Why do these curly brackets make a difference?

[ihazguitar]

This is in VC++.

First, here is the code, it is to check whether a number is prime or not.

#include "stdafx.h"
#include "iostream"
#include "math.h"

using namespace std;

int main() {
	int n;
	int i;
	int is_prime;

	is_prime = true;

	cout << "Input a number and press ENTER: ";
	cin >> n;

	i = 2;
	while (i <= sqrt(static_cast<double>(n))) {
		if (n % i == 0) 
			is_prime = false;
			++i;
		
	}

		if (is_prime) {
			cout << "The number is prime." << endl << endl;
		}
		else {
			cout << "The number is not prime." << endl << endl;
		}

		return 0; 
}

When I add curly brackets to this:

if (n % i == 0) 
	is_prime = false;
	++i;

so it becomes this:

if (n % i == 0) {
	is_prime = false;
        ++i;
}

the code compiles normally but when I execute it the program only works for some numbers, like 4 and 6, but not for others, like 10 and 15.

Why is that?

Thanks.

[Disch]

C++ doesn't care about indentation. It has no way of knowing that the ++i is part of the if block without those braces.

if/while/for/etc only have an effect on the next block. That means up until the next semicolon, or the next set of {braces}.

// "yes" = part of the if block
// "no" = not part of the if block
// remember:  white space / newlines / indentation does not matter at all

if(blah) yes; no;
if(blah) { yes; yes; yes; } no;
if(blah); no; no;

[Ryhnn]

Looks like someone has been coding in Python. C++ defines code blocks with {} it allows the programmer to format they way they find best and gives a clear representation of what happens where.

[ihazguitar]

Thanks, it makes sense now. But now I have an unrelated question, about a modified version of this code.

// Optimized prime1.cpp code.  Checks a number to see whether it is prime or not.

#include "stdafx.h"
#include "iostream"
#include "math.h"
using namespace std;

int main() {
	int n;
	
	// Prompt user for number.
	cout << "Input a number and press ENTER: ";
	cin >> n;
	
	// Initiating variables.
	int i = 2;
	int is_prime = true;
	double rt = sqrt((static_cast<double>( n )));
	
	if ( n == 0 || 1 ) {
		cout << "The number is not prime." << endl << endl;
		goto end1;
	}

	// Test for prime number.
	while ( i <= rt) {
		if ( n % i == 0 ) {
		is_prime = false;
		break;
		}
		i++;
		}
		
		if ( is_prime ) {
		cout << "The number is prime." << endl << endl;
		}
		
		else {
		cout << "The number is not prime." << endl << endl;
		}
		
		end1:

		return 0;
}

In this section:

if ( n == 0 || 1 ) {
		cout << "The number is not prime." << endl << endl;
		goto end1;
	}

I made it work so if the number is a 1 or a zero, the result is not prime. But when i reverse the order of the 1 and zero, like this:

	if ( n == 1 || 0 ) {
		cout << "The number is not prime." << endl << endl;
		goto end1;
	} 

it says the number is prime even if i enter a 0. Does the order matter for the || operator?

[Disch]

|| does not work that way. || examines the left and right side, and if either one of them is true (nonzero), then it results as true.

if(n == 0 || 1)

This will always be true, because '1' is nonzero. The || operator will look at 'n==0', then it will look at '1'. Since one of those things (in this case, the '1') is nonzero, the overall expression results in 'true'.


In this:
if ( n == 1 || 0 )

the '|| 0' is basically ignored because '0' will never be nonzero.


What you want is this:

if( n == 1 || n == 0 )

[ihazguitar]

Yeah, that code worked, thanks.

[S6k9GNh0]

This topic should be sticky'd for how well the question was laid out and for the use of codetags. All of the responses were clear cut and resolving. I haven't seen a post like this ever on this forum, I couldn't leave it alone. Good job everyone. :D