Much frustrated gnashing from undefined function return type

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-1

My experience with recursive calls is not deep or long, but I thought this problem would make for a good evaluation. Most of it appears to work, but the function returns undefined:

const target = 1234; // count the number of digits



function numDigits(num, count=0){

     let div10 = count;

     div10 += 1;



     if(Math.floor(num / 10) < 1){

          console.log(`Entered the end and div10 is ${div10}`);

          return div10; // returning undefined

     }



     let curr = Math.floor(num / 10);

     console.log(`Bottom of call and dividing ${curr}`);



     numDigits(curr, div10);



}



numDigits(target);

asked Jan 3 at 3:25

jobechoi

295

add a comment |

-1

My experience with recursive calls is not deep or long, but I thought this problem would make for a good evaluation. Most of it appears to work, but the function returns undefined:

const target = 1234; // count the number of digits



function numDigits(num, count=0){

     let div10 = count;

     div10 += 1;



     if(Math.floor(num / 10) < 1){

          console.log(`Entered the end and div10 is ${div10}`);

          return div10; // returning undefined

     }



     let curr = Math.floor(num / 10);

     console.log(`Bottom of call and dividing ${curr}`);



     numDigits(curr, div10);



}



numDigits(target);

asked Jan 3 at 3:25

jobechoi

295

add a comment |

-1

My experience with recursive calls is not deep or long, but I thought this problem would make for a good evaluation. Most of it appears to work, but the function returns undefined:

const target = 1234; // count the number of digits



function numDigits(num, count=0){

     let div10 = count;

     div10 += 1;



     if(Math.floor(num / 10) < 1){

          console.log(`Entered the end and div10 is ${div10}`);

          return div10; // returning undefined

     }



     let curr = Math.floor(num / 10);

     console.log(`Bottom of call and dividing ${curr}`);



     numDigits(curr, div10);



}



numDigits(target);

asked Jan 3 at 3:25

jobechoi

295

My experience with recursive calls is not deep or long, but I thought this problem would make for a good evaluation. Most of it appears to work, but the function returns undefined:

const target = 1234; // count the number of digits



function numDigits(num, count=0){

     let div10 = count;

     div10 += 1;



     if(Math.floor(num / 10) < 1){

          console.log(`Entered the end and div10 is ${div10}`);

          return div10; // returning undefined

     }



     let curr = Math.floor(num / 10);

     console.log(`Bottom of call and dividing ${curr}`);



     numDigits(curr, div10);



}



numDigits(target);

javascript recursion undefined

asked Jan 3 at 3:25

jobechoi

295

asked Jan 3 at 3:25

jobechoi

295

asked Jan 3 at 3:25

jobechoi

295

asked Jan 3 at 3:25

jobechoi

295

asked Jan 3 at 3:25

jobechoi

295

add a comment |

2 Answers
2

active

oldest

votes

You miss a return when calling recursively numDigits:

const target = 1234; // count the number of digits



function numDigits(num, count=0){

     let div10 = count;

     div10 += 1;



     if(Math.floor(num / 10) < 1){

          console.log(`Entered the end and div10 is ${div10}`);

          return div10; // returning undefined

     }



     let curr = Math.floor(num / 10);

     console.log(`Bottom of call and dividing ${curr}`);



     return numDigits(curr, div10); // HERE



}



console.log(numDigits(target));

answered Jan 3 at 3:30

quirimmo

7,72811536

Thanks! Not sure I understand what is happening, as up until the 3rd recursive call, things appeared to work, and that was without the return. Can you elaborate on why that made it work?

– jobechoi
Jan 3 at 3:34

calls heppen, they do. But if you don't execute return as I added, then your func will not return anything. Does it make sense now?

– quirimmo
Jan 3 at 3:35

return isn't called (it's not a function), it's executed. ;-)

– RobG
Jan 3 at 3:38

@RobG fair enough :D thanks, edited :)

– quirimmo
Jan 3 at 3:38

I sort of get it: the recursive call at some point has to return something. In my initial code, that wasn't happening. Thanks.

– jobechoi
Jan 3 at 3:42

|
show 1 more comment

Recursion is a functional heritage and so using it with functional style yields the best results. There's no "missing return"; there's little need for return in functional style as it is a side effect, it does not produce a value, it cannot be called like a function, and it cannot be combined with other expressions. Other imperative style statements like the reassignment/mutation div10 += 1 are also a recipe for a migraine when used recursively.

Your program can be dramatically simplified –

const numDigits = n =>

  n < 10

    ? 1

    : 1 + numDigits (Math .floor (n / 10))

        

console .log

  ( numDigits (1)         // 1

  , numDigits (22)        // 2

  , numDigits (333)       // 3

  , numDigits (999999999) // 9

  )

The count variable can still be used if you prefer. This would be a practical step toward making the program stack-safe –

const numDigits = (n, count = 1) =>

  n < 10

    ? count

    : numDigits

        ( Math .floor (n / 10)

        , count + 1

        )

        

console .log

  ( numDigits (1)         // 1

  , numDigits (22)        // 2

  , numDigits (333)       // 3

  , numDigits (999999999) // 9

  )

edited Jan 3 at 7:12

answered Jan 3 at 6:59

user633183

72k21143184

"There's no "missing return"" might be a reasonable statement if you hadn't added an implicit return through use of an arrow function. There is indeed a missing return statement, your refactored code is effectively if (n < 10) {return count} else {return numDigits(Math.floor (n / 10), count + 1)}. Removing the redundant else but keeping the return statement in the block is effectively identical to the OP's code. QED.

– RobG
Jan 3 at 8:52

The ternary is simple and succinct. As stated: new to recursion. I will keep this model in mind now. I see that my departure happened much earlier, even before the code was written: I never thought to check whether the integer was less than 10. Thanks, user633183.

– jobechoi
Jan 3 at 11:25

@RobG the difference lies between statements and expressions – statements like if, else, foreach, switch, while, and return are side effects, do not produce a value, and cannot be composed with other statements. Worse, if allows you to write conditionals without an else branch – ternary expressions using ?: evaluate to a value, can be composed with other expressions, and cannot be written without an "else" branch. It looks like I never got around to finishing this answer but I think you will find it useful.

– user633183
Jan 3 at 21:47

@RobG and for more about discussion about statements vs. expressions, I wish to share this. Any further discussion is welcomed and encouraged. The topics here are among my favourite :D

– user633183
Jan 3 at 22:11

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

You miss a return when calling recursively numDigits:

const target = 1234; // count the number of digits



function numDigits(num, count=0){

     let div10 = count;

     div10 += 1;



     if(Math.floor(num / 10) < 1){

          console.log(`Entered the end and div10 is ${div10}`);

          return div10; // returning undefined

     }



     let curr = Math.floor(num / 10);

     console.log(`Bottom of call and dividing ${curr}`);



     return numDigits(curr, div10); // HERE



}



console.log(numDigits(target));

answered Jan 3 at 3:30

quirimmo

7,72811536

Thanks! Not sure I understand what is happening, as up until the 3rd recursive call, things appeared to work, and that was without the return. Can you elaborate on why that made it work?

– jobechoi
Jan 3 at 3:34

calls heppen, they do. But if you don't execute return as I added, then your func will not return anything. Does it make sense now?

– quirimmo
Jan 3 at 3:35

return isn't called (it's not a function), it's executed. ;-)

– RobG
Jan 3 at 3:38

@RobG fair enough :D thanks, edited :)

– quirimmo
Jan 3 at 3:38

I sort of get it: the recursive call at some point has to return something. In my initial code, that wasn't happening. Thanks.

– jobechoi
Jan 3 at 3:42

|
show 1 more comment

You miss a return when calling recursively numDigits:

const target = 1234; // count the number of digits



function numDigits(num, count=0){

     let div10 = count;

     div10 += 1;



     if(Math.floor(num / 10) < 1){

          console.log(`Entered the end and div10 is ${div10}`);

          return div10; // returning undefined

     }



     let curr = Math.floor(num / 10);

     console.log(`Bottom of call and dividing ${curr}`);



     return numDigits(curr, div10); // HERE



}



console.log(numDigits(target));

answered Jan 3 at 3:30

quirimmo

7,72811536

Thanks! Not sure I understand what is happening, as up until the 3rd recursive call, things appeared to work, and that was without the return. Can you elaborate on why that made it work?

– jobechoi
Jan 3 at 3:34

calls heppen, they do. But if you don't execute return as I added, then your func will not return anything. Does it make sense now?

– quirimmo
Jan 3 at 3:35

return isn't called (it's not a function), it's executed. ;-)

– RobG
Jan 3 at 3:38

@RobG fair enough :D thanks, edited :)

– quirimmo
Jan 3 at 3:38

I sort of get it: the recursive call at some point has to return something. In my initial code, that wasn't happening. Thanks.

– jobechoi
Jan 3 at 3:42

|
show 1 more comment

You miss a return when calling recursively numDigits:

const target = 1234; // count the number of digits



function numDigits(num, count=0){

     let div10 = count;

     div10 += 1;



     if(Math.floor(num / 10) < 1){

          console.log(`Entered the end and div10 is ${div10}`);

          return div10; // returning undefined

     }



     let curr = Math.floor(num / 10);

     console.log(`Bottom of call and dividing ${curr}`);



     return numDigits(curr, div10); // HERE



}



console.log(numDigits(target));

answered Jan 3 at 3:30

quirimmo

7,72811536

You miss a return when calling recursively numDigits:

const target = 1234; // count the number of digits



function numDigits(num, count=0){

     let div10 = count;

     div10 += 1;



     if(Math.floor(num / 10) < 1){

          console.log(`Entered the end and div10 is ${div10}`);

          return div10; // returning undefined

     }



     let curr = Math.floor(num / 10);

     console.log(`Bottom of call and dividing ${curr}`);



     return numDigits(curr, div10); // HERE



}



console.log(numDigits(target));

const target = 1234; // count the number of digits



function numDigits(num, count=0){

     let div10 = count;

     div10 += 1;



     if(Math.floor(num / 10) < 1){

          console.log(`Entered the end and div10 is ${div10}`);

          return div10; // returning undefined

     }



     let curr = Math.floor(num / 10);

     console.log(`Bottom of call and dividing ${curr}`);



     return numDigits(curr, div10); // HERE



}



console.log(numDigits(target));

const target = 1234; // count the number of digits



function numDigits(num, count=0){

     let div10 = count;

     div10 += 1;



     if(Math.floor(num / 10) < 1){

          console.log(`Entered the end and div10 is ${div10}`);

          return div10; // returning undefined

     }



     let curr = Math.floor(num / 10);

     console.log(`Bottom of call and dividing ${curr}`);



     return numDigits(curr, div10); // HERE



}



console.log(numDigits(target));

answered Jan 3 at 3:30

quirimmo

7,72811536

answered Jan 3 at 3:30

quirimmo

7,72811536

answered Jan 3 at 3:30

quirimmo

7,72811536

answered Jan 3 at 3:30

quirimmo

7,72811536

Thanks! Not sure I understand what is happening, as up until the 3rd recursive call, things appeared to work, and that was without the return. Can you elaborate on why that made it work?

– jobechoi
Jan 3 at 3:34

calls heppen, they do. But if you don't execute return as I added, then your func will not return anything. Does it make sense now?

– quirimmo
Jan 3 at 3:35

return isn't called (it's not a function), it's executed. ;-)

– RobG
Jan 3 at 3:38

@RobG fair enough :D thanks, edited :)

– quirimmo
Jan 3 at 3:38

I sort of get it: the recursive call at some point has to return something. In my initial code, that wasn't happening. Thanks.

– jobechoi
Jan 3 at 3:42

|
show 1 more comment

Thanks! Not sure I understand what is happening, as up until the 3rd recursive call, things appeared to work, and that was without the return. Can you elaborate on why that made it work?

– jobechoi
Jan 3 at 3:34

calls heppen, they do. But if you don't execute return as I added, then your func will not return anything. Does it make sense now?

– quirimmo
Jan 3 at 3:35

return isn't called (it's not a function), it's executed. ;-)

– RobG
Jan 3 at 3:38

@RobG fair enough :D thanks, edited :)

– quirimmo
Jan 3 at 3:38

I sort of get it: the recursive call at some point has to return something. In my initial code, that wasn't happening. Thanks.

– jobechoi
Jan 3 at 3:42

Thanks! Not sure I understand what is happening, as up until the 3rd recursive call, things appeared to work, and that was without the return. Can you elaborate on why that made it work?

– jobechoi
Jan 3 at 3:34

calls heppen, they do. But if you don't execute return as I added, then your func will not return anything. Does it make sense now?

– quirimmo
Jan 3 at 3:35

return isn't called (it's not a function), it's executed. ;-)

– RobG
Jan 3 at 3:38

@RobG fair enough :D thanks, edited :)

– quirimmo
Jan 3 at 3:38

I sort of get it: the recursive call at some point has to return something. In my initial code, that wasn't happening. Thanks.

– jobechoi
Jan 3 at 3:42

|
show 1 more comment

Your program can be dramatically simplified –

const numDigits = n =>

  n < 10

    ? 1

    : 1 + numDigits (Math .floor (n / 10))

        

console .log

  ( numDigits (1)         // 1

  , numDigits (22)        // 2

  , numDigits (333)       // 3

  , numDigits (999999999) // 9

  )

The count variable can still be used if you prefer. This would be a practical step toward making the program stack-safe –

const numDigits = (n, count = 1) =>

  n < 10

    ? count

    : numDigits

        ( Math .floor (n / 10)

        , count + 1

        )

        

console .log

  ( numDigits (1)         // 1

  , numDigits (22)        // 2

  , numDigits (333)       // 3

  , numDigits (999999999) // 9

  )

edited Jan 3 at 7:12

answered Jan 3 at 6:59

user633183

72k21143184

"There's no "missing return"" might be a reasonable statement if you hadn't added an implicit return through use of an arrow function. There is indeed a missing return statement, your refactored code is effectively if (n < 10) {return count} else {return numDigits(Math.floor (n / 10), count + 1)}. Removing the redundant else but keeping the return statement in the block is effectively identical to the OP's code. QED.

– RobG
Jan 3 at 8:52

The ternary is simple and succinct. As stated: new to recursion. I will keep this model in mind now. I see that my departure happened much earlier, even before the code was written: I never thought to check whether the integer was less than 10. Thanks, user633183.

– jobechoi
Jan 3 at 11:25

@RobG the difference lies between statements and expressions – statements like if, else, foreach, switch, while, and return are side effects, do not produce a value, and cannot be composed with other statements. Worse, if allows you to write conditionals without an else branch – ternary expressions using ?: evaluate to a value, can be composed with other expressions, and cannot be written without an "else" branch. It looks like I never got around to finishing this answer but I think you will find it useful.

– user633183
Jan 3 at 21:47

@RobG and for more about discussion about statements vs. expressions, I wish to share this. Any further discussion is welcomed and encouraged. The topics here are among my favourite :D

– user633183
Jan 3 at 22:11

add a comment |

Your program can be dramatically simplified –

const numDigits = n =>

  n < 10

    ? 1

    : 1 + numDigits (Math .floor (n / 10))

        

console .log

  ( numDigits (1)         // 1

  , numDigits (22)        // 2

  , numDigits (333)       // 3

  , numDigits (999999999) // 9

  )

The count variable can still be used if you prefer. This would be a practical step toward making the program stack-safe –

const numDigits = (n, count = 1) =>

  n < 10

    ? count

    : numDigits

        ( Math .floor (n / 10)

        , count + 1

        )

        

console .log

  ( numDigits (1)         // 1

  , numDigits (22)        // 2

  , numDigits (333)       // 3

  , numDigits (999999999) // 9

  )

edited Jan 3 at 7:12

answered Jan 3 at 6:59

user633183

72k21143184

"There's no "missing return"" might be a reasonable statement if you hadn't added an implicit return through use of an arrow function. There is indeed a missing return statement, your refactored code is effectively if (n < 10) {return count} else {return numDigits(Math.floor (n / 10), count + 1)}. Removing the redundant else but keeping the return statement in the block is effectively identical to the OP's code. QED.

– RobG
Jan 3 at 8:52

The ternary is simple and succinct. As stated: new to recursion. I will keep this model in mind now. I see that my departure happened much earlier, even before the code was written: I never thought to check whether the integer was less than 10. Thanks, user633183.

– jobechoi
Jan 3 at 11:25

@RobG the difference lies between statements and expressions – statements like if, else, foreach, switch, while, and return are side effects, do not produce a value, and cannot be composed with other statements. Worse, if allows you to write conditionals without an else branch – ternary expressions using ?: evaluate to a value, can be composed with other expressions, and cannot be written without an "else" branch. It looks like I never got around to finishing this answer but I think you will find it useful.

– user633183
Jan 3 at 21:47

@RobG and for more about discussion about statements vs. expressions, I wish to share this. Any further discussion is welcomed and encouraged. The topics here are among my favourite :D

– user633183
Jan 3 at 22:11

add a comment |

Your program can be dramatically simplified –

const numDigits = n =>

  n < 10

    ? 1

    : 1 + numDigits (Math .floor (n / 10))

        

console .log

  ( numDigits (1)         // 1

  , numDigits (22)        // 2

  , numDigits (333)       // 3

  , numDigits (999999999) // 9

  )

The count variable can still be used if you prefer. This would be a practical step toward making the program stack-safe –

const numDigits = (n, count = 1) =>

  n < 10

    ? count

    : numDigits

        ( Math .floor (n / 10)

        , count + 1

        )

        

console .log

  ( numDigits (1)         // 1

  , numDigits (22)        // 2

  , numDigits (333)       // 3

  , numDigits (999999999) // 9

  )

edited Jan 3 at 7:12

answered Jan 3 at 6:59

user633183

72k21143184

Your program can be dramatically simplified –

const numDigits = n =>

  n < 10

    ? 1

    : 1 + numDigits (Math .floor (n / 10))

        

console .log

  ( numDigits (1)         // 1

  , numDigits (22)        // 2

  , numDigits (333)       // 3

  , numDigits (999999999) // 9

  )

The count variable can still be used if you prefer. This would be a practical step toward making the program stack-safe –

const numDigits = (n, count = 1) =>

  n < 10

    ? count

    : numDigits

        ( Math .floor (n / 10)

        , count + 1

        )

        

console .log

  ( numDigits (1)         // 1

  , numDigits (22)        // 2

  , numDigits (333)       // 3

  , numDigits (999999999) // 9

  )

const numDigits = n =>

  n < 10

    ? 1

    : 1 + numDigits (Math .floor (n / 10))

        

console .log

  ( numDigits (1)         // 1

  , numDigits (22)        // 2

  , numDigits (333)       // 3

  , numDigits (999999999) // 9

  )

const numDigits = n =>

  n < 10

    ? 1

    : 1 + numDigits (Math .floor (n / 10))

        

console .log

  ( numDigits (1)         // 1

  , numDigits (22)        // 2

  , numDigits (333)       // 3

  , numDigits (999999999) // 9

  )

const numDigits = (n, count = 1) =>

  n < 10

    ? count

    : numDigits

        ( Math .floor (n / 10)

        , count + 1

        )

        

console .log

  ( numDigits (1)         // 1

  , numDigits (22)        // 2

  , numDigits (333)       // 3

  , numDigits (999999999) // 9

  )

const numDigits = (n, count = 1) =>

  n < 10

    ? count

    : numDigits

        ( Math .floor (n / 10)

        , count + 1

        )

        

console .log

  ( numDigits (1)         // 1

  , numDigits (22)        // 2

  , numDigits (333)       // 3

  , numDigits (999999999) // 9

  )

edited Jan 3 at 7:12

answered Jan 3 at 6:59

user633183

72k21143184

edited Jan 3 at 7:12

answered Jan 3 at 6:59

user633183

72k21143184

answered Jan 3 at 6:59

user633183

72k21143184

answered Jan 3 at 6:59

user633183

72k21143184

"There's no "missing return"" might be a reasonable statement if you hadn't added an implicit return through use of an arrow function. There is indeed a missing return statement, your refactored code is effectively if (n < 10) {return count} else {return numDigits(Math.floor (n / 10), count + 1)}. Removing the redundant else but keeping the return statement in the block is effectively identical to the OP's code. QED.

– RobG
Jan 3 at 8:52

The ternary is simple and succinct. As stated: new to recursion. I will keep this model in mind now. I see that my departure happened much earlier, even before the code was written: I never thought to check whether the integer was less than 10. Thanks, user633183.

– jobechoi
Jan 3 at 11:25

@RobG the difference lies between statements and expressions – statements like if, else, foreach, switch, while, and return are side effects, do not produce a value, and cannot be composed with other statements. Worse, if allows you to write conditionals without an else branch – ternary expressions using ?: evaluate to a value, can be composed with other expressions, and cannot be written without an "else" branch. It looks like I never got around to finishing this answer but I think you will find it useful.

– user633183
Jan 3 at 21:47

@RobG and for more about discussion about statements vs. expressions, I wish to share this. Any further discussion is welcomed and encouraged. The topics here are among my favourite :D

– user633183
Jan 3 at 22:11

add a comment |

"There's no "missing return"" might be a reasonable statement if you hadn't added an implicit return through use of an arrow function. There is indeed a missing return statement, your refactored code is effectively if (n < 10) {return count} else {return numDigits(Math.floor (n / 10), count + 1)}. Removing the redundant else but keeping the return statement in the block is effectively identical to the OP's code. QED.

– RobG
Jan 3 at 8:52

The ternary is simple and succinct. As stated: new to recursion. I will keep this model in mind now. I see that my departure happened much earlier, even before the code was written: I never thought to check whether the integer was less than 10. Thanks, user633183.

– jobechoi
Jan 3 at 11:25

@RobG the difference lies between statements and expressions – statements like if, else, foreach, switch, while, and return are side effects, do not produce a value, and cannot be composed with other statements. Worse, if allows you to write conditionals without an else branch – ternary expressions using ?: evaluate to a value, can be composed with other expressions, and cannot be written without an "else" branch. It looks like I never got around to finishing this answer but I think you will find it useful.

– user633183
Jan 3 at 21:47

@RobG and for more about discussion about statements vs. expressions, I wish to share this. Any further discussion is welcomed and encouraged. The topics here are among my favourite :D

– user633183
Jan 3 at 22:11

"There's no "missing return"" might be a reasonable statement if you hadn't added an implicit return through use of an arrow function. There is indeed a missing return statement, your refactored code is effectively if (n < 10) {return count} else {return numDigits(Math.floor (n / 10), count + 1)}. Removing the redundant else but keeping the return statement in the block is effectively identical to the OP's code. QED.

– RobG
Jan 3 at 8:52

The ternary is simple and succinct. As stated: new to recursion. I will keep this model in mind now. I see that my departure happened much earlier, even before the code was written: I never thought to check whether the integer was less than 10. Thanks, user633183.

– jobechoi
Jan 3 at 11:25

@RobG the difference lies between statements and expressions – statements like if, else, foreach, switch, while, and return are side effects, do not produce a value, and cannot be composed with other statements. Worse, if allows you to write conditionals without an else branch – ternary expressions using ?: evaluate to a value, can be composed with other expressions, and cannot be written without an "else" branch. It looks like I never got around to finishing this answer but I think you will find it useful.

– user633183
Jan 3 at 21:47

@RobG and for more about discussion about statements vs. expressions, I wish to share this. Any further discussion is welcomed and encouraged. The topics here are among my favourite :D

– user633183
Jan 3 at 22:11

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