函數作為參數進行傳遞

1 參數函數沒有參數，按照正常流程執行

demo：

function fun1(someFun){

console.log('this is from fun1');

someFun();

}

function fun2(){

console.log('this is from fun2');

}

fun1(fun2)

 

 

 1 參數函數帶參數，參數函數直接就執行了，在主函數中再次調用就會undifined

demo：

function fun1(someFun){

console.log('this is from fun1');

someFun();

}

function fun2(num){

console.log('this is from fun2');

console.log(num);

}

fun1(fun2(100))

 解決辦法，直接把參數和函數名作為兩個參數進行傳遞：

demo:

function fun1(someFun,somedata){

console.log('this is from fun1');

someFun(somedata);

}

function fun2(num){

console.log('this is from fun2');

console.log(num);

}
fun1(fun2,1000)