Scheme - eq? compare between 2 strings?

I have a problem in my program.

I have a condition that compare between 2 string:开发者_如何学Go

(if (eq? (exp1) (exp2)))

When exp1 give me a string, and exp2 give me a string. To be sure, when I change the "eq?" to "=", it give me the next problem:

=: expects type <number> as 2nd argument, given: ie; other arguments were: ie.

When I'm running the program, the function doesnt enter to the first expression in the "if" function, and enter to the second one (meaning the condition is false).

What can I do?

Thank you.

According to the Equivalence predicates section of R6RS, you should be using equal?, not eq?, which instead tests whether its two arguments are exactly the same object (not two objects with the same value).

(eq? "a" "a")                           ; unspecified
(equal? "abc" "abc")                    ; #t

As knivil notes in a comment, the Strings section also mentions string=?, specifically for string comparisons, which probably avoids doing a type check.

I wrote a little helper function for this problem.

; test if eq?
(define ==
  (lambda (x y)
    (if (and (string? x) (string? y))
      (string=? x y)
      (if (or (string? x) (string? y))
            (= 1 0) ;return false
            (equal? x y)))))
(define a "aString")
(define l '("aString" "aOtherString"))
(== (car l) a) ; true
(== 1 1) ; true
(== 1 0) ; false
(== "a" 1) ; false diff. type
(== "a" "b") ; false
(== "a" "a") ; true
(== '("a" "b") '("a" "b"))