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up vote 25 down vote favorite
6

How does the below code execute?

if(a=2 && (b=8))
{
    console.log(a)
}

OUTPUT

a=8
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7  
Why would you put b=8 in parens but not a=2? It looks almost as if this is a trolling attempt. –  Niklas B. yesterday
    
When you saw b=, why didn't you bother to check the value of b? That would have given enough clues... –  rvighne yesterday
    
@NiklasB. If I don't put b=8 into brackets then it gives me an error because it executes it someting like a = (2 && b=8). I am not clear about this error but the reason to put b=8 into brackets is only give high precedence to b=8 or you can simply say, to avoid error. –  Jay Shukla 17 hours ago
    
@Jay So you already knew that the code parses as a = (2 && (b=8)). This pretty much would have allowed to figure out the problem. At the very least I'd expect a minimal amount of debugging before posting a question on Stack Overflow –  Niklas B. 6 hours ago
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6 Answers

up vote 32 down vote accepted

It has nothing to do with the if statement, but:

if(a=2 && (b=8))

Here the last one, (b=8), actually returns 8 as assigning always returns the assigned value, so it's the same as writing

a = 2 && 8;

And 2 && 8 returns 8, as 2 is truthy, so it's the same as writing a = 8.

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Perfect. Thank you so much for nice explanation. –  Jay Shukla yesterday
3  
With the side effect that b will also be 8, of course. –  CompuChip yesterday
    
@CompuChip - Yep, the assigment assigns 8 to b, but assigning will return the assigned value as well, so that's correct. –  adeneo yesterday
    
"assigning always returns the assigned value" - didn't know this one, thank you. –  Valdas 16 hours ago
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up vote 16 down vote

It's generally a bad idea to do variable assignment inside of an if statement like that. However, in this particular case you're essentially doing this:

if(a = (2 && (b = 8)));

The (b = 8) part returns 8, so we can rewrite it as:

if(a = (2 && 8))

The && operator returns the value of the right hand side if the left hand side is considered true, so 2 && 8 returns 8, so we can rewrite again as:

if(a = 8)
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is it because anything greater than 0 is considered true? –  staticx yesterday
    
@staticx Any non-zero integer (so positive or negative) is considered true when converted to a boolean. –  Anthony Grist yesterday
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up vote 8 down vote

It is called operator precedence

(a=2 && (b=8))

In the above example. then results are evaluated against the main && sign.

(a=2 && (b=8)) evaluated to a = 2 && 8

So 2 && 8 return a = 8

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up vote 5 down vote

You're setting (not comparing) a to 2 && (b=8). Since 2 is tru-ish the second half of the expression will be executed, i.e. a = true && (b = 8), i.e. a = (b = 8), i.e. a = 8.

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up vote 4 down vote

Your statement is interpreted like

a = (2 && (b=8))

when you uses && statement, then last true statement value will be returned. Here (b=8) will becomes value 8 which is true and last statement.

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up vote 3 down vote

To understand what is going on, refer to the operator precedence and associativity chart. The expression a = 2 && (b = 8) is evaluated as follows:

  • && operator is evaluated before = since it has higher priority
    • the left hand expression 2 is evaluated which is truthy
    • the right hand expression b = 8 is evaluated (b becomes 8 and 8 is returned)
    • 8 is returned
  • a = 8 is evaluated (a becomes 8 and 8 is returned)

Finally, the if clause is tested for 8.

Note that 2 does not play any role in this example. However, if we use some falsy value then the result will be entirely different. In that case a will contain that falsy value and b will remain untouched.

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What makes 2 truthy? –  staticx yesterday
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