Why doesn't a string based template variable change in Kotlin?

Question

I know I can embed a var such as $aa in the string k expression, so line 3 displays the result 10.

I think the var k will change when I change the value of aa, so I think the line 5 will display the result 15, but in fact it displays 10. why?

1  var aa=10
2  var k="$aa"
3  toast(k)  //This is  10

4  aa=aa+5
5  toast(k)   //This is  10, I think it should be 15

6  k="$aa"
7  toast(k)   //This is  15

Added Content

I think the system will recalculate line 5 in the following code, so I think the new value aa 15 will be embed into k, but in fact , I get the result "10 bb", why?

1 var aa=10
2 var k="$aa"
3 toast(k)    //This is  10

4  aa=aa+5
5  k=k+" bb"   //I think system will recalculate, so I think new value 15 will be embed into k
6  toast(k)   //This is  "10 bb", I think it should be "15 bb"

What's hard to understand in below answer? Strings are immutable. Period. The system itself cannot detect you updated a value. The string value of 10 is assigned to k, it's not always a templated variable — OneCricketeer
– OneCricketeer, Commented Jul 1, 2018 at 3:55

stephenjfox · Accepted Answer · 2018-07-01 02:15:52Z

2

Strings are immutable and the reference k points to the specific string "10". That string is made up of the value of the content of your variable aa.

If k were to change with mutation to aa would demand a few things:

Strings are partially mutable (which has significant speed impacts)
every templated string would have to be recomputed on nearly every access
the scope of its creation would have to be closed around (see "closure") so the variables could be kept in reference
Subsequently, there would be phantom objects on the heap.

That is a lot of computational overhead to display a string.

To connect my two points, the last line of your example evaluates as you expect because your variable now points to a different string "15", because the value that the variable aa holds is different than previously at string construction time.

answered Jul 1, 2018 at 2:15

stephenjfox

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4 Comments

HelloCW Over a year ago

Thanks! but what you mean is hard to understand.

HelloCW Over a year ago

Would you please see my added content in my question?

stephenjfox Over a year ago

Immutable strings simply won't be recomputed. I was long-winded in answering.

stephenjfox Over a year ago

If you want a template to be regularly regenerated, evaluate a method or property accessor

Geno Chen · Accepted Answer · 2018-07-01 06:02:23Z

2

Kotlin doesn't have "lazy evaluation", and var k = "$aa" is not a function call, it's just an evaluation and instantly assignment. Then when you ask var k = "$aa", "$aa" is evaluated, and k is assigned.

After that, k will "forget" about how it was evaluated, so you do aa = aa + 5 will have no effect on k.

For the "Added Content", regarding the same thought of above, k has been "forgot" how it was evaluated, assigning k = k + " bb" will just directly add the existing k (program level joining, not memory level concatenate) with string " bb", not re-evaluate k with "$aa".

answered Jul 1, 2018 at 6:02

Geno Chen

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Comments

zsmb13 · Accepted Answer · 2018-07-01 05:28:38Z

When you write down this line:

var k = "$aa"

... the string template on the right side of the assignment is evaluated immediately, and what gets assigned to k is the result of the template itself, the literal "10" string. That line is the same as if you had and of these in Java - and of course these won't change automatically later as aa gets updated either:

String k = "" + aa;

String k = String.valueOf(aa);

String k = new StringBuilder().append(aa).toString();

To recompute this value every time, you could introduce a local function that you call over and over to create new strings every time, with aa's then current value inserted into the template:

var aa = 10
fun k() = "$aa" // same as: fun k(): String { return "$aa" }
println(k()) // 10

aa += 5
println(k()) // 15

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