Why do variable in an inner function return nan when there is the same variable name at the inner function...

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Why do variable in an inner function return nan when there is the same variable name at the inner function declared after log

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Why do variable in an inner function return nan when there is the same variable name at the inner function declared after log



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6















What's happening here? I get a different result if I declare a variable after console.log in the inner function



I understand that var has a functional scope and inner function can access the variable from their parent






function outer() {
var a = 2;

function inner() {
a++;
console.log(a) //log NaN
var a = 8
}
inner()
}
outer()








function outer() {
var a = 2;

function inner() {
a++;
console.log(a) //log 3
var b = 8
}
inner()
}
outer()





The log returns NaN in the first example and log 3 in the second example










share|improve this question





























    6















    What's happening here? I get a different result if I declare a variable after console.log in the inner function



    I understand that var has a functional scope and inner function can access the variable from their parent






    function outer() {
    var a = 2;

    function inner() {
    a++;
    console.log(a) //log NaN
    var a = 8
    }
    inner()
    }
    outer()








    function outer() {
    var a = 2;

    function inner() {
    a++;
    console.log(a) //log 3
    var b = 8
    }
    inner()
    }
    outer()





    The log returns NaN in the first example and log 3 in the second example










    share|improve this question



























      6












      6








      6


      1






      What's happening here? I get a different result if I declare a variable after console.log in the inner function



      I understand that var has a functional scope and inner function can access the variable from their parent






      function outer() {
      var a = 2;

      function inner() {
      a++;
      console.log(a) //log NaN
      var a = 8
      }
      inner()
      }
      outer()








      function outer() {
      var a = 2;

      function inner() {
      a++;
      console.log(a) //log 3
      var b = 8
      }
      inner()
      }
      outer()





      The log returns NaN in the first example and log 3 in the second example










      share|improve this question
















      What's happening here? I get a different result if I declare a variable after console.log in the inner function



      I understand that var has a functional scope and inner function can access the variable from their parent






      function outer() {
      var a = 2;

      function inner() {
      a++;
      console.log(a) //log NaN
      var a = 8
      }
      inner()
      }
      outer()








      function outer() {
      var a = 2;

      function inner() {
      a++;
      console.log(a) //log 3
      var b = 8
      }
      inner()
      }
      outer()





      The log returns NaN in the first example and log 3 in the second example






      function outer() {
      var a = 2;

      function inner() {
      a++;
      console.log(a) //log NaN
      var a = 8
      }
      inner()
      }
      outer()





      function outer() {
      var a = 2;

      function inner() {
      a++;
      console.log(a) //log NaN
      var a = 8
      }
      inner()
      }
      outer()





      function outer() {
      var a = 2;

      function inner() {
      a++;
      console.log(a) //log 3
      var b = 8
      }
      inner()
      }
      outer()





      function outer() {
      var a = 2;

      function inner() {
      a++;
      console.log(a) //log 3
      var b = 8
      }
      inner()
      }
      outer()






      javascript function






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 34 mins ago









      Nick Parsons

      10.3k2926




      10.3k2926










      asked 43 mins ago









      ClaudeClaude

      426




      426
























          3 Answers
          3






          active

          oldest

          votes


















          9














          This is due to hoisting



          The declaration of a in the inner function is hoisted to the top of the function, overriding the outer function's a, so a is undefined



          undefined++ returns NaN, hence your result.



          Your code is equivalent to:



          function outer() {
          var a=2;

          function inner() {
          var a;
          a++;
          console.log(a); //log NaN
          a = 8;
          }

          inner();
          }

          outer();


          Rewriting your code in this way makes it easy to see what's going on.






          share|improve this answer

































            1














            Because var is hoisted through the function, you're essentially running undefined++ which is NaN. If you remove var a = 8 in inner, the code works as expected:






            function outer() {
            var a = 2;

            function inner() {
            a++;
            console.log(a);
            }
            inner();
            }
            outer();








            share|improve this answer































              -1














              var a=0;
              function outer(){
              a=2;
              function inner(){
              a=a+1;
              console.log(a)
              a = 8
              }
              inner()
              }
              outer()





              share|improve this answer



















              • 3





                How does this piece of code explains the issue? Can you provide an explanation of the code you have posted?

                – Shidersz
                27 mins ago











              • They can’t access the inner function value so we have to defined globally. After globally you can use A value anywhere in the code

                – Darshit Shah
                25 mins ago












              Your Answer






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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              9














              This is due to hoisting



              The declaration of a in the inner function is hoisted to the top of the function, overriding the outer function's a, so a is undefined



              undefined++ returns NaN, hence your result.



              Your code is equivalent to:



              function outer() {
              var a=2;

              function inner() {
              var a;
              a++;
              console.log(a); //log NaN
              a = 8;
              }

              inner();
              }

              outer();


              Rewriting your code in this way makes it easy to see what's going on.






              share|improve this answer






























                9














                This is due to hoisting



                The declaration of a in the inner function is hoisted to the top of the function, overriding the outer function's a, so a is undefined



                undefined++ returns NaN, hence your result.



                Your code is equivalent to:



                function outer() {
                var a=2;

                function inner() {
                var a;
                a++;
                console.log(a); //log NaN
                a = 8;
                }

                inner();
                }

                outer();


                Rewriting your code in this way makes it easy to see what's going on.






                share|improve this answer




























                  9












                  9








                  9







                  This is due to hoisting



                  The declaration of a in the inner function is hoisted to the top of the function, overriding the outer function's a, so a is undefined



                  undefined++ returns NaN, hence your result.



                  Your code is equivalent to:



                  function outer() {
                  var a=2;

                  function inner() {
                  var a;
                  a++;
                  console.log(a); //log NaN
                  a = 8;
                  }

                  inner();
                  }

                  outer();


                  Rewriting your code in this way makes it easy to see what's going on.






                  share|improve this answer















                  This is due to hoisting



                  The declaration of a in the inner function is hoisted to the top of the function, overriding the outer function's a, so a is undefined



                  undefined++ returns NaN, hence your result.



                  Your code is equivalent to:



                  function outer() {
                  var a=2;

                  function inner() {
                  var a;
                  a++;
                  console.log(a); //log NaN
                  a = 8;
                  }

                  inner();
                  }

                  outer();


                  Rewriting your code in this way makes it easy to see what's going on.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 24 mins ago









                  Shidersz

                  9,3112933




                  9,3112933










                  answered 37 mins ago









                  jrojro

                  562113




                  562113

























                      1














                      Because var is hoisted through the function, you're essentially running undefined++ which is NaN. If you remove var a = 8 in inner, the code works as expected:






                      function outer() {
                      var a = 2;

                      function inner() {
                      a++;
                      console.log(a);
                      }
                      inner();
                      }
                      outer();








                      share|improve this answer




























                        1














                        Because var is hoisted through the function, you're essentially running undefined++ which is NaN. If you remove var a = 8 in inner, the code works as expected:






                        function outer() {
                        var a = 2;

                        function inner() {
                        a++;
                        console.log(a);
                        }
                        inner();
                        }
                        outer();








                        share|improve this answer


























                          1












                          1








                          1







                          Because var is hoisted through the function, you're essentially running undefined++ which is NaN. If you remove var a = 8 in inner, the code works as expected:






                          function outer() {
                          var a = 2;

                          function inner() {
                          a++;
                          console.log(a);
                          }
                          inner();
                          }
                          outer();








                          share|improve this answer













                          Because var is hoisted through the function, you're essentially running undefined++ which is NaN. If you remove var a = 8 in inner, the code works as expected:






                          function outer() {
                          var a = 2;

                          function inner() {
                          a++;
                          console.log(a);
                          }
                          inner();
                          }
                          outer();








                          function outer() {
                          var a = 2;

                          function inner() {
                          a++;
                          console.log(a);
                          }
                          inner();
                          }
                          outer();





                          function outer() {
                          var a = 2;

                          function inner() {
                          a++;
                          console.log(a);
                          }
                          inner();
                          }
                          outer();






                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered 34 mins ago









                          Jack BashfordJack Bashford

                          13.8k31848




                          13.8k31848























                              -1














                              var a=0;
                              function outer(){
                              a=2;
                              function inner(){
                              a=a+1;
                              console.log(a)
                              a = 8
                              }
                              inner()
                              }
                              outer()





                              share|improve this answer



















                              • 3





                                How does this piece of code explains the issue? Can you provide an explanation of the code you have posted?

                                – Shidersz
                                27 mins ago











                              • They can’t access the inner function value so we have to defined globally. After globally you can use A value anywhere in the code

                                – Darshit Shah
                                25 mins ago
















                              -1














                              var a=0;
                              function outer(){
                              a=2;
                              function inner(){
                              a=a+1;
                              console.log(a)
                              a = 8
                              }
                              inner()
                              }
                              outer()





                              share|improve this answer



















                              • 3





                                How does this piece of code explains the issue? Can you provide an explanation of the code you have posted?

                                – Shidersz
                                27 mins ago











                              • They can’t access the inner function value so we have to defined globally. After globally you can use A value anywhere in the code

                                – Darshit Shah
                                25 mins ago














                              -1












                              -1








                              -1







                              var a=0;
                              function outer(){
                              a=2;
                              function inner(){
                              a=a+1;
                              console.log(a)
                              a = 8
                              }
                              inner()
                              }
                              outer()





                              share|improve this answer













                              var a=0;
                              function outer(){
                              a=2;
                              function inner(){
                              a=a+1;
                              console.log(a)
                              a = 8
                              }
                              inner()
                              }
                              outer()






                              share|improve this answer












                              share|improve this answer



                              share|improve this answer










                              answered 34 mins ago









                              Darshit ShahDarshit Shah

                              53




                              53








                              • 3





                                How does this piece of code explains the issue? Can you provide an explanation of the code you have posted?

                                – Shidersz
                                27 mins ago











                              • They can’t access the inner function value so we have to defined globally. After globally you can use A value anywhere in the code

                                – Darshit Shah
                                25 mins ago














                              • 3





                                How does this piece of code explains the issue? Can you provide an explanation of the code you have posted?

                                – Shidersz
                                27 mins ago











                              • They can’t access the inner function value so we have to defined globally. After globally you can use A value anywhere in the code

                                – Darshit Shah
                                25 mins ago








                              3




                              3





                              How does this piece of code explains the issue? Can you provide an explanation of the code you have posted?

                              – Shidersz
                              27 mins ago





                              How does this piece of code explains the issue? Can you provide an explanation of the code you have posted?

                              – Shidersz
                              27 mins ago













                              They can’t access the inner function value so we have to defined globally. After globally you can use A value anywhere in the code

                              – Darshit Shah
                              25 mins ago





                              They can’t access the inner function value so we have to defined globally. After globally you can use A value anywhere in the code

                              – Darshit Shah
                              25 mins ago


















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