为什么当我尝试将函数的返回值与字符串连接时我的代码不起作用？

Question

So, in this code I have a string of 0's and 1's and the length of the string is 32, which will be split in 6 equal parts but the last part will have the length of 2 so I will add (4) 0's after that which will make its length 6. So I wrote a function that will add the remaining 0's which is padding(num). And that function will be invoked in side the slicing(str) function. But the code breaks when I try to do execute. Any help? Thanks.

// This code works.


function padding0s(num) {
  let s = "";
  for (i = 0; i < 6 - num; i++) {
    s += "0";
  }
  return s;
}

function slicing(str) {
  let k = 6;
  let res = [];
  let temp1 = 0;
  let f = padding0s(2);
  for (i = 0; i < str.length; ) {
    res.push(str.slice(i, k));
    i += 6;
    k += 6;
    if (res[temp1].length !== 6) {
      res[temp1] += f;
    }
    temp1++;
  }
  console.log(res);
}

slicing("01000011010011110100010001000101");
    

// But this does not..


function padding0s(num) {
  let s = "";
  for (i = 0; i < 6 - num; i++) {
    s += "0";
  }
  return s;
}

function slicing(str) {
  let k = 6;
  let res = [];
  let temp1 = 0;
  
  for (i = 0; i < str.length; ) {
    res.push(str.slice(i, k));
    i += 6;
    k += 6;
    if (res[temp1].length !== 6) {
     let f = padding0s(res[temp1].length);
      res[temp1] += f;
    }
    temp1++;
  }
  console.log(res);
}

slicing("01000011010011110100010001000101");

Answer 1

Always define variables before using them

Not doing so can result in undefined behaviour, which is exactly what is happening in your second case. Here is how:

for (i = 0; i < str.length; ) {...}
  // ^ Assignment to undefined variable i

In the above for-loop, by using i before you define it, you are declaring it as a global variable. But so far, so good, as it doesn't matter, if not for this second problem. The real problem is the call to padding0s() in your loop. Let's look at padding0s :

function padding0s(num) {
  ...
  for (i = 0; i < 6 - num; i++) {
    s += "0";
  }
}

This is another loop using i without defining it. But since i was already defined as a global variable in the parent loop, this loop will be setting its value. So in short, the value of i is always equal to 6 - num in the parent loop. Since your exit condition is i < str.length , with a string of length 32 the loop will run forever.

You can get around this in many ways, one of which you've already posted. The other way would be to use let i or var i instead of i in the parent loop. Even better is to write something like this (but beware that padEnd may not work on old browsers ):

 function slicing(str) { return str.match(/.{1,6}/g).map((item) => { return item.padEnd(6, "0"); }); } console.log(slicing("01000011010011110100010001000101"));