Re: Convert loop and if constructs from C to Perl

Would a kind monk explain what rules C uses to terminate a block

If there's a block, then the rules are the same as in Perl. But if there's no block, then it's a bit more complicated.

Basically, C expects to see either a statement, or a block. A C statement is either something straightfoward (typically an expression, but it could even be nothing) that ends in a semicolon, or it could be a if or a for control structure, or other similar control structures such those with if/else and while. If those use a block for the body, then they don't require a semicolon.

So, if nested braceless control structures end, they all end in the same place!

I'll not fix your code, but I'll comment it.

01 sub standarization { # was standarization(void)
02   my(i,j,k);
03   for (i=1; i<=10; i++) # where does "for" end?
04     if (d[i][0]!=0)     # where does "if" end?
05       for (k=1; k<24; k++)
06       {
07         d[0][k] += d[i][k];
08         d[i][k] /= d[i][0];
09       }  
         # end of "for" statement on line 05
       # end of "if" on line 04
    # end of "for" on line 03
10
11   for (i=9; i<=18; i++) # where does "for" end?
12     if (psb[i][0]!=0)   # where does "if" end?
13       for (k=11; k<=BUST; k++)
14       { # new
15         psb[i][k] /= psb[i][0];
16       } # new
       # end of "if" on line 12
    # end of "for" on line 11
17
18   for (j=1; j<=10; j++) # where does "for" end?
19     if (s[j][0][0]!=0) {
20       for (i=12; i<=18; i++) {
21         if (s[j][i][0]!=0) # where does "if" end?
22           for (k=12; k<=BUST; k++)
23           { # new
24             s[j][i][k] /= s[j][i][0];
25           } # new
           # end of "if" on line 21
26
27         if (s[j+10][i][0]!=0) # where does "if" end?
28           for (k=12; k<=BUST; k++)
29           { # new
30             s[j+10][i][k] /= s[j+10][i][0];
31           } # new
           # end of "if" on line 27
32         s[j][i][0]    /= s[j][0][0];
33         s[j][i][1]    /= s[j][0][0];
34         s[j+10][i][0] /= s[j][0][0];
35         s[j+10][i][1] /= s[j][0][0];
36       } # end "for" line 20
37
38       for (k=17; k<=BUST; k++) {
39         s[j][0][k]    /= s[j][0][0];
40         s[j+10][0][k] /= s[j][0][0];
41       }
42     } # end of "if" line 19
43 } # end of sub
[download]

You'll probably agree by now, that Perl's enforcement of use of braces for control structures, was a sane decision!

Comment on Re: Convert loop and if constructs from C to Perl Select or Download Code

Replies are listed 'Best First'.
Re^2: Convert loop and if constructs from C to Perl by syphilis (Archbishop) on Dec 16, 2006 at 10:33 UTC
You'll probably agree by now, that Perl's enforcement of use of braces for control structures, was a sane decision! That was definitely a sane decision ... but I still find it hard to accept afterthought conditions (aka "modifiers") as valid constructs. Yeah ... I know, I know ... it's just me :-) (I think in terms of "if condition, do" rather than "do, if condition" .... and nothing's gunna change that :-) Cheers, Rob	[reply]
Re^3: Convert loop and if constructs from C to Perl by Anonymous Monk on Dec 16, 2006 at 11:10 UTC
Maybe penicillin will change your mind? /duck	[reply]
Re^4: Convert loop and if constructs from C to Perl by syphilis (Archbishop) on Dec 16, 2006 at 11:26 UTC
Maybe penicillin will change your mind? /duck Heh ... DigitalKitty (or was it one of those other spunky perlmonk chicks ?) made a similar remark on the CB a while back. As I explained then, I've tried smoking penicillin ... and it didn't help. (Tasted bloody awful, too :-) Cheers, Rob	[reply]
Re^2: Convert loop and if constructs from C to Perl by Anonymous Monk on Dec 16, 2006 at 10:06 UTC
Got it! Thanks so much for explaining the rules along with your code comments. Fortunately for me, the other subs I need to convert follow a similar and consistent coding style. I'll certainly apply the rules And, I learned a little more about C and a lot about the wisdom behind Perl. Best regards, Bernie	[reply]