Basic data type, java Basic Data Type

Basic data type, java Basic Data Type

Byte tt = (byte) 130 is equal to-126. Byte occupies one byte, 8 bits. The first digit is the symbol bit. 0 indicates a positive number, and 1 indicates a negative number. Therefore, the value range of byte is [-128,127].

The binary value of 130 is 00000000000000000000000010000010, Which is truncated to byte type. Only the last 8 digits are retained, that is, 10000010. the first digit is 1 or negative, the negative number stored by the computer is a supplementary code. The other bits must be reversed and added with 1. After the value is reversed and 1, this value is-1111110. The binary number is converted to an integer of-126.

Byte a = (byte) 0x00000111; hexadecimal 16*16 + 16 + 1 = 256 + 17 = 0000001 00010001. If the last 8 digits are intercepted, a = 17.

Byte c = (byte) 0xFF; hexadecimal system 15*16 + 15 = 255 = 11111111, the first 1 indicates a negative number, and the bucket stores a supplemental code, if the last 7 digits are reversed and 1 is equal to 1, then c =-1. System. out. println (c); input-1.

The octal value must start with 0, and the decimal value cannot start with 0 (except 0 ).

Byte a = 00000111; For octal 64 + 8 + 1, a = 73.

Byte a = (byte) 00001111; octal 8*8*8 + 64 + 8 + 1 = 00000010 01001001, converted to byte, take the last 8 bits, then a = 73. Byte a = 00001111, an error is returned out of the range [-128,127].

 

Byte occupies 8 places. Short occupies 16 bits. value range: [-32768,327 67]

Short s1 = 1; s1 = s1 + 1; an error is returned. Because 1 is of the int type and the calculation result is of the int type, it must be forcibly converted to the short type and written as s1 = (short) (s1 + 1); or s1 + = 1; this method contains implicit forced type conversion. Equivalent to (short) (s1 + 1)

 

The int type occupies 32 bits in memory and 4 bytes. Value Range: [-2147483648,214 7483647].

2 ^ 31 = 2147483648

 

If the long value is out of the int value range, you must add L or l after the number. Indicates that the value is a long integer.

Long n = 2147483650L + 4, then the n value is 2147483654.

Long m = (long) (2147483647 + 4); computing process: 2147483647 storage is: 01111111 11111111 11111111 11111111 plus 00000000 00000000 00000000 equals 00000100 10000000 00000000 00000000.

The first 1 indicates a negative number. The negative number stores the complement code, and the other bits are reversed with 1. The source code is 11111111 11111111 11111111 11111101. Then m is equal to-(2 ^ 32-1-2) =-2147483645.