如何将 arguments 从 javascript function 传递给另一个 ZC1C425268E68385D1AB5074C17A94F 而不调用它？

Question

我试图在计算后将变量从 function 传递到另一个 function，方法是将其作为参数传递。 但是当我这样做时，function 会立即被调用。

 let myarr = [2,4,6,8]; function checkaverage() { let sum = 0; for (let i=0; i<myarr.length; ++i) { sum = sum + myarr[i]; } let averageis = sum/(myarr.length) averagealerter(averageis); } checkaverage(); function averagealerter(averageis) { alert("the average is "+ averageis) };

在这里，我将“averageis”变量作为参数传递给 function“averagealerter”以供以后使用。 但它会立即被调用。 我想要完成的是稍后使用“averagealerter()”来调用它。

Answer 1

let myarr = [2,4,6,8];

function checkaverage() {
  let sum = 0;
  for (let i=0; i<myarr.length; ++i) {
     sum = sum + myarr[i];
     } 
 let averageis = sum/(myarr.length)

 return averageis;
}

function averagealerter() {
  alert("the average is "+ checkaverage())
}

// call the averagealerter() function when required.

Answer 2

您可以将.bind变量绑定到averageis并返回averagealerter的副本。

然后，调用 function 来实现结果：

 let myarr = [2,4,6,8]; function checkaverage() { let sum = 0; for (let i=0; i<myarr.length; ++i) { sum = sum + myarr[i]; } let averageis = sum/(myarr.length) return averagealerter.bind(undefined,averageis); } function averagealerter(averageis) { alert("the average is "+ averageis) }; let myaveragealerter = checkaverage(); myaveragealerter();

Answer 3

您需要做的就是使 checkAverage function 返回一个值，以便该值可以通过调用 function 来用于其他函数，如下所示

 let myarr = [2,4,6,8]; function checkaverage() { let sum = 0; for (let i=0; i<myarr.length; ++i) { sum = sum + myarr[i]; } let averageis = sum/(myarr.length) return averageis; } function averagealerter() { alert("the average is "+ checkaverage()) }; averagealerter()