Scheme - difference between if and or
does somebody know what the difference between if and or in Scheme is or why I can't use if to replace or as an identifier? I posted some code below.
Thank's!
;if definition
(define heiner-or
(lambda (test-1 test-2)
(if test-1
#t
test-2)))
;or definition
> (heiner-or (= 10 10) (> 2 5))
#t
> (heiner-or (> 23 42) (< 5 2))
#f
functional-programming scheme
asked Nov 16 '18 at 8:29
user6825883user6825883
177
add a comment |
does somebody know what the difference between if and or in Scheme is or why I can't use if to replace or as an identifier? I posted some code below.
Thank's!
;if definition
(define heiner-or
(lambda (test-1 test-2)
(if test-1
#t
test-2)))
;or definition
> (heiner-or (= 10 10) (> 2 5))
#t
> (heiner-or (> 23 42) (< 5 2))
#f
functional-programming scheme
asked Nov 16 '18 at 8:29
user6825883user6825883
177
add a comment |
does somebody know what the difference between if and or in Scheme is or why I can't use if to replace or as an identifier? I posted some code below.
Thank's!
;if definition
(define heiner-or
(lambda (test-1 test-2)
(if test-1
#t
test-2)))
;or definition
> (heiner-or (= 10 10) (> 2 5))
#t
> (heiner-or (> 23 42) (< 5 2))
#f
functional-programming scheme
asked Nov 16 '18 at 8:29
user6825883user6825883
177
does somebody know what the difference between if and or in Scheme is or why I can't use if to replace or as an identifier? I posted some code below.
Thank's!
;if definition
(define heiner-or
(lambda (test-1 test-2)
(if test-1
#t
test-2)))
;or definition
> (heiner-or (= 10 10) (> 2 5))
#t
> (heiner-or (> 23 42) (< 5 2))
#f
functional-programming scheme
functional-programming scheme
asked Nov 16 '18 at 8:29
user6825883user6825883
177
asked Nov 16 '18 at 8:29
user6825883user6825883
177
edited Nov 16 '18 at 8:36
user6825883
asked Nov 16 '18 at 8:29
user6825883user6825883
177
asked Nov 16 '18 at 8:29
user6825883user6825883
177
asked Nov 16 '18 at 8:29
user6825883user6825883
177
177
add a comment |
add a comment |
1 Answer
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oldest
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Take a look at the documentation for if and or. if is used for testing a condition and executing only one of two possible values - if the condition is true the first part will get executed, if it's false the second part will get executed:
(if (= 1 0)
"not executed"
"executed")
=> "executed"
or is a logical connector that can receive multiple arguments, it'll return the first non-false value it finds, or false if all values are false. Typically you'd use or to connect boolean expressions, but that's not always the case:
(or #f 7 10)
=> 7
(or (= 1 0) (= 1 2))
=> #f
Also bear in mind that in Scheme all values are considered true, except #f, which is false. Regarding the last part of your question: yes, you can use if to simulate an or, it would be something like this:
; equivalent to (or value1 value2)
(if value1
value1
value2)
The above is a simplification, because ideally we should evaluate value1 exactly once, and also because this version doesn't support multiple arguments. But beware: the moment you try to write it as a procedure you'll run into trouble!
(define (my-or value1 value2)
(if value1
value1
value2))
That will not work as an or, the procedure evaluates both arguments before evaluating the if, whereas a real or only evaluates the needed expressions until it finds a true value - that's what we call short-circuit evaluation. To see what I mean, try this:
(or 42 (/ 1 0))
=> 42
(my-or 42 (/ 1 0))
=> /: division by zero
We can't truly implement or, if as procedures, they're special forms and need to be implemented at the interpreter level, with special evaluation rules.
answered Nov 16 '18 at 9:07
Óscar LópezÓscar López
176k24225322
add a comment |
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1 Answer
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votes
Take a look at the documentation for if and or. if is used for testing a condition and executing only one of two possible values - if the condition is true the first part will get executed, if it's false the second part will get executed:
(if (= 1 0)
"not executed"
"executed")
=> "executed"
or is a logical connector that can receive multiple arguments, it'll return the first non-false value it finds, or false if all values are false. Typically you'd use or to connect boolean expressions, but that's not always the case:
(or #f 7 10)
=> 7
(or (= 1 0) (= 1 2))
=> #f
Also bear in mind that in Scheme all values are considered true, except #f, which is false. Regarding the last part of your question: yes, you can use if to simulate an or, it would be something like this:
; equivalent to (or value1 value2)
(if value1
value1
value2)
The above is a simplification, because ideally we should evaluate value1 exactly once, and also because this version doesn't support multiple arguments. But beware: the moment you try to write it as a procedure you'll run into trouble!
(define (my-or value1 value2)
(if value1
value1
value2))
That will not work as an or, the procedure evaluates both arguments before evaluating the if, whereas a real or only evaluates the needed expressions until it finds a true value - that's what we call short-circuit evaluation. To see what I mean, try this:
(or 42 (/ 1 0))
=> 42
(my-or 42 (/ 1 0))
=> /: division by zero
We can't truly implement or, if as procedures, they're special forms and need to be implemented at the interpreter level, with special evaluation rules.
answered Nov 16 '18 at 9:07
Óscar LópezÓscar López
176k24225322
add a comment |
Take a look at the documentation for if and or. if is used for testing a condition and executing only one of two possible values - if the condition is true the first part will get executed, if it's false the second part will get executed:
(if (= 1 0)
"not executed"
"executed")
=> "executed"
or is a logical connector that can receive multiple arguments, it'll return the first non-false value it finds, or false if all values are false. Typically you'd use or to connect boolean expressions, but that's not always the case:
(or #f 7 10)
=> 7
(or (= 1 0) (= 1 2))
=> #f
Also bear in mind that in Scheme all values are considered true, except #f, which is false. Regarding the last part of your question: yes, you can use if to simulate an or, it would be something like this:
; equivalent to (or value1 value2)
(if value1
value1
value2)
The above is a simplification, because ideally we should evaluate value1 exactly once, and also because this version doesn't support multiple arguments. But beware: the moment you try to write it as a procedure you'll run into trouble!
(define (my-or value1 value2)
(if value1
value1
value2))
That will not work as an or, the procedure evaluates both arguments before evaluating the if, whereas a real or only evaluates the needed expressions until it finds a true value - that's what we call short-circuit evaluation. To see what I mean, try this:
(or 42 (/ 1 0))
=> 42
(my-or 42 (/ 1 0))
=> /: division by zero
We can't truly implement or, if as procedures, they're special forms and need to be implemented at the interpreter level, with special evaluation rules.
answered Nov 16 '18 at 9:07
Óscar LópezÓscar López
176k24225322
add a comment |
Take a look at the documentation for if and or. if is used for testing a condition and executing only one of two possible values - if the condition is true the first part will get executed, if it's false the second part will get executed:
(if (= 1 0)
"not executed"
"executed")
=> "executed"
or is a logical connector that can receive multiple arguments, it'll return the first non-false value it finds, or false if all values are false. Typically you'd use or to connect boolean expressions, but that's not always the case:
(or #f 7 10)
=> 7
(or (= 1 0) (= 1 2))
=> #f
Also bear in mind that in Scheme all values are considered true, except #f, which is false. Regarding the last part of your question: yes, you can use if to simulate an or, it would be something like this:
; equivalent to (or value1 value2)
(if value1
value1
value2)
The above is a simplification, because ideally we should evaluate value1 exactly once, and also because this version doesn't support multiple arguments. But beware: the moment you try to write it as a procedure you'll run into trouble!
(define (my-or value1 value2)
(if value1
value1
value2))
That will not work as an or, the procedure evaluates both arguments before evaluating the if, whereas a real or only evaluates the needed expressions until it finds a true value - that's what we call short-circuit evaluation. To see what I mean, try this:
(or 42 (/ 1 0))
=> 42
(my-or 42 (/ 1 0))
=> /: division by zero
We can't truly implement or, if as procedures, they're special forms and need to be implemented at the interpreter level, with special evaluation rules.
answered Nov 16 '18 at 9:07
Óscar LópezÓscar López
176k24225322
Take a look at the documentation for if and or. if is used for testing a condition and executing only one of two possible values - if the condition is true the first part will get executed, if it's false the second part will get executed:
(if (= 1 0)
"not executed"
"executed")
=> "executed"
or is a logical connector that can receive multiple arguments, it'll return the first non-false value it finds, or false if all values are false. Typically you'd use or to connect boolean expressions, but that's not always the case:
(or #f 7 10)
=> 7
(or (= 1 0) (= 1 2))
=> #f
Also bear in mind that in Scheme all values are considered true, except #f, which is false. Regarding the last part of your question: yes, you can use if to simulate an or, it would be something like this:
; equivalent to (or value1 value2)
(if value1
value1
value2)
The above is a simplification, because ideally we should evaluate value1 exactly once, and also because this version doesn't support multiple arguments. But beware: the moment you try to write it as a procedure you'll run into trouble!
(define (my-or value1 value2)
(if value1
value1
value2))
That will not work as an or, the procedure evaluates both arguments before evaluating the if, whereas a real or only evaluates the needed expressions until it finds a true value - that's what we call short-circuit evaluation. To see what I mean, try this:
(or 42 (/ 1 0))
=> 42
(my-or 42 (/ 1 0))
=> /: division by zero
We can't truly implement or, if as procedures, they're special forms and need to be implemented at the interpreter level, with special evaluation rules.
answered Nov 16 '18 at 9:07
Óscar LópezÓscar López
176k24225322
edited Nov 16 '18 at 9:26
answered Nov 16 '18 at 9:07
Óscar LópezÓscar López
176k24225322
answered Nov 16 '18 at 9:07
Óscar LópezÓscar López
176k24225322
answered Nov 16 '18 at 9:07
Óscar LópezÓscar López
176k24225322
176k24225322
add a comment |
add a comment |
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