If block does not get executed on truthy condition. JS

Question

I stumbled upon unexpected behaviour in JS. If I log regexpr.test(inp) , console outputs true, yet if this code is used as condition 'if block' console.log('in if'); is not executed.

If I assign this code to a variable and then use it as a condition 'if block' gets executed as expected.

Tested it in chrome console and node.

var str = '()'

function validateInput (inp) {

  let regexpr = /^(?:\(|\)|\s)+$/gim
  console.log(regexpr.test(inp));   // logs true
  if (regexpr.test(inp)) {          // staments in if are not executed
    console.log('in if');
  }
  console.log('after if');
  return false
}

function validateInput2 (inp) {

  let regexpr = /^(?:\(|\)|\s)+$/gim
  let t1 = regexpr.test(inp);
  console.log(t1);                  // logs true
  if (t1) {                         // staments in if are executed
    console.log('in if');
  }
  console.log('after if');
  return false
}

console.log(validateInput(str));
console.log('------')
console.log(validateInput2(str));

true
after if
false
------
true
in if
after if
false

Don't use the global flag unless you actually need the global flag — CertainPerformance, Jun 29 '19 at 08:04

score -1 · Accepted Answer · answered Jun 29 '19 at 08:10

-1

You're matching the string from start to end - you don't need to use the g (global) flag at all.

let regexpr = /^(?:\(|\)|\s)+$/im;

answered Jun 29 '19 at 08:10

Jack Bashford

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If block does not get executed on truthy condition. JS

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