My simple C program isn't compiling

Question

I'm doing a small exercise for my C class and I'm getting difficulties I know shouldn't really be happening as these are supposed to take like 30 minutes max. Here is my program so far:

#include <stdio.h>
#include <stdbool.h>
#define LIMIT 1000000;

bool isPrime( int num ) {
    for ( int factor = 2; factor * factor <= num; factor++ )
      if ( num % factor == 0 )
        return false;

    return true;
  }

int main() {
  for ( int num = 2; num <= LIMIT; num++ ) {
    if ( isPrime( num ) ) {
      printf( num );
    }
  }
  return 0;
}

Here is the error I'm getting:

primes.c: In function “main”:
primes.c:14: error: expected expression before “;” token
primes.c:16: warning: passing argument 1 of “printf” makes pointer from integer without a cast
/usr/include/stdio.h:361: note: expected “const char * restrict” but argument is of type “int”

Answer 1

As @ Inspired said there is an extra semicolon in the LIMIT macro definition, that semicolon will be expanded by the preprocessor making this line

for ( int num = 2; num <= LIMIT; num++ ) {

like this

for ( int num = 2; num <= LIMIT;; num++ ) {
                            /* ^^ 2 semicolons, now the num++ is extra */

but your program has yet another problem

printf(num);

will not work, printf() expects a format string and then the arguments so it should be

printf("%d\n", num);

read this

Answer 2

You have an extra ; in #define LIMIT 1000000; .

When handling #define , compiler just performs a text substitution: it replaces LIMIT with 1000000; . So your for loop looks like

for (int num=2; num < 1000000 ;; num++) 
                              ^^

The second error happens because indeed printf expects the first argument to be a string (format string), not an integer. Eg printf("%d is prime.\\n", num); ( %d is a placeholder for an integer value, and \\n is end-of-line).

Answer 3

No semi-colon after LIMIT define. Processor directives don't get them so it is literally copying "100000;" into the for loop.

First argument of printf should be format string "%d" so printf("%d\\n", num)

Simple stuff you'll get used to (and still mess up when not thinking), but if you are just learning, it looks great. Far better than my first C programs.