I am reading SICP in JS about a non-terminating example of ternary conditions:

function is_even(n) { return n % 2 === 0; } function expmod(base, exp, m) { const half_exp = expmod(base, exp / 2, m); return exp === 0 ? 1 : is_even(exp) ? half_exp * half_exp % m : base * expmod(base, exp - 1, m) % m; } console.log(expmod(4, 3, 5))

It explains that:

This would make the function not just inefficient, but actually non-terminating! The problem is that the constant declaration appears outside the conditional expression, which means that it is executed even when the base case exp === 0 is met.

I just cannot get its idea, when exp === 0, it terminate with 1 but why half_exp executed?

## Answer

The part you’ve misunderstood is how and when the variables are initialized, and not how the ternary works. The ternary would work as you thought, *if the interpreter had reached it*.

You’ve put the `half_exp`

variable in a conditional expression and expected it to not evaluate its initializer until it is used.

**However, that’s not how it works.**

All variable initialization statements (both `var`

, `let`

and `const`

) **evaluate their initializer immediately** when the control reaches the statement, without checking whether the variable is used later; and store the * value of the initializer* to the variable.

You can see it by running the following snippet:

const foo = console.log("I'm executed!") //`foo` is never used, but the code will print "I'm executed!" anyway

You can also confirm this by looking at the ECMAScript Specification.

LexicalBinding:BindingIdentifierInitializer

Let

bindingIdbe StringValue ofBindingIdentifier.Let

lhsbe ResolveBinding(bindingId).If IsAnonymousFunctionDefinition(

Initializer) istrue, thena. Let

valuebe NamedEvaluation ofInitializerwith argumentbindingId.Else,

a.

Let.rhsbe the result of evaluatingInitializer *

b. Letvaluebe ? GetValue(rhs).Return InitializeReferencedBinding(

lhs,value).

_{*: Emphasis mine.}

So, as you can see, the interpreter won’t wait for the variable to be used.

This means that in your code:

// v-------------------------------------------+ function expmod(base, exp, m) { // | const half_exp = expmod(base, exp / 2, m); // ---+ // ^^^^^^--- This will always be called // This line is not even reached! return exp === 0 ? 1 : is_even(exp) ? half_exp * half_exp % m : base * expmod(base, exp - 1, m) % m; }

…you have infinite recursion.

To get around the issue, we have to move that call into a conditional part. In your code, that’s easy, as instead of writing a multiplication with itself, we can raise the value to its second power, eliminating one of the references:

function is_even(n) { return n % 2 === 0; } function expmod(base, exp, m) { return exp === 0 ? 1 : is_even(exp) ? expmod(base, exp / 2, m) ** 2 % m : base * expmod(base, exp - 1, m) % m; } console.log(expmod(4, 3, 5)) //4

In other cases, where there’s no such straightforward way, we could’ve refactored the code otherwise, for example, using `if`

s:

function is_even(n) { return n % 2 === 0; } function expmod(base, exp, m) { if(exp === 0) return 1; if(is_even(exp)){ // We are in a conditional statement, so it's safe to call: const half_exp = expmod(base, exp / 2, m) return half_exp * half_exp % m } return base * expmod(base, exp - 1, m) % m; } console.log(expmod(4, 3, 5)) //4