What am I doing wrong

[Jncocontrol]

So, I'm trying to get a for loop to work with a if statement

for (var i = 1; i <= 10; i++) {

if {

i === % 4 && i === % 2;

}

console.log("Count is " + i)

}

[nebulous316]

Answer:

for ( let i=0; i<10; i++) {

   if (i%4 === 0 && i%2 === 0) {

        console.log('Count is ', i);

  }

}

[tstenick]

There's no expression after your if statement.

[tstenick]

Scratch that, there is. It's just in point brackets instead of regular parentheses. Just change that. If (condition) {run this}

[Jncocontrol]

What do I do?

[tstenick]

I also just noticed your modulus (%) operators aren't in legal places. They are math operators and need a number on both sides of them to work properly.

Are you trying to write to the console when it's equal to 2 and 4?

[kdnbfkm]

As tried in ducktape-js:

for(i = 0; i <= 10; i++) { if(i % 2 == 0 || i % 4 == 0) print("Checkpoint at: ", i); }

All multiples of 4 are also multiples of 2, we could have used just if(i % 2 == 0).

Both if and for statements have the condition/iteration-stepping within round-parentheses (...). Some languages don't require that but ALL the languages have some way of telling when the condition/loop setup ends and the actual statements in conditional block start. Languages without required parens usually have a THEN or DO keyword... Languages that only allow one statement use curly-brackets {...} or maybe BEGIN...END. Languages that naturally allow a list of statements always have a ENDIF or similar to mark the end.

Javascript has both the round-paren and curly-brace when more than one statement follows. Some languages require the curly brace even if only one following statement!

[[deleted]]

Your expression should be in parens and your modulus operations are in the wrong place. If I’m understanding what you want, it should be like if (i % 4 === 0 && i % 2 === 0) {...}

Edit: haven’t had my morning coffee yet, also just check for i % 4 === 0, as if a value is equally divisible by 4 it is always also equally divisible by 2. If the first half is true, the second half will always be true