我有两个文本框。每个都将输入多达千位数。
现在我想添加这两个数字。我的问题是我应该使用什么数据类型来存储结果？
我试过这个：
<script>
var x = 'Thousand digit of number'
var y = 'Thousand digit of number'
var z = x + y
</script>
但我得到指数形式的结果。结果如何存储并显示？
【问题讨论】：
标签： javascript
<body>
<p>Click the button to calculate x.</p>
<button onclick="myFunction()">Try it</button>
<br/>
<br/>Enter first number:
<input type="text" id="txt1" name="text1">Enter second number:
<input type="text" id="txt2" name="text2">
<p id="demo"></p>
<script>
function myFunction() {
var y = document.getElementById("txt1").value;
var z = document.getElementById("txt2").value;
var x = +y + +z;
document.getElementById("demo").innerHTML = x;
}
</script>
【讨论】：
+y 和+z 会有所作为吗？
input up to thousand digits
要么使用像 https://mathjs.org/docs/datatypes/bignumbers.html 这样的大数字库，或者你可以使用像 http://www.discoversdk.com/knowledge-base/arbitrary-length-integer-addition-in-javascript 这样更轻量（但易于理解）的东西
【讨论】：
将数字输入为字符串并将每个字符相互添加为数组，如下所示：
function add() {
document.getElementById("demo").innerHTML = "";
var x = document.getElementById("txt1").value;
var y = document.getElementById("txt2").value;
var len;
var lenx = x.length;
var leny = y.length;
var x1,y1,rem,div=0;
if(lenx>leny) len = lenx; else len = leny;
for(var i=0;i<len;i++){
if(i>=lenx) x1 = 0;
else x1 = parseInt(x[lenx-i-1]);
if(i>=leny) y1 = 0;
else y1 = parseInt(y[leny-i-1]);
rem = (x1+y1+div)%10;
div = Math.floor((x1 + y1+div)/10);
document.getElementById("demo").innerHTML = rem + document.getElementById("demo").innerHTML;
}
if(div>0){
document.getElementById("demo").innerHTML = div + document.getElementById("demo").innerHTML;
}
}
这里的代码：https://jsfiddle.net/mtsL1k2x/5/
注意：这仅适用于自然数。您可以根据您的输入进行修改
【讨论】：
这是另一种解决方案，与您在互联网上找到的其他解决方案没有太大区别（请注意它不适用于负数！）：
function sums(arg1, arg2) {
var sum = "";
var r = 0;
var a1, a2, i;
// Pick the shortest string as first parameter and the longest as second parameter in my algorithm
if (arg1.length < arg2.length) {
a1 = arg1;
a2 = arg2;
}
else {
a1 = arg2;
a2 = arg1;
}
a1 = a1.split("").reverse();
a2 = a2.split("").reverse();
// Sum a1 and a2 digits
for (i = 0; i < a2.length; i++) {
var t = ((i < a1.length) ? parseInt(a1[i]) : 0) + parseInt(a2[i]) + r;
sum += t % 10;
r = t < 10 ? 0 : Math.floor(t / 10);
}
// Append the last remain
if (r > 0)
sum += r;
sum = sum.split("").reverse();
// Trim the leading "0"
while (sum[0] == "0")
sum.shift();
return sum.length > 0 ? sum.join("") : "0";
}
// Test
function testEquals(expected, actual) {
if (expected == actual)
console.log("OK: " + expected);
else
console.error("ERROR: " + expected + " != " + actual);
}
testEquals("100", sums("99", "1"));
testEquals("100", sums("00099", "0001"));
testEquals("10000000000", sums("9999999999", "1"));
testEquals("10000010101", sums("9999999999", "10102"));
testEquals("0", sums("0", "0"));
testEquals("1", sums("0", "1"));
testEquals("9", sums("8", "1"));
testEquals("9", sums("1", "8"));
testEquals("10000000000000000000000000000000000000000", sums("9999999999999999999999999999999999999999", "1"));【讨论】：
好吧，如果您想在不使用 BigInt 或任何第三方库的情况下执行此操作，那么我认为您不需要转换为数组，您可以使用 charAt() 函数在每个字符处添加单个字符点在字符串中。您将不得不使用 for 循环，从最大值开始并减少到最小值。代码sn-p如下；
function add(a, b) {
let sum='';
let z,x;
let r=0;
if (a.length>=b.length){
z=a;
x=b;
}
else{
z=b;
x=a;
};
let p=x.length;
for (let i=z.length;i>0;i--){
let t=((p>0)?parseInt(x.charAt(p-1)):0)+parseInt(z.charAt(i-1))+r;
sum=(t%10)+sum;
r=t<10?0:Math.floor(t/10);
p=p-1;
};
if (r>0){sum=r+sum};
return sum;
};
【讨论】：
function add(x, y) {
//this function adds two extremely large numbers, negative and/or positive
var temp, borrow=false, bothNeg=false, oneNeg=false, neg=false;
if (x < 0 && y < 0) { bothNeg = true; x = -x; y = -y; }
else if (x < 0 || y < 0) {
oneNeg = true;
if (Math.abs(x) == Math.abs(y)) { x = 0; y = 0; }
else if (x < 0 && Math.abs(x) > Math.abs(y)) { neg = true; x = -x; y = -y; }
else if (x < 0 && Math.abs(x) < Math.abs(y)) { temp = y; y = x; x = temp; }
else if (y < 0 && Math.abs(x) < Math.abs(y)) { neg = true; temp = y; y = -x; x = -temp; }
}
x = parseInt(x*1000000000/10).toString();
y = parseInt(y*1000000000/10).toString();
var lenx=x.length, leny=y.length, len=(lenx>leny)?lenx:leny, sum="", div=0, x1, y1, rem;
for (var i = 0; i < len; i++) {
x1 = (i >= lenx) ? 0 : parseInt(x[lenx-i-1]);
y1 = (i >= leny) ? 0 : parseInt(y[leny-i-1]);
y1 = (isNaN(y1)) ? 0 : y1;
if (oneNeg) y1 = -y1;
if (borrow) x1 = x1 - 1;
if (y < 0 && x1 > 0 && Math.abs(x1) >= Math.abs(y1)) { borrow=false; div=0; }
if (y < 0 && y1 <= 0 && (x1 < 0 || Math.abs(x1) < Math.abs(y1))) { borrow=true; rem=(x1+y1+div+10)%10; div=10; }
else { rem=(x1+y1+div)%10; div=Math.floor((x1+y1+div)/10); }
sum = Math.abs(rem).toString() + sum;
}
if (div > 0) sum = div.toString() + sum;
sum = parseFloat(sum*10/1000000000);
if (bothNeg || neg) sum = -sum;
return sum;
}
【讨论】：
add(9999999999999999999999999999999999999999999999999999999999999999999999999999,999999999999999999999999999999999999999)
另一种解决方案，因为它更快更干净。
function add(A, B) {
const AL = A.length
const BL = B.length
const ML = Math.max(AL, BL)
let carry = 0, sum = ''
for (let i = 1; i <= ML; i++) {
let a = +A.charAt(AL - i)
let b = +B.charAt(BL - i)
let t = carry + a + b
carry = t/10 |0
t %= 10
sum = (i === ML && carry)
? carry*10 + t + sum
: t + sum
}
return sum
}
> add(
'9999999999999999999999999999999999999999999999999999999999999999999999999999',
'999999999999999999999999999999999999999'
)
> "10000000000000000000000000000000000000999999999999999999999999999999999999998"
【讨论】：
0？
按照此处所述使用 BigInt：https://stackoverflow.com/a/56370672/641913
const z = BigInt(x) + BigInt(y);
console.log(z.toString());
【讨论】：