Function that given a number returns the number within a range

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3
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I am trying to write a function that given a number, it returns the number within the range of 1 to 12.

So, for example, if the input is 1, the output is 1, if the input is 12, the output is 12, if the input is 13, the output is 1, if the input is 14, the output is 2, if the input is 24, the output is 12.

I've tried this so far:

function toRange(number) {

  if (number > 12) { 

    return number % 12

  }



  return number

}

But I am wondering if there is a way that I can solve this without doing conditionals. I thought of doing return (number + 12) % 12, but that wouldn't work if number is 12, because it would return 0.

asked Nov 8 at 11:23

Hommer Smith

8,61937110203

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up vote
3
down vote

favorite

I am trying to write a function that given a number, it returns the number within the range of 1 to 12.

I've tried this so far:

function toRange(number) {

  if (number > 12) { 

    return number % 12

  }



  return number

}

But I am wondering if there is a way that I can solve this without doing conditionals. I thought of doing return (number + 12) % 12, but that wouldn't work if number is 12, because it would return 0.

asked Nov 8 at 11:23

Hommer Smith

8,61937110203

add a comment |

up vote
3
down vote

favorite

I am trying to write a function that given a number, it returns the number within the range of 1 to 12.

I've tried this so far:

function toRange(number) {

  if (number > 12) { 

    return number % 12

  }



  return number

}

But I am wondering if there is a way that I can solve this without doing conditionals. I thought of doing return (number + 12) % 12, but that wouldn't work if number is 12, because it would return 0.

asked Nov 8 at 11:23

Hommer Smith

8,61937110203

I am trying to write a function that given a number, it returns the number within the range of 1 to 12.

I've tried this so far:

function toRange(number) {

  if (number > 12) { 

    return number % 12

  }



  return number

}

But I am wondering if there is a way that I can solve this without doing conditionals. I thought of doing return (number + 12) % 12, but that wouldn't work if number is 12, because it would return 0.

javascript modulo

asked Nov 8 at 11:23

Hommer Smith

8,61937110203

asked Nov 8 at 11:23

Hommer Smith

8,61937110203

asked Nov 8 at 11:23

Hommer Smith

8,61937110203

asked Nov 8 at 11:23

Hommer Smith

8,61937110203

asked Nov 8 at 11:23

Hommer Smith

8,61937110203

add a comment |

3 Answers
3

active

oldest

votes

up vote
2
down vote

accepted

You could use an adjustment for a zero based value and add the adustment at the end.

function toRange(number) {

    return (number - 1) % 12 + 1;

}



var i;



for (i = 1; i <= 24; i++) console.log(i, toRange(i));

.as-console-wrapper { max-height: 100% !important; top: 0; }

answered Nov 8 at 11:27

Nina Scholz

168k1382145

add a comment |

up vote
0
down vote

maybe something like this:

function toRange(number) {

  if (number > 12) { 

    if (number % 12 === 0)

      return number - 12;

    return number % 12

  }

  return number

}

answered Nov 8 at 11:28

simonecosci

56128

add a comment |

up vote
0
down vote

Can also be thought of like - Get the remainder of number by 12 and return 12 in case remainder is 0.

function toRange(number) {

    return number % 12 || 12;

}



for (var i = 1; i <= 24; i++) console.log(i, toRange(i));

answered Nov 8 at 12:21

Nitish Narang

1,40169

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
2
down vote

accepted

You could use an adjustment for a zero based value and add the adustment at the end.

function toRange(number) {

    return (number - 1) % 12 + 1;

}



var i;



for (i = 1; i <= 24; i++) console.log(i, toRange(i));

.as-console-wrapper { max-height: 100% !important; top: 0; }

answered Nov 8 at 11:27

Nina Scholz

168k1382145

add a comment |

up vote
2
down vote

accepted

You could use an adjustment for a zero based value and add the adustment at the end.

function toRange(number) {

    return (number - 1) % 12 + 1;

}



var i;



for (i = 1; i <= 24; i++) console.log(i, toRange(i));

.as-console-wrapper { max-height: 100% !important; top: 0; }

answered Nov 8 at 11:27

Nina Scholz

168k1382145

add a comment |

up vote
2
down vote

accepted

You could use an adjustment for a zero based value and add the adustment at the end.

function toRange(number) {

    return (number - 1) % 12 + 1;

}



var i;



for (i = 1; i <= 24; i++) console.log(i, toRange(i));

.as-console-wrapper { max-height: 100% !important; top: 0; }

answered Nov 8 at 11:27

Nina Scholz

168k1382145

You could use an adjustment for a zero based value and add the adustment at the end.

function toRange(number) {

    return (number - 1) % 12 + 1;

}



var i;



for (i = 1; i <= 24; i++) console.log(i, toRange(i));

.as-console-wrapper { max-height: 100% !important; top: 0; }

function toRange(number) {

    return (number - 1) % 12 + 1;

}



var i;



for (i = 1; i <= 24; i++) console.log(i, toRange(i));

.as-console-wrapper { max-height: 100% !important; top: 0; }

function toRange(number) {

    return (number - 1) % 12 + 1;

}



var i;



for (i = 1; i <= 24; i++) console.log(i, toRange(i));

.as-console-wrapper { max-height: 100% !important; top: 0; }

answered Nov 8 at 11:27

Nina Scholz

168k1382145

answered Nov 8 at 11:27

Nina Scholz

168k1382145

answered Nov 8 at 11:27

Nina Scholz

168k1382145

answered Nov 8 at 11:27

Nina Scholz

168k1382145

add a comment |

up vote
0
down vote

maybe something like this:

function toRange(number) {

  if (number > 12) { 

    if (number % 12 === 0)

      return number - 12;

    return number % 12

  }

  return number

}

answered Nov 8 at 11:28

simonecosci

56128

add a comment |

up vote
0
down vote

maybe something like this:

function toRange(number) {

  if (number > 12) { 

    if (number % 12 === 0)

      return number - 12;

    return number % 12

  }

  return number

}

answered Nov 8 at 11:28

simonecosci

56128

add a comment |

up vote
0
down vote

maybe something like this:

function toRange(number) {

  if (number > 12) { 

    if (number % 12 === 0)

      return number - 12;

    return number % 12

  }

  return number

}

answered Nov 8 at 11:28

simonecosci

56128

maybe something like this:

function toRange(number) {

  if (number > 12) { 

    if (number % 12 === 0)

      return number - 12;

    return number % 12

  }

  return number

}

answered Nov 8 at 11:28

simonecosci

56128

answered Nov 8 at 11:28

simonecosci

56128

answered Nov 8 at 11:28

simonecosci

56128

answered Nov 8 at 11:28

simonecosci

56128

add a comment |

up vote
0
down vote

Can also be thought of like - Get the remainder of number by 12 and return 12 in case remainder is 0.

function toRange(number) {

    return number % 12 || 12;

}



for (var i = 1; i <= 24; i++) console.log(i, toRange(i));

answered Nov 8 at 12:21

Nitish Narang

1,40169

add a comment |

up vote
0
down vote

Can also be thought of like - Get the remainder of number by 12 and return 12 in case remainder is 0.

function toRange(number) {

    return number % 12 || 12;

}



for (var i = 1; i <= 24; i++) console.log(i, toRange(i));

answered Nov 8 at 12:21

Nitish Narang

1,40169

add a comment |

up vote
0
down vote

Can also be thought of like - Get the remainder of number by 12 and return 12 in case remainder is 0.

function toRange(number) {

    return number % 12 || 12;

}



for (var i = 1; i <= 24; i++) console.log(i, toRange(i));

answered Nov 8 at 12:21

Nitish Narang

1,40169

Can also be thought of like - Get the remainder of number by 12 and return 12 in case remainder is 0.

function toRange(number) {

    return number % 12 || 12;

}



for (var i = 1; i <= 24; i++) console.log(i, toRange(i));

function toRange(number) {

    return number % 12 || 12;

}



for (var i = 1; i <= 24; i++) console.log(i, toRange(i));

function toRange(number) {

    return number % 12 || 12;

}



for (var i = 1; i <= 24; i++) console.log(i, toRange(i));

answered Nov 8 at 12:21

Nitish Narang

1,40169

answered Nov 8 at 12:21

Nitish Narang

1,40169

answered Nov 8 at 12:21

Nitish Narang

1,40169

answered Nov 8 at 12:21

Nitish Narang

1,40169

add a comment |

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