Also this code gives the same result if you enter
>>> 32. So unless gcc is psychic the '1' result is calculated at
>>> runtime, not compile time:
>>
>>It depends on the optimization level.  Experiment shows that gcc
>>generates code to compute `1 << s` at run time if optimization is not
>>enabled, but computes it at compile time with -O1 or higher.
>
> How can it compute it at compile time if it doesn't know what the
> value of 's' is going to be?

It can't, of course.  In the example I was referring to, the value of s
*can* be determined at compile time, and gcc does so with "-O1" or
higher.

>>> int main()
>>> {
>>>         char str[10];
>>>         int s;
>>>         gets(str);
>>>         s = atoi(str);
>>>         printf("%d\n",(int)1 << s);
>>>         return 0;
>>> }
>>
>>Why are you calling gets()?
>
> Why in this example or why at all because its an unsafe function?
>
> Its an easy way to get user input in a simple example to demonstrate that the
> output from printf can't possibly be computed at compile time since the user
> enters the shift value at runtime.

Ok.  Using gets() obscured the point you were making.  It's not merely
unsafe; it's been removed from the language as of the 2011 standard.

-- 
Keith Thompson (The_Other_Keith) kst-u@mib.org  <http://www.ghoti.net/~kst>
Working, but not speaking, for JetHead Development, Inc.
"We must do something.  This is something.  Therefore, we must do this."
    -- Antony Jay and Jonathan Lynn, "Yes Minister"