NetNews Usenet Archive 1992 #27/NN_1992_27.iso/spool/comp/lang/c/17068

Text File | 1992-11-24 | 3.2 KB | 84 lines

Newsgroups: comp.lang.c
Path: sparky!uunet!math.fu-berlin.de!hamel!rene
From: [e-mail] (Rene Mueller)
Subject: Re: If statement question
Message-ID: <[e-mail]>
Sender: rene@hamel (Rene Mueller)
Organization: Free University of Berlin, Germany
References:  <[e-mail]>
Distribution: na
Date: Mon, 23 Nov 1992 23:16:24 GMT
Lines: 71

In article <[e-mail]>, [e-mail] (G. Lee Owen) writes:
|>
|>
|>     Sorry, took me a day extra to get this out, but here we go...
|>
|> From [e-mail] (Chee K Koo) on 20 Nov 92:
|> CKK> I am new to C programming, while i was doing a numerical analysis
|> CKK> program which uses many for loop
|> CKK> here is the loop looks like
|> CKK> for(i=1;i<=n;i++);
|> CKK> {
|> CKK> }
|> CKK> i put the ";" after the loop header, when the program is excuted, the
|> CKK> loop is skipped, however, it works fine when the ";" is removed.
|>
|>   When you say 'for(i=1;i<=n;i++);' you are really telling the
|> compiler 'for(i=1;i<=n;i++) /* DO NOTHING */;'.  The for loop will
|> work on the (single) statement following it OR on the block following
|> it, where { and } mark off the block.
|>   If you were writing a loop that would execute one statement, it
|> would be like this, right?
|>     for(i=1;i<=n;i++) printf("Hi #%d\n", i);
|> Notice how there is no ; between the end of the loop parens and the
|> printf statement.  Well, the same thing goes for a for loop with a
|> block following it.  Say you had to execute two statements, you would
|> put them in a block:
|>     for(i=1;i<=n;i++)
|>         {
|>         printf("Hi ");
|>         printf("#%d\n", i);
|>     }
|> like such. The output of this would be (n == 5 here)
|> Hi 1
|> Hi 2
|> Hi 3
|> Hi 4
|> Hi 5
|>   Notice that, without the blocks, like this:
|>     for(i=1;i<=n;i++)
|>         printf("Hi ");
|>         printf("#%d\n", i);
|> it will print "Hi " n times and then print "#i" where i has the same
Ok so far...
|> value as n. If n equalled 5 you would get "Hi Hi Hi Hi Hi 5".
That was wrong!
Your condition was i<=n, so i==n will be executed, then i++ and THEN the loop
will be breaked...
That means the output is:
"HI HI HI HI HI 6"!!!

|>
|>     I'm afraid I find this difficult to explain succinctly; I hope
|> these examples illustrate what you are seeing.  Perhaps you should dig
|> in your textbook a bit more...
|>
|> CKK> I just want to find out why the compiler does not catch it during
|> CKK> compilation. It is compiled with UNIX gcc.
|>   Well, frankly, it's not the compiler's job.  More importantly,
|> neither would it always be correct to flag that.  Consider this:
|>   for (i=0; ((i<ARRAY_MAX)&&(array[i]!=42)); i++)  /* DO NOTHING */ ;
|>     printf("The Answer to Life is at array position %d\n", i);
|>   This fragment uses the loop to find the position in the array which
|> holds the value 42; the loop need not execute a statement, and the
|> side effect is that when the loop terminates, i has the value we are
|> interested in.
Sorry, but here is the same mistake...
You must FIRST decrement(increment?? grmbl) SUBSTRACT 1 from i to get the
correct array position:
printf("The Answer to Life is at array position %d\n", --i);

Hope there will be nobody confused to much...