如何在 Java 中使用具有长值的 printf() 方法？
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How to use printf() method with long values in Java?
提问by Karthik Cherukuri
I remember that I did this in C but I couldn't get it work in Java.
我记得我是用 C 完成的，但我无法在 Java 中使用它。
How to print long values with printf() method in Java?
如何在 Java 中使用 printf() 方法打印长值？
I tried with below code what I actually need is to print the long value in hexadecimal way like below
我尝试使用下面的代码，我实际需要的是以十六进制方式打印长值，如下所示
long l = 32L -----> 0000000000000022
长 l = 32L -----> 0000000000000022
If I use %d then it prints integer value which I don't want...
如果我使用 %d 那么它会打印我不想要的整数值......
class TestPrintf()
{
public static void main(String[] args)
{
long l = 100L;
System.out.printf(“%l”+l); // Error unknown format exception
System.out.printf(“%d”+l); // Prints 100
System.out.printf(“%f”+l); // Unknown illegal format conversion float != java.lang.long
}
}
采纳答案by Andy Turner
If you want a 16-character zero-padded string, with A-Fin uppercase, prefixed by 0x, you should use:
如果你想要一个 16 个字符的零填充字符串，A-F大写，前缀为0x，你应该使用：
System.out.printf("0x%016X", l);
回答by Crazyjavahacking
You need to put the actual argument to print as the next argument of a printf()methods. Concatenating will not work.
您需要将要打印的实际参数作为方法的下一个参数printf()。连接将不起作用。
System.out.printf("%d%n", 123); // for ints and longs
System.out.printf("%d%n", 12345L); // for ints and longs
System.out.printf("%f%n", (double) 12345L); // for floating point numbers
Full documentation in java.util.Formatter
完整文档在 java.util.Formatter
