Harnessing Spread Syntax with Map and Filter Operations

Question

Harnessing Spread Syntax with Map and Filter Operations

Recently stumbled upon a fascinating problem that I couldn't wait to share with all of you. Here is the question at hand:

[...[0,1,...[-1,0,1].map((x)=> x+1)].filter((x)=>x)),7]

I managed to successfully solve the initial section up to the filter part resulting in

[0,1,(-1+1),(0+1),(1+1)] = [0,1,0,1,2]

.

However, the unexpected appearance of 7 at the end caught me off guard. At first, I assumed it might be a mistake but after copying and pasting the problem into the console, the output was [1,1,2,7]. This left me pondering two key questions.

Why were the 0's omitted from the filter?
What purpose does the 7 serve within this problem?

javascript arrays ecmascript-6 functional-programming spread-syntax

Answer 1

Answer №1

Starting off with the operation Array#map [-1, 0, 1].map(x => x + 1), where each element in the array has 1 added to it resulting in [0, 1, 2].
Following that is the Array#filter operation,
```
[0, 1, ...[0, 1, 2]].filter(x => x)
```
, which produces a new array without any falsy values like false, 0, undefined, null, or an empty string.
The final operation involves [...[1, 1, 2], 7] and utilizes the spread operator to remove the nested array, resulting in [1, 1, 2, 7].

Answer 2

Starting off with the operation Array#map [-1, 0, 1].map(x => x + 1), where each element in the array has 1 added to it resulting in [0, 1, 2].
Following that is the Array#filter operation,
```
[0, 1, ...[0, 1, 2]].filter(x => x)
```
, which produces a new array without any falsy values like false, 0, undefined, null, or an empty string.
The final operation involves [...[1, 1, 2], 7] and utilizes the spread operator to remove the nested array, resulting in [1, 1, 2, 7].

Answer 3

Answer №2

Explaining the breakdown of [0,1,...[-1,0,1].map((x)=> x+1)].filter((x)=>x),7:

[-1,0,1].map((x)=> x+1) // [0,1,2]

[0,1,...[-1,0,1].map((x)=> x+1)] // [0,1,0,1,2]

[0,1,...[-1,0,1].map((x)=> x+1)].filter((x)=> x) // [1,1,2]

[...[0,1,...[-1,0,1].map((x)=> x+1)].filter((x)=> x),7] // [1,1,2,7]

Answer 4

Explaining the breakdown of [0,1,...[-1,0,1].map((x)=> x+1)].filter((x)=>x),7:

[-1,0,1].map((x)=> x+1) // [0,1,2]

[0,1,...[-1,0,1].map((x)=> x+1)] // [0,1,0,1,2]

[0,1,...[-1,0,1].map((x)=> x+1)].filter((x)=> x) // [1,1,2]

[...[0,1,...[-1,0,1].map((x)=> x+1)].filter((x)=> x),7] // [1,1,2,7]

Answer 5

Answer №3

Running the code snippet [-1,0,1].map((x)=> x+1) transforms the array into [0,1,2]. Subsequently, applying this logic to another array

[0,1,...[-1,0,1].map((x)=> x+1)]

yields [0,1,1,2]. Removing the element 0 after a filtering process leaves us with [1,1,2], and the final element of the updated list is 7. Therefore, the end result is [1,1,2,7].

Answer 6

Running the code snippet [-1,0,1].map((x)=> x+1) transforms the array into [0,1,2]. Subsequently, applying this logic to another array

[0,1,...[-1,0,1].map((x)=> x+1)]

yields [0,1,1,2]. Removing the element 0 after a filtering process leaves us with [1,1,2], and the final element of the updated list is 7. Therefore, the end result is [1,1,2,7].

Answer 7

Answer №4

The code undergoes evaluation through the following steps:

[...[0, 1, ...[-1, 0, 1].map((x)=>x+1)].filter((x)=>x)), 7] // mapping
[...[0, 1, ...[(x=>x+1)(-1), (x=>x+1)(0), (x=>x+1)(1)]].filter((x)=>x)), 7] // function application
[...[0, 1, ...[0, 1, 2]].filter((x)=>x)), 7] // spreading
[...[0, 1, 0, 1, 2].filter((x)=>x)), 7] // filtering
[...[...(x=>x)(0)?[0]:[], ...(x=>x)(1)?[1]:[], ...(x=>x)(0)?[0]:[], ...(x=>x)(1)?[1]:[], ...(x=>x)(2)?[2]:[]], 7] // function application
[...[...0?[0]:[], ...1?[1]:[], ...0?[0]:[], ...1?[1]:[], ...2?[2]:[]], 7] // conditional statement
[...[...[], ...[1], ...[], ...[1], ...[2]], 7] // spreading (from filtering)
[...[1, 1, 2], 7] // final spread
[1, 1, 2, 7]

Answer 8

The code undergoes evaluation through the following steps:

[...[0, 1, ...[-1, 0, 1].map((x)=>x+1)].filter((x)=>x)), 7] // mapping
[...[0, 1, ...[(x=>x+1)(-1), (x=>x+1)(0), (x=>x+1)(1)]].filter((x)=>x)), 7] // function application
[...[0, 1, ...[0, 1, 2]].filter((x)=>x)), 7] // spreading
[...[0, 1, 0, 1, 2].filter((x)=>x)), 7] // filtering
[...[...(x=>x)(0)?[0]:[], ...(x=>x)(1)?[1]:[], ...(x=>x)(0)?[0]:[], ...(x=>x)(1)?[1]:[], ...(x=>x)(2)?[2]:[]], 7] // function application
[...[...0?[0]:[], ...1?[1]:[], ...0?[0]:[], ...1?[1]:[], ...2?[2]:[]], 7] // conditional statement
[...[...[], ...[1], ...[], ...[1], ...[2]], 7] // spreading (from filtering)
[...[1, 1, 2], 7] // final spread
[1, 1, 2, 7]

Harnessing Spread Syntax with Map and Filter Operations

Answer №1

Answer №2

Answer №3

Answer №4

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