有没有办法只将一个参数传入args并让其他值在TypeScript中默认?我不想要" args = {}"并且由于intellisense而声明函数内的默认值。
function generateBrickPattern (
wallWidth: number,
wallHeight: number,
args = {
maxBrickWidth: 100,
maxBrickHeight: 50,
minBrickWidth: 50,
minBrickHeight: 25
}) {}
generateBrickPattern(500,500,{maxBrickWidth: 75}) //Prefered
generateBrickPattern(500,500,{maxBrickWidth: 75,
maxBrickHeight: 50,
minBrickWidth: 50,
minBrickHeight: 25}) //Not wanted
首选语法会出现以下错误。
类型的参数' {maxBrickWidth:number; }'不能转让给 参数类型' {maxBrickWidth:number; maxBrickHeight:number; minBrickWidth:number; minBrickHeight:number; } ...'
答案 0 :(得分:1)
您必须实际定义最后一个参数的类型,而不是让TypeScript推断它。
试试这个:
interface BrickPatternOptions {
maxBrickWidth?: number;
maxBrickHeight?: number;
minBrickWidth?: number;
minBrickHeight?: number;
}
function generateBrickPattern (
wallWidth: number,
wallHeight: number,
args: BrickPatternOptions = {
maxBrickWidth: 100,
maxBrickHeight: 50,
minBrickWidth: 50,
minBrickHeight: 25
}) {}
或者,如果您想要,也可以内联它:
function generateBrickPattern (
wallWidth: number,
wallHeight: number,
args: {
maxBrickWidth?: number,
maxBrickHeight?: number,
minBrickWidth?: number,
minBrickHeight?: number
} = {
maxBrickWidth: 100,
maxBrickHeight: 50,
minBrickWidth: 50,
minBrickHeight: 25
}) {}
答案 1 :(得分:0)
您可以通过使用默认值解析args
来执行此操作:
function generateBrickPattern (
wallWidth: number,
wallHeight: number,
{
maxBrickWidth: maxBrickWidth = 100,
maxBrickHeight: maxBrickHeight = 50,
minBrickWidth: minBrickWidth = 50,
minBrickHeight: minBrickHeight = 25
} = {}
) {
console.log(maxBrickWidth);
}
如果您不想进行结构化,则可以将提供的args
与默认值合并为:
interface BrickPatternOptions {
maxBrickWidth: number;
maxBrickHeight: number;
minBrickWidth: number;
minBrickHeight: number;
}
function generateBrickPattern (
wallWidth: number,
wallHeight: number,
args: Partial<BrickPatternOptions> = {}
) {
const options: BrickPatternOptions = {
maxBrickWidth: 100,
maxBrickHeight: 50,
minBrickWidth: 50,
minBrickHeight: 25,
...args
};
console.log(options.maxBrickWidth);
}