use*_*151 5 java string hex biginteger parseint
给出以下字符串:
3132333435363738396162636465666768696a6b6c6d6e6f70
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我将字符串转换为十六进制,现在我想将其写为十六进制而不是字符串.我尝试将其转换为int但Integer.parseInt仅转换为4,如果超出此范围,则会产生错误.
你尝试过BigInteger构造函数的字符串和基数吗?
BigInteger value = new BigInteger(hex, 16);
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示例代码:
import java.math.BigInteger;
public class Test {
public static void main(String[] args) {
String hex = "3132333435363738396162636465666768696a6b6c6d6e6f70";
BigInteger number = new BigInteger(hex , 16);
System.out.println(number); // As decimal...
}
}
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输出:
308808885829455478403317837970537433512288994552567292653424
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