Best way to approximate in direction to zero in C++

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I have some troubles explaining clearly but it's rather simple..

I have a double named value in my C++ program and I want to Floor it if it's a positive value and Ceil if it's a negative value, the precision is given by an external variable.

An example:

precision is of 1000
value is 0.2659 so approximated value is 0.265

value is -0.2659 so approximated value is -0.265

I have wrote a simple code but I was wondering if there was a more simple or/and way to do it.

Here what I have so far:

void NodeImpl::approximation(double& value, const double& precision)

 if (value > 0.0)
 
 value = std::floor(value * precision);
 
 else
 
 value = std::ceil(value * precision);
 
 value /= precision;

I have wrote a simple code but I was wondering if there was a more simple or/and way to do it.

c++ double approximation

share|improve this question

asked 5 hours ago

tony497

999

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Looks ok to me but do note that the final value may have joke digits after the decimal 15th significant figure, e.g. 0.265 is actually 0.26500000000000001332267629550187848508358001708984375. Out of interest, why are you passing precision by const reference?
  – Bathsheba
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Unrelated to your problem, but you don't need to pass precision by constant reference. Just pass by value, or possibly constant value (i.e. plain double precision or const double precision). I would also personally rather return the new value instead of using a reference argument.
  – Some programmer dude
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  be warned that your intended operation cannot be performed exactly: the number 0.265, for example, has no finite floating-point representation, so you must not rely in your code on any exactness.
  – Walter
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  @Bathsheba "joke digits" 😂
  – Lightness Races in Orbit
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Considering that references usually is implemented as pointers, you're still passing 64 bits of data, which also happens to happens to be the common size for the double type. In short, there's nothing gained. Generally, for primitive types (int, double, etc.) don't pass by reference unless you need to modify the value inside the function.
  – Some programmer dude
  3 hours ago
  

  
  
  
  

 | 
show 7 more comments

up vote
9
down vote

favorite

I have some troubles explaining clearly but it's rather simple..

I have a double named value in my C++ program and I want to Floor it if it's a positive value and Ceil if it's a negative value, the precision is given by an external variable.

An example:

precision is of 1000
value is 0.2659 so approximated value is 0.265

value is -0.2659 so approximated value is -0.265

I have wrote a simple code but I was wondering if there was a more simple or/and way to do it.

Here what I have so far:

void NodeImpl::approximation(double& value, const double& precision)

 if (value > 0.0)
 
 value = std::floor(value * precision);
 
 else
 
 value = std::ceil(value * precision);
 
 value /= precision;

I have wrote a simple code but I was wondering if there was a more simple or/and way to do it.

c++ double approximation

share|improve this question

asked 5 hours ago

tony497

999

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Looks ok to me but do note that the final value may have joke digits after the decimal 15th significant figure, e.g. 0.265 is actually 0.26500000000000001332267629550187848508358001708984375. Out of interest, why are you passing precision by const reference?
  – Bathsheba
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Unrelated to your problem, but you don't need to pass precision by constant reference. Just pass by value, or possibly constant value (i.e. plain double precision or const double precision). I would also personally rather return the new value instead of using a reference argument.
  – Some programmer dude
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  be warned that your intended operation cannot be performed exactly: the number 0.265, for example, has no finite floating-point representation, so you must not rely in your code on any exactness.
  – Walter
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  @Bathsheba "joke digits" 😂
  – Lightness Races in Orbit
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Considering that references usually is implemented as pointers, you're still passing 64 bits of data, which also happens to happens to be the common size for the double type. In short, there's nothing gained. Generally, for primitive types (int, double, etc.) don't pass by reference unless you need to modify the value inside the function.
  – Some programmer dude
  3 hours ago
  

  
  
  
  

 | 
show 7 more comments

up vote
9
down vote

favorite

up vote
9
down vote

favorite

I have some troubles explaining clearly but it's rather simple..

I have a double named value in my C++ program and I want to Floor it if it's a positive value and Ceil if it's a negative value, the precision is given by an external variable.

An example:

precision is of 1000
value is 0.2659 so approximated value is 0.265

value is -0.2659 so approximated value is -0.265

I have wrote a simple code but I was wondering if there was a more simple or/and way to do it.

Here what I have so far:

void NodeImpl::approximation(double& value, const double& precision)

 if (value > 0.0)
 
 value = std::floor(value * precision);
 
 else
 
 value = std::ceil(value * precision);
 
 value /= precision;

I have wrote a simple code but I was wondering if there was a more simple or/and way to do it.

c++ double approximation

share|improve this question

asked 5 hours ago

tony497

999

I have some troubles explaining clearly but it's rather simple..

I have a double named value in my C++ program and I want to Floor it if it's a positive value and Ceil if it's a negative value, the precision is given by an external variable.

An example:

precision is of 1000
value is 0.2659 so approximated value is 0.265

value is -0.2659 so approximated value is -0.265

I have wrote a simple code but I was wondering if there was a more simple or/and way to do it.

Here what I have so far:

void NodeImpl::approximation(double& value, const double& precision)

 if (value > 0.0)
 
 value = std::floor(value * precision);
 
 else
 
 value = std::ceil(value * precision);
 
 value /= precision;

I have wrote a simple code but I was wondering if there was a more simple or/and way to do it.

c++ double approximation

c++ double approximation

share|improve this question

asked 5 hours ago

tony497

999

share|improve this question

asked 5 hours ago

tony497

999

share|improve this question

share|improve this question

asked 5 hours ago

tony497

999

asked 5 hours ago

tony497

999

asked 5 hours ago

tony497

999

999

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Looks ok to me but do note that the final value may have joke digits after the decimal 15th significant figure, e.g. 0.265 is actually 0.26500000000000001332267629550187848508358001708984375. Out of interest, why are you passing precision by const reference?
  – Bathsheba
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Unrelated to your problem, but you don't need to pass precision by constant reference. Just pass by value, or possibly constant value (i.e. plain double precision or const double precision). I would also personally rather return the new value instead of using a reference argument.
  – Some programmer dude
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  be warned that your intended operation cannot be performed exactly: the number 0.265, for example, has no finite floating-point representation, so you must not rely in your code on any exactness.
  – Walter
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  @Bathsheba "joke digits" 😂
  – Lightness Races in Orbit
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Considering that references usually is implemented as pointers, you're still passing 64 bits of data, which also happens to happens to be the common size for the double type. In short, there's nothing gained. Generally, for primitive types (int, double, etc.) don't pass by reference unless you need to modify the value inside the function.
  – Some programmer dude
  3 hours ago
  

  
  
  
  

 | 
show 7 more comments

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Looks ok to me but do note that the final value may have joke digits after the decimal 15th significant figure, e.g. 0.265 is actually 0.26500000000000001332267629550187848508358001708984375. Out of interest, why are you passing precision by const reference?
  – Bathsheba
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Unrelated to your problem, but you don't need to pass precision by constant reference. Just pass by value, or possibly constant value (i.e. plain double precision or const double precision). I would also personally rather return the new value instead of using a reference argument.
  – Some programmer dude
  5 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  be warned that your intended operation cannot be performed exactly: the number 0.265, for example, has no finite floating-point representation, so you must not rely in your code on any exactness.
  – Walter
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  @Bathsheba "joke digits" 😂
  – Lightness Races in Orbit
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Considering that references usually is implemented as pointers, you're still passing 64 bits of data, which also happens to happens to be the common size for the double type. In short, there's nothing gained. Generally, for primitive types (int, double, etc.) don't pass by reference unless you need to modify the value inside the function.
  – Some programmer dude
  3 hours ago
  

  
  
  
  

2

2

Looks ok to me but do note that the final value may have joke digits after the decimal 15th significant figure, e.g. 0.265 is actually 0.26500000000000001332267629550187848508358001708984375. Out of interest, why are you passing precision by const reference?
– Bathsheba
5 hours ago

Looks ok to me but do note that the final value may have joke digits after the decimal 15th significant figure, e.g. 0.265 is actually 0.26500000000000001332267629550187848508358001708984375. Out of interest, why are you passing precision by const reference?
– Bathsheba
5 hours ago

3

3

Unrelated to your problem, but you don't need to pass precision by constant reference. Just pass by value, or possibly constant value (i.e. plain double precision or const double precision). I would also personally rather return the new value instead of using a reference argument.
– Some programmer dude
5 hours ago

Unrelated to your problem, but you don't need to pass precision by constant reference. Just pass by value, or possibly constant value (i.e. plain double precision or const double precision). I would also personally rather return the new value instead of using a reference argument.
– Some programmer dude
5 hours ago

2

2

be warned that your intended operation cannot be performed exactly: the number 0.265, for example, has no finite floating-point representation, so you must not rely in your code on any exactness.
– Walter
4 hours ago

be warned that your intended operation cannot be performed exactly: the number 0.265, for example, has no finite floating-point representation, so you must not rely in your code on any exactness.
– Walter
4 hours ago

4

4

@Bathsheba "joke digits" 😂
– Lightness Races in Orbit
4 hours ago

@Bathsheba "joke digits" 😂
– Lightness Races in Orbit
4 hours ago

2

2

Considering that references usually is implemented as pointers, you're still passing 64 bits of data, which also happens to happens to be the common size for the double type. In short, there's nothing gained. Generally, for primitive types (int, double, etc.) don't pass by reference unless you need to modify the value inside the function.
– Some programmer dude
3 hours ago

Considering that references usually is implemented as pointers, you're still passing 64 bits of data, which also happens to happens to be the common size for the double type. In short, there's nothing gained. Generally, for primitive types (int, double, etc.) don't pass by reference unless you need to modify the value inside the function.
– Some programmer dude
3 hours ago

 | 
show 7 more comments

1 Answer
1

active

oldest

votes

up vote
9
down vote

You can use std::trunc:

return std::trunc(value * precision) / precision;

share|improve this answer

edited 4 hours ago

ash108

1,2851318

answered 4 hours ago

user2079303

73.5k553111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user463035818 it return nearest integer not greater in magnitude to the argument i.e. rounds towards zero.
  – user2079303
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  well reading till the end of the sentence helps sometimes ;)
  – user463035818
  4 hours ago
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
9
down vote

You can use std::trunc:

return std::trunc(value * precision) / precision;

share|improve this answer

edited 4 hours ago

ash108

1,2851318

answered 4 hours ago

user2079303

73.5k553111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user463035818 it return nearest integer not greater in magnitude to the argument i.e. rounds towards zero.
  – user2079303
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  well reading till the end of the sentence helps sometimes ;)
  – user463035818
  4 hours ago
  

  
  
  
  

add a comment | 

up vote
9
down vote

You can use std::trunc:

return std::trunc(value * precision) / precision;

share|improve this answer

edited 4 hours ago

ash108

1,2851318

answered 4 hours ago

user2079303

73.5k553111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user463035818 it return nearest integer not greater in magnitude to the argument i.e. rounds towards zero.
  – user2079303
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  well reading till the end of the sentence helps sometimes ;)
  – user463035818
  4 hours ago
  

  
  
  
  

add a comment | 

up vote
9
down vote

up vote
9
down vote

You can use std::trunc:

return std::trunc(value * precision) / precision;

share|improve this answer

edited 4 hours ago

ash108

1,2851318

answered 4 hours ago

user2079303

73.5k553111

You can use std::trunc:

return std::trunc(value * precision) / precision;

share|improve this answer

edited 4 hours ago

ash108

1,2851318

answered 4 hours ago

user2079303

73.5k553111

share|improve this answer

share|improve this answer

edited 4 hours ago

ash108

1,2851318

edited 4 hours ago

ash108

1,2851318

edited 4 hours ago

ash108

1,2851318

1,2851318

answered 4 hours ago

user2079303

73.5k553111

answered 4 hours ago

user2079303

73.5k553111

answered 4 hours ago

user2079303

73.5k553111

73.5k553111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user463035818 it return nearest integer not greater in magnitude to the argument i.e. rounds towards zero.
  – user2079303
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  well reading till the end of the sentence helps sometimes ;)
  – user463035818
  4 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user463035818 it return nearest integer not greater in magnitude to the argument i.e. rounds towards zero.
  – user2079303
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  well reading till the end of the sentence helps sometimes ;)
  – user463035818
  4 hours ago
  

  
  
  
  

@user463035818 it return nearest integer not greater in magnitude to the argument i.e. rounds towards zero.
– user2079303
4 hours ago

@user463035818 it return nearest integer not greater in magnitude to the argument i.e. rounds towards zero.
– user2079303
4 hours ago

well reading till the end of the sentence helps sometimes ;)
– user463035818
4 hours ago

well reading till the end of the sentence helps sometimes ;)
– user463035818
4 hours ago

add a comment | 

 

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