ES6 Array Destructuring: Why Doesn\'t It Work As Expected?

ES6 Array Destructuring: Unforeseen Behavior

In ES6, destructuring assignment for arrays can lead to unexpected results, leaving programmers puzzled. One such instance is illustrated by the following code:

<code class="js">let a, b, c
[a, b] = ['A', 'B']
[b, c] = ['BB', 'C']
console.log(`a=${a} b=${b} c=${c}`)</code>

Intended Output:
a=A b=BB c=C

Actual Output:
a=BB b=C c=undefined

Explanation:

Contrary to expectations, this code does not yield the desired output. Instead, it swaps the values of b and c, leaving c undefined. To understand why this happens, we need to examine the code closely.

Parsing and Evaluation:

In JavaScript, semicolons are optional to delimit statements. Without explicit semicolons, the code is parsed as a single statement:

<code class="js">let a = undefined, b = undefined, c = undefined;
[a, b] = (['A', 'B']
[(b, c)] = ['BB', 'C']);
console.log(`a=${a} b=${b} c=${c}`);</code>

Breakdown of the Statement:

• [a, b] = (['A', 'B'] is a destructuring assignment, similar to the one in the original code.
• [(b, c)] = ['BB', 'C'] is an assignment expression that assigns the array ['BB', 'C'] to the left-hand operand. This expression evaluates to the same array.
• ['A', 'B'][…] is a property reference on an array literal, which evaluates to undefined.
• (b, c) uses the comma operator, which evaluates to the last operand (c), which is undefined.

Implications:

Therefore, the code assigns undefined to both a and c, while b correctly receives the value 'C'. To avoid this behavior, programmers should explicitly use semicolons or begin every line with an operator that requires a semicolon to be automatically inserted (e.g., (, [, /, , -, or `).

This understanding ensures that destructuring assignments in ES6 operate as expected, preventing unexpected value swaps and undefined assignments.

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