我想简化与ImmutableList.of()功能相关的现有代码
我已经尝试通过消除“new ...”构造函数来优化第二个List的创建,但当然我无法通过调用.add()来扩展不可变列表。
当前代码:
static final ImmutableList<ProductCodeEnum> PRODUCTS = ImmutableList.of(ProductCodeEnum.A, ProductCodeEnum.B, ProductCodeEnum.C);
static final ImmutableList<ProductCodeEnum> PRODUCTS_EXTENDED_LIST = new ImmutableList.Builder<ProductCodeEnum>().addAll(PRODUCTS)
.add(ProductCodeEnum.D)
.add(ProductCodeEnum.E)
.build();
预期代码如:
static final ImmutableList<ProductCodeEnum> PRODUCTS = ImmutableList.of(ProductCodeEnum.A, ProductCodeEnum.B, ProductCodeEnum.C);
static final ImmutableList<ProductCodeEnum> PRODUCTS_EXTENDED = PRODUCTS + ImmutableList.of(ProductCodeEnum.D, ProductCodeEnum.E);
我想你用的是Guava ImmutableList?
在这种情况下,您的代码将如下所示:
static final ImmutableList<ProductCodeEnum> PRODUCTS = ImmutableList.of(ProductCodeEnum.A, ProductCodeEnum.B, ProductCodeEnum.C);
static final ImmutableList<ProductCodeEnum> PRODUCTS_EXTENDED = ImmutableList.builder().addAll(PRODUCTS).add(ProductCodeEnum.D, ProductCodeEnum.E).build();