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What are the characteristics of a typeless programming language?

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3

$begingroup$

Assume that we have an imaginary programming language that allows you to assign a Literal to a variable, but does not allow you to set the the data type of the variable, for example

Allocate4Bytes an_int_variable 123456;
Allocate2Bytes a_short_int_variable 123;
Allocate4Bytes a_float_variable 2.1;

And this programming language also provides different operators to work with different data types, for example:

• The + operator is used to add an int to a short int.


• The #+ operator is used to add a float to an int.


• the = operator is used to assign an int to an int.


• the #= operator is used to assign a short int to a short int.


• etc.

So it is the job of the programmer to keep track of the data type of each variable and use the appropriate operator on it.

Is this programming language considered to be a typeless programming language, or can we say that this programming language have data types (even though it does not have operator overloading and type safety, etc.)?

programming-languages

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asked 5 hours ago

user4582812user4582812

161

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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's simple: the variables have no type. An example of a typed language is C/C++ (you have a char, an int, a float,...), while not typed languages don't have it (e.g., Mathematica).
  $endgroup$
  – Iago Carvalho
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your language, on the other hand, looks like to be typed (even if not explicitly defined). However, I'm not an expert on this subject and I'm not 100% sure about it... Please, someone correct me if I'm wrong.
  $endgroup$
  – Iago Carvalho
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Iago Carvalho You said that a typeless language means that "the variables have no type", and then you said "Your language, on the other hand, looks like to be typed". How is my language looks like to be typed if my variables don't have data types?
  $endgroup$
  – user4582812
  4 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

Assume that we have an imaginary programming language that allows you to assign a Literal to a variable, but does not allow you to set the the data type of the variable, for example

Allocate4Bytes an_int_variable 123456;
Allocate2Bytes a_short_int_variable 123;
Allocate4Bytes a_float_variable 2.1;

And this programming language also provides different operators to work with different data types, for example:

• The + operator is used to add an int to a short int.


• The #+ operator is used to add a float to an int.


• the = operator is used to assign an int to an int.


• the #= operator is used to assign a short int to a short int.


• etc.

So it is the job of the programmer to keep track of the data type of each variable and use the appropriate operator on it.

Is this programming language considered to be a typeless programming language, or can we say that this programming language have data types (even though it does not have operator overloading and type safety, etc.)?

programming-languages

share|cite|improve this question

asked 5 hours ago

user4582812user4582812

161

New contributor

user4582812 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's simple: the variables have no type. An example of a typed language is C/C++ (you have a char, an int, a float,...), while not typed languages don't have it (e.g., Mathematica).
  $endgroup$
  – Iago Carvalho
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your language, on the other hand, looks like to be typed (even if not explicitly defined). However, I'm not an expert on this subject and I'm not 100% sure about it... Please, someone correct me if I'm wrong.
  $endgroup$
  – Iago Carvalho
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Iago Carvalho You said that a typeless language means that "the variables have no type", and then you said "Your language, on the other hand, looks like to be typed". How is my language looks like to be typed if my variables don't have data types?
  $endgroup$
  – user4582812
  4 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Assume that we have an imaginary programming language that allows you to assign a Literal to a variable, but does not allow you to set the the data type of the variable, for example

Allocate4Bytes an_int_variable 123456;
Allocate2Bytes a_short_int_variable 123;
Allocate4Bytes a_float_variable 2.1;

And this programming language also provides different operators to work with different data types, for example:

• The + operator is used to add an int to a short int.


• The #+ operator is used to add a float to an int.


• the = operator is used to assign an int to an int.


• the #= operator is used to assign a short int to a short int.


• etc.

So it is the job of the programmer to keep track of the data type of each variable and use the appropriate operator on it.

Is this programming language considered to be a typeless programming language, or can we say that this programming language have data types (even though it does not have operator overloading and type safety, etc.)?

programming-languages

share|cite|improve this question

asked 5 hours ago

user4582812user4582812

161

New contributor

user4582812 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

Allocate4Bytes an_int_variable 123456;
Allocate2Bytes a_short_int_variable 123;
Allocate4Bytes a_float_variable 2.1;

share|cite|improve this question

asked 5 hours ago

user4582812user4582812

161

New contributor

user4582812 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked 5 hours ago

user4582812user4582812

161

New contributor

user4582812 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 5 hours ago

user4582812user4582812

161

New contributor

user4582812 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 5 hours ago

user4582812user4582812

161

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4

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It depends on what happens if the programmer tries to do something like add two things with the wrong operator. If it causes a compile-time error, the language is probably statically typed. If it throws a compile-time error complaining that the two have the wrong type, the language is probably dynamically typed. If it doesn't throw an error and just tries to add those two things (possibly resulting in gibberish), it might be untyped.

To put it another way, it depends how those operations are implemented. If they are implemented to keep track of the type of the variables and check that those types match what is expected, it is typed. If it doesn't keep track of types, then it is untyped or weakly typed.

See https://en.wikipedia.org/wiki/Type_system and https://www.sitepoint.com/typing-versus-dynamic-typing/.

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answered 4 hours ago

D.W.♦D.W.

104k14131299

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4

$begingroup$

It depends on what happens if the programmer tries to do something like add two things with the wrong operator. If it causes a compile-time error, the language is probably statically typed. If it throws a compile-time error complaining that the two have the wrong type, the language is probably dynamically typed. If it doesn't throw an error and just tries to add those two things (possibly resulting in gibberish), it might be untyped.

To put it another way, it depends how those operations are implemented. If they are implemented to keep track of the type of the variables and check that those types match what is expected, it is typed. If it doesn't keep track of types, then it is untyped or weakly typed.

See https://en.wikipedia.org/wiki/Type_system and https://www.sitepoint.com/typing-versus-dynamic-typing/.

share|cite|improve this answer

answered 4 hours ago

D.W.♦D.W.

104k14131299

$endgroup$

add a comment | 

4

$begingroup$

It depends on what happens if the programmer tries to do something like add two things with the wrong operator. If it causes a compile-time error, the language is probably statically typed. If it throws a compile-time error complaining that the two have the wrong type, the language is probably dynamically typed. If it doesn't throw an error and just tries to add those two things (possibly resulting in gibberish), it might be untyped.

To put it another way, it depends how those operations are implemented. If they are implemented to keep track of the type of the variables and check that those types match what is expected, it is typed. If it doesn't keep track of types, then it is untyped or weakly typed.

See https://en.wikipedia.org/wiki/Type_system and https://www.sitepoint.com/typing-versus-dynamic-typing/.

share|cite|improve this answer

answered 4 hours ago

D.W.♦D.W.

104k14131299

$endgroup$

add a comment | 

$begingroup$

It depends on what happens if the programmer tries to do something like add two things with the wrong operator. If it causes a compile-time error, the language is probably statically typed. If it throws a compile-time error complaining that the two have the wrong type, the language is probably dynamically typed. If it doesn't throw an error and just tries to add those two things (possibly resulting in gibberish), it might be untyped.

To put it another way, it depends how those operations are implemented. If they are implemented to keep track of the type of the variables and check that those types match what is expected, it is typed. If it doesn't keep track of types, then it is untyped or weakly typed.

See https://en.wikipedia.org/wiki/Type_system and https://www.sitepoint.com/typing-versus-dynamic-typing/.

share|cite|improve this answer

answered 4 hours ago

D.W.♦D.W.

104k14131299

$endgroup$

share|cite|improve this answer

answered 4 hours ago

D.W.♦D.W.

104k14131299

answered 4 hours ago

D.W.♦D.W.

104k14131299

answered 4 hours ago

D.W.♦D.W.

104k14131299