如何在JavaScript中将字符串转换为整数?


当前回答

对于C和JavaScript之间的绑定,我实际上需要将字符串“保存”为整数,因此我将字符串转换为整数值:

/*
    Examples:
        int2str( str2int("test") ) == "test" // true
        int2str( str2int("t€st") ) // "t¬st", because "€".charCodeAt(0) is 8364, will be AND'ed with 0xff
    Limitations:
        maximum 4 characters, so it fits into an integer
*/
function str2int(the_str) {
    var ret = 0;
    var len = the_str.length;
    if (len >= 1) ret += (the_str.charCodeAt(0) & 0xff) <<  0;
    if (len >= 2) ret += (the_str.charCodeAt(1) & 0xff) <<  8;
    if (len >= 3) ret += (the_str.charCodeAt(2) & 0xff) << 16;
    if (len >= 4) ret += (the_str.charCodeAt(3) & 0xff) << 24;
    return ret;
}

function int2str(the_int) {
    var tmp = [
        (the_int & 0x000000ff) >>  0,
        (the_int & 0x0000ff00) >>  8,
        (the_int & 0x00ff0000) >> 16,
        (the_int & 0xff000000) >> 24
    ];
    var ret = "";
    for (var i=0; i<4; i++) {
        if (tmp[i] == 0)
            break;
        ret += String.fromCharCode(tmp[i]);
    }
    return ret;
}

其他回答

前面的所有答案都是正确的。请先通过“typeot x==='number'”确认这是字符串中的数字。否则,它将返回NaN。

 var num = "fsdfsdf242342";
 typeof num => 'string';

 var num1 = "12423";
 typeof num1 => 'number';
 +num1 = > 12423`

将数字的乘法与各自的十次幂相加:

即:123=100+20+3=1100+2+10+31=1*(10^2)+2*(10^1)+3*(10^0)

function atoi(array) {

    // Use exp as (length - i), other option would be
    // to reverse the array.
    // Multiply a[i] * 10^(exp) and sum

    let sum = 0;

    for (let i = 0; i < array.length; i++) {
        let exp = array.length - (i+1);
        let value = array[i] * Math.pow(10, exp);
        sum += value;
    }

    return sum;
}

我用这个

String.prototype.toInt = function (returnval) {
    var i = parseInt(this);
     return isNaN(i) ? returnval !== undefined ? returnval : - 1  :      i;
}

var str = "7";
var num = str.toInt(); // outputs 7, if not str outputs -1
//or
var num = str.toInt(0); // outputs 7, if not str outputs 0

这样我总能得到一个整数。

我们可以使用+(stringOfNumber)而不是parseInt(stringOfNumber)。

示例:+(“21”)返回int值为21,类似于parseInt(“21)。

我们也可以使用这个一元“+”运算符来解析浮点数。。。

确保获得有效整数的最安全方法:

let integer = (parseInt(value, 10) || 0);

示例:

// Example 1 - Invalid value:
let value = null;
let integer = (parseInt(value, 10) || 0);
// => integer = 0
// Example 2 - Valid value:
let value = "1230.42";
let integer = (parseInt(value, 10) || 0);
// => integer = 1230
// Example 3 - Invalid value:
let value = () => { return 412 };
let integer = (parseInt(value, 10) || 0);
// => integer = 0