Compiler wants ";" - I don't know why (in C) [closed] - c

Question

Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 8 years ago.
Improve this question
Working on a project Euler problem (#10), I've an interesting compile error. My compiler tells me I need a semicolon. The exact error is 49: ';' expected.
Now I know this is not unusual with C code, but the problem is I've triple checked all semicolons and parentheses and brackets, and none are missing. Not even in loops. Chances are I'm just missing it.
HOWEVER
The compiler tells me 49: ';' expected - but the code on line 49 is just one bracket ({).
Would anyone have any insight into what might be wrong here?
Code is as follows:
// this is a program to find the sum of all primes below 2 million
// uses the Seive of Eratosthenes
#include <stdio.h>
int main() {
long int sum = 0;
long int s = 2000000; // size of array
long int i; // 'for' loops
long int n; // number to test
long int a; // prime checker
long int multiples;
long int* array; // define pointer for calloc int array
// calloc the array
array = (int*)calloc(s, sizeof(int)); // allocates 2,000,000 spots (s) of
// size "int" to array "array"
// set all array values equal to 1
for (i = 0; i < 2000000; i++) {
array[i] = 1;
}
/*
VALUES IN ARRAY
The values of the array indicates whether number-
(0) IS PRIME
(1) UNTESTED
(2) IS COMPOSITE
*/
// implement prime finder
for (n = 0; n < 2000000; n++) {
if (array[n] == 0) // IF n IS PRIME
{
multiples = n;
multiples += n;
while (multiples < 2000000) {
array[multiples] = 2;
multiples += n;
}
} else
(array[n] == 2) // IF n IS COMPOSITE (is a multiple)
{ // THIS IS LINE 49
continue;
}
else(array[n] == 1) // UNTESTED
{
for (a = 2; a < n; a++) // double checks for composites
{
// tests factors
if (n % a == 0) {
printf("ERROR");
goto end;
}
}
array[n] = 0;
multiples = n;
multiples += n;
while (multiples < 2000000) {
array[multiples] = 2;
multiples += n;
}
}
}
// read array and sum primes
for (n = 0; n < 2000000; n++) {
if (array[n] == 0) // IF n IS PRIME
{
sum += n;
} else
(array[n] == 2) // IF n IS COMPOSITE
{
continue;
}
else(array[n] == 1) // IF n MAY/MAY NOT BE PRIME
{
printf("ERROR");
goto end;
}
}
printf("The sum of all primes < 2,000,000 is... %ld!\n", sum);
end:
getchar();
return 0;
}

Answer 1

Change the else to else if and you are fine.
And do that with all the else.

Answer 2

else( array[n] == 2) // IF n IS COMPOSITE (is a multiple)
{ //THIS IS LINE 49
else doesn't take a condition. The else should be else if.
The compiler's syntax recovery tries to suggest the simplest change that would result in a syntactically valid program. Often the suggested change doesn't make much sense. In this case, adding a ; after the else would turn the following ( array[n] == 2) into the beginning of an expression statement (which would have to be followed by another semicolon).
If your code produces a syntax error message, the best approach is often to ignore the suggestion it makes and just look at the line it refers to and the line(s) preceding it, and figure out what's wrong with the syntax at that point. Then fix the error and recompile.

Answer 3

You're really missing is some ifs to go with your elses:
if( array[n] == 0) // IF n IS PRIME
{
sum += n;
}
else if( array[n] == 2) // IF n IS COMPOSITE
{
continue;
}
else if( array[n] == 1) // IF n MAY/MAY NOT BE PRIME
{
printf("ERROR");
goto end;
}
It's pretty easy to confuse a compiler like this, and when you do, some of its error messages can get hard to understand.