Need help with parse error problem

bliss89 · 10-17-2007, 08:52 PM

Hey,
I writing a simple unix program, however, ever time I try to compile my program the same message pops up: "parse error before "printf." I can't figure out what I've done wrong.

Here is my program,

#include <stdio.h>
#include <math.h>
int main()
{
int k, b_k;
b_k =0

printf("\n\n");
printf("Part (a):");
printf("\n\n");
printf(" i b_k=k^3-1 ");
printf("\n____________________ ");
for(k=1;k<=20;k++){
b_k = (k)^3-1
printf(" %2d %6d", k, b_k);}
}

prg.c: In function `main':
prg.c:8: error: parse error before "printf"
prg.c:15: error: parse error before "printf"

Hivemind · 10-17-2007, 09:19 PM

Missing semicolon after

Code:

b_k =0

With an editor that supports automatic indentation like emacs, such things are easy to spot. Also, post code inside code tags next time.

bigrigdriver · 10-17-2007, 09:23 PM

Quote:

b_k =0

missing semicolon

Quote:

b_k = (k)^3-1

missing semicolon

Quote:

printf(" i b_k=(k^3-1 ");

missing right parenthesis around k^3.

Beyond that, the expressions k=k^3-l and k=(k)^3-1 generate warnings. Put k^3 in parentheses and the warning go away.

End result:

Quote:

1 #include <stdio.h>
2 #include <math.h>
3 int main()
4 {
5 int k, b_k;
6 b_k =0;
7
8 printf("\n\n");
9 printf("Part (a):");
10 printf("\n\n");
11 printf(" i b_k=(k^3)-1 ");
12 printf("\n____________________ ");
13 for(k=1;k<=20;k++){
14 b_k = (k^3)-1;
15 printf(" %2d %6d", k, b_k);}
16 }

Only one warning left:

Quote:

prog.c: In function `main':
prog.c:16: warning: control reaches end of non-void function

bliss89 · 10-17-2007, 09:53 PM

I completely missed that, now I feel a bit stupid. Thanks, i guess it's obvious that I'm new at programming. Also, when I executed the program I got something that was crazy. I'll see if I can fix it, if not I'll post it here. Again, thanks for the help.