You didn't specify where in the loop, but I'm going to assume the obvious :)
in a for loop, you are creating a variable to count the iterations, and this variable is going to change, thus it can't be a constant value.
for(leti=0;i<5;i++){}
if i were a const, it would throw an error at the end of the first iteration, when i++ was executed.
on for ... of loops, you create a variable to keep the value of the iterable (list) to use during that iteration of the loop... when it loops over, a new variable will be created, with the same name and different value. it does not try to reassign, it just releases the variable that was used in the last iteration and creates a whole new one.
for(constvalueof[1,2,3,4]){}
the value variable will only exist until the end of each iteration.
A new lexical scope is created for each iteration and each lexical closure has its own copy of i.
for ([initialization]; [condition]; [final-expression])
statement
initialization initializes the loop environment‡.
final-expression runs at the beginning of each iteration except the first one.
before condition is checked a new lexical environment is created for the iteration and the current loop environment (i) is copied into that then condition is checked. Continue if satisfied, break if not.
statement is executed.
the lexical environment is copied back to the loop environment.
"final-expression runs at the beginning of each iteration except the first one. "
mdn says: final-expression (called afterthought in their docs) evaluates at the end of each loop iteration not at the beginning of each iteration.
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You didn't specify where in the loop, but I'm going to assume the obvious :)
in a
forloop, you are creating a variable to count the iterations, and this variable is going to change, thus it can't be a constant value.if
iwere aconst, it would throw an error at the end of the first iteration, wheni++was executed.on
for ... ofloops, you create a variable to keep the value of the iterable (list) to use during that iteration of the loop... when it loops over, a new variable will be created, with the same name and different value. it does not try to reassign, it just releases the variable that was used in the last iteration and creates a whole new one.the
valuevariable will only exist until the end of each iteration.Thanks Guilherme
The
letin the modernforloop is actually a bit disingenuous.The early version of the
forloop worked like this:Unintuitive but correct as by the time
console.log()executediwas2.Typically the intention is this:
The
letversion automatically works that wayA new lexical scope is created for each iteration and each lexical closure has its own copy of
i.initializationinitializes the loop environment‡.final-expressionruns at the beginning of each iteration except the first one.conditionis checked a new lexical environment is created for the iteration and the current loop environment (i) is copied into that thenconditionis checked. Continue if satisfied, break if not.statementis executed.‡
initializationhas its own lexical scope."final-expression runs at the beginning of each iteration except the first one. "
mdn says: final-expression (called afterthought in their docs) evaluates at the end of each loop iteration not at the beginning of each iteration.