如何在Javascript中使用在参数中创建的函数？

Question

我试图将此deepMap([4,5, [3,4,[2]]], x => x + 5)传递给此函数：

const deepMap = (arr, fn) => {
  return arr.reduce((first, second) => first.concat(Array.isArray(second) ? [deepMap(second)] : fn), []);  
}

之前它的工作原理如下：

const deepMap = (arr, fn) => {
  return arr.reduce((first, second) => first.concat(Array.isArray(second) ? [deepMap(second)] : second + 5), []);  
}

但这并不允许我在第二个参数中使用该功能。我知道它需要像第二个例子一样工作，但我能想到改变它的唯一方法就像在第一个例子中一样。我通过实验尝试了很多变化，但我一直收到错误或错误答案。

Answer 1

你应该放置函数的返回值而不是函数本身。像这样：

const deepMap = (arr, fn) => {
  return arr.reduce((first, second) => first.concat(Array.isArray(second) ? [deepMap(second, fn)] : fn(second)), []);  
}

Answer 2

将数组和函数传递给递归// // NSDecimalNumberBugTests.m // // Created by Lane Roathe on 6/1/17. // For Quicken, Inc. // #import <XCTest/XCTest.h> @interface NSDecimalNumberBugTests : XCTestCase @end @implementation NSDecimalNumberBugTests - (void)setUp { [super setUp]; // Put setup code here. This method is called before the invocation of each test method in the class. } - (void)tearDown { // Put teardown code here. This method is called after the invocation of each test method in the class. [super tearDown]; } - (void)testBug { // Use XCTAssert and related functions to verify your tests produce the correct results. NSDecimalNumber* decimalLength; NSUInteger interval; // Start with a number that requires 65+ bits // This FAILS (interval is zero) decimalLength = [NSDecimalNumber decimalNumberWithString:@"1.8446744073709551616"]; interval = decimalLength.unsignedIntegerValue; XCTAssert(interval == 1); // This Works, interval is 1 interval = decimalLength.unsignedIntValue; XCTAssert(interval == 1); // Now test with a number that fits in 64 bits // This WORKS (interval is 1) decimalLength = [NSDecimalNumber decimalNumberWithString:@"1.8446744073709551615"]; interval = decimalLength.unsignedIntegerValue; XCTAssert(interval == 1); } @end 调用并调用fn（cur）而不是仅指向它。

deepMap