模块内部动态定义的 function 调用

Question

'interpret'中有一个代码，问题：如何从'func'内部调用任何给定的function，而不仅仅是'sum3'？

const sum = (...args) => {
    return args.reduce((prev, curr) => prev + curr)}

const defn = (functionName, args, body) => {
    let [func, ...params] = body;
    let result = `var ${functionName} = function(${args.join(',')}) { return ${func.name}(${params.join(',')})}`
    return result;
}

const interpret = (...code) => {
    let [dfStr, callString] = code;
    let [func, ...args] = callString;
    eval(dfStr[0](dfStr[1], dfStr[2], dfStr[3] ));
    return sum3.apply(this, args);
    }


const result = interpret(
    [defn, "sum3", ['a', 'b', 'c'], [sum, 'a', 'b', 'c']],
    ['sum3', 10, 20, 30]
)

Answer 1

请注意，您评估的是 function 定义，而不是相应的调用，

看一下这个：

 const sum = (...args) => { return args.reduce((prev, curr) => prev + curr) } const defn = (functionName, args, body) => { let [func, ...params] = body; let result = `var ${functionName} = function(${args.join(',')}) { return ${func.name}(${params.join(',')})}` return result; } const interpret = (...code) => { let [dfStr, callString] = code; let [func, ...args] = callString; eval(dfStr[0](dfStr[1], dfStr[2], dfStr[3])); return eval(`${func}.apply(this, args)`); } const result = interpret( [defn, "sum3", ['a', 'b', 'c'], [sum, 'a', 'b', 'c'] ], ['sum3', 10, 20, 30] ) console.log(result)

TBH 我不知道您要做什么，但请记住，使用 eval 是一种邪恶的做法。 也许你可以想出一个更清洁的解决方案来满足这个目的。

更新：

 const sum = (...args) => { return args.reduce((prev, curr) => prev + curr) } const defn = (functionName, args, body) => { let [func, ...params] = body; let result = `(${args.join(',')}) =>${func.name}(${params.join(',')})` return result; } const interpret = (...code) => { let [dfStr, callString] = code; let [func, ...args] = callString; return eval(dfStr.shift()(...dfStr)).apply(this, args) } const result = interpret( [defn, "sum3", ['a', 'b', 'c'], [sum, 'a', 'b', 'c'] ], ['sum3', 10, 20, 30] ) console.log(result)