I'm just trying to learn to use the debugger.
Here's a function:
(dolist (i (explode "string"))
(println {i is } i)
(for (j 0 3)
(println { j is } j)
(if (= j 2)
(trace true))
))
)
If i just run this, it does this:
i is s
j is 0
j is 1
j is 2
j is 3
i is t
j is 0
j is 1
j is 2
j is 3
i is r
j is 0
j is 1
j is 2
j is 3
i is i
j is 0
j is 1
j is 2
j is 3
i is n
j is 0
j is 1
j is 2
j is 3
i is g
j is 0
j is 1
j is 2
j is 3
nil
0>
- I was expecting it to stop a bit earlier, when j was 2. Of course, I hadn't started it using (debug ...). Is that required?
But if I go (debug (t)), and then press c (for cont) it runs through without stopping. Again, I was expecting it to stop earlier. Is my understanding of what it does wrong?
probably you have hit n/next instead of s/step. Also do this in an interactive termial. When using s/step these are the steps:
> (debug (t))
-----
(define (t )
#(dolist (i (explode "string"))
(println "i is " i)
(for (j 0 3)
(println " j is " j)
(if (= j 2)
(trace true))))#)
[-> 3 ] s|tep n|ext c|ont q|uit > s
-----
(define (t )
(dolist (i #(explode "string")#)
(println "i is " i)
(for (j 0 3)
(println " j is " j)
(if (= j 2)
(trace true)))))
[-> 4 ] s|tep n|ext c|ont q|uit > s
-----
(define (t )
(dolist (i #(explode "string")#)
(println "i is " i)
(for (j 0 3)
(println " j is " j)
(if (= j 2)
(trace true)))))
RESULT: ("s" "t" "r" "i" "n" "g")
[<- 4 ] s|tep n|ext c|ont q|uit > s
-----
(define (t )
(dolist (i (explode "string"))
#(println "i is " i)#
(for (j 0 3)
(println " j is " j)
(if (= j 2)
(trace true)))))
[-> 4 ] s|tep n|ext c|ont q|uit > s
i is s
-----
(define (t )
(dolist (i (explode "string"))
#(println "i is " i)#
(for (j 0 3)
(println " j is " j)
(if (= j 2)
(trace true)))))
RESULT: "s"
[<- 4 ] s|tep n|ext c|ont q|uit > s
-----
(define (t )
(dolist (i (explode "string"))
(println "i is " i)
#(for (j 0 3)
(println " j is " j)
(if (= j 2)
(trace true)))#))
[-> 4 ] s|tep n|ext c|ont q|uit > s
-----
(define (t )
(dolist (i (explode "string"))
(println "i is " i)
(for (j 0 3)
#(println " j is " j)#
(if (= j 2)
(trace true)))))
[-> 5 ] s|tep n|ext c|ont q|uit > s
j is 0
-----
(define (t )
(dolist (i (explode "string"))
(println "i is " i)
(for (j 0 3)
#(println " j is " j)#
(if (= j 2)
(trace true)))))
RESULT: 0
[<- 5 ] s|tep n|ext c|ont q|uit > s
-----
(define (t )
(dolist (i (explode "string"))
(println "i is " i)
(for (j 0 3)
(println " j is " j)
#(if (= j 2)
(trace true))#)))
[-> 5 ] s|tep n|ext c|ont q|uit > s
-----
(define (t )
(dolist (i (explode "string"))
(println "i is " i)
(for (j 0 3)
(println " j is " j)
(if #(= j 2)#
(trace true)))))
[-> 6 ] s|tep n|ext c|ont q|uit > s
-----
(define (t )
(dolist (i (explode "string"))
(println "i is " i)
(for (j 0 3)
(println " j is " j)
(if #(= j 2)#
(trace true)))))
RESULT: nil
[<- 6 ] s|tep n|ext c|ont q|uit > s
-----
(define (t )
(dolist (i (explode "string"))
(println "i is " i)
(for (j 0 3)
(println " j is " j)
#(if (= j 2)
(trace true))#)))
RESULT: nil
[<- 5 ] s|tep n|ext c|ont q|uit > s
-----
(define (t )
(dolist (i (explode "string"))
(println "i is " i)
(for (j 0 3)
#(println " j is " j)#
(if (= j 2)
(trace true)))))
[-> 5 ] s|tep n|ext c|ont q|uit >
The arrows -> or <- show if you are entering or exiting the code segment marked by hash symbols #
Lutz
Hi Lutz!
What I was trying to do was to run for a while normally and then stop at a 'breakpoint'. Reducing it down to this simple example - how do i get it to run at full speed until the moment when trace is set, when j takes a certain value?
Edit: ah, I think I see the issue: the (trace true) drops you into the next function, rather than stopping straight away. Is that a fair assumption?
Quote
Yes, this is how it is supposed to work, the documentation is misleading.
Lutz