如何将任意基数的字符串版本转换为整数？

Question

how to convert string to integer??

for ex:

• "5328764",to int base 10
• "AB3F3A", to int base 16

any code will be helpfull

Answer 1

Assuming arbitrary base (not 16, 10, 8, 2):

In C (C++), use strtol

return strtol("AB3F3A", NULL, 16);

In Javascript, use parseInt .

return parseInt("AB3F3A", 16);

In Python, use int(string, base) .

return int("AB3F3A", 16)

In Java, use Integer.parseInt (thanks Michael.)

return Integer.parseInt("AB3F3A", 16);

In PHP, use base_convert .

return intval(base_convert('AB3F3A', 16, 10));

In Ruby, use to_i

"AB3F3A".to_i(16)

In C#, write one yourself.

Answer 2

9999 is really 9000 + 900 + 90 + 9 So, start at the right hand side of the string, and pick off the numbers one at a time. Each character number has an ASCII code, which can be translated to the number, and multiplied by the appropriate amount.

Answer 3

在C＃中，我认为它是：Convert.ToInt64（value，base）并且基数必须为2、8、10或16

Answer 4

Two functions in java, in both directions: "code" parameter represent the numerical system: "01" for base 2, "0123456789" for base 10, "0123456789abcdef" for hexdecimal and so on...

public String convert(long num, String code) {
    final int base = code.length();
    String text = "";
    while (num > 0) {
        text = code.charAt((int) (num%base)) + text;
        num /= base;
    }
    return text;
}

public long toLong(String text, String code) {
    final long base = code.length();
    long num = 0;
    long pow = 1;
    int len = text.length();
    for(int i = 0; i < len; i++) {
        num += code.indexOf(text.charAt(len - i - 1)) * pow;
        pow *= base;
    }
    return num;
}

println(convert(9223372036854775807L,"0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"));
println(convert(9223372036854775807L,"0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ@=-+*/^%$#&()!?.,:;[]"));
println(toLong("Ns8T$87=uh","0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ@=-+*/^%$#&()!?.,:;[]"));```

in your example:

toLong("5328764", "0123456789") = 5328764
toLong("AB3F3A", "0123456789ABCDEF") = 11222842