sprinng

integer to String :

int i = 42;
String str = Integer.toString(i);
or
String str = "" + i

double to String :
String str = Double.toString(i);

long to String :
String str = Long.toString(l);
float to String :
String str = Float.toString(f);
String to integer :
str = "25";
int i = Integer.valueOf(str).intValue();
or
int i = Integer.parseInt(str);
String to double :
double d = Double.valueOf(str).doubleValue();

String to long :
long l = Long.valueOf(str).longValue();
or
long l = Long.parseLong(str);

String to float :
float f = Float.valueOf(str).floatValue();
decimal to binary :
int i = 42;
String binstr = Integer.toBinaryString(i);
decimal to hexadecimal :
int i = 42;
String hexstr = Integer.toString(i, 16);
or
String hexstr = Integer.toHexString(i);
hexadecimal (String) to integer :
int i = Integer.valueOf("B8DA3", 16).intValue();
or
int i = Integer.parseInt("B8DA3", 16);
ASCII code to i = 64;
String aChar = new Character((char)i).toString();
integer to ASCII code c = \'A\';
int i = (int) c; // i will have the value 65 decimal
To extract Ascii codes from a test = "ABCD";
for ( int i = 0; i < test.length(); ++i ) {
char c = test.charAt( i );
int i = (int) c;
System.out.println(i);
}

integer to boolean :
b = (i != 0);
boolean to =
note :
To catch illegal number conversion, try using the try/catch mechanism.
try{
i = Integer.parseInt(aString);
}
catch(NumberFormatException e)
{
}

分类:

技术点:

相关文章:

  • 2021-09-13
  • 2022-03-06
  • 2021-08-26
  • 2021-11-04
猜你喜欢
  • 2021-11-18
  • 2022-12-23
  • 2021-05-22
  • 2022-12-23
  • 2021-11-18
  • 2021-09-13
相关资源
相似解决方案