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if-statementfunctional-programmingschemeconditional-statements
URL87
URL87

Reputation: 11022

Scheme operations within if condition

Generally in Scheme - syntax: (if test consequent alternate) .

I trying to do some operations within the consequent section , for example : if true then set sum1 on 44 and return 1 .
In Scheme code its - 

(if #t ( 
          (set! sum1 44)
          (if #t 1)
          ))

The above code doesn't works out and prompts - 

application: not a procedure;
 expected a procedure that can be applied to arguments
  given: #<void>
  arguments...:
   1

Upvotes: 2

Views: 705

Answers (3)

&#211;scar L&#243;pez
&#211;scar L&#243;pez

Reputation: 236034

Some interpreters (Racket, for instance) complain if you write an if without an else part. For dealing with this, and for avoiding the need to explicitly use a begin when more than one expression goes in the consequent part, it's better to use the when special form (if available, because it's non-standard):

(when #t
  (set! sum1 44)
  (when #t 1))

In general, this is the structure of a when and its sibling unless:

(when <condition>    ; equivalent to (if <condition> (begin ...))
  <exp1>             ; consequent, no alternative
  <exp2>)

(unless <condition>  ; equivalent to (if (not <condition>) (begin ...))
  <exp1>             ; consequent, no alternative
  <exp2>)

If both the consequent and the alternative have more than one expression, you could use if with a begin in each part:

(if <condition>
    (begin     ; consequent
      <exp1>
      <exp2>)
    (begin     ; alternative
      <exp3>
      <exp4>))

... But it'd be more practical to use a cond, which implicitly uses a begin in each of its clauses:

(cond (<condition>
       <exp1>  ; consequent
       <exp2>)
      (else    ; alternative
       <exp3>
       <exp4>))

Upvotes: 3

Dirk
Dirk

Reputation: 31053

(if #t (begin 
          (set! sum1 44)
          (if #t 1)))

Upvotes: 3

John Bartholomew
John Bartholomew

Reputation: 6596

In Scheme, parentheses are always used for application; you can't add extra parentheses for grouping.

Your consequent here is:

((set! sum1 44) (if #t 1))

The outer pair of parentheses makes Scheme attempt to use the result of (set! sum1 44) as a procedure, applying it to the result of (if #t 1).

What (I think) you want is to evaluate the two expressions in sequence, and then return the result of the last one. The form to do this is begin, so it should be:

(begin (set! sum1 44) (if #t 1))

Upvotes: 4

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