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'타입변환'에 해당되는 글 1

  1. 2010.12.08 이해안가던 자바퍼즐러6 Multicast
2010.12.08 11:39

이해안가던 자바퍼즐러6 Multicast Study/Java2010.12.08 11:39

이해 안가서 차례대로 변환해봤다.

..생략..
System.out.println((byte)-1);
System.out.println((char)(byte)-1);
System.out.println((int)(char)(byte)-1);

결과

-1
?
65535

일단 int - > byte 타입은 축소기본타입변환인데.. 하위 8비트를 제외하고 모두 잘라낸다. <JLS 5.1.3>

<JLS 발췌>
5.1.3 Narrowing Primitive Conversions
The following 22 specific conversions on primitive types are called the narrowing
primitive conversions:
• short to byte or char
• char to byte or short
• int to byte, short, or char
• long to byte, short, char, or int
• float to byte, short, char, int, or long
• double to byte, short, char, int, long, or float
Narrowing conversions may lose information about the overall magnitude of a
numeric value and may also lose precision.
A narrowing conversion of a signed integer to an integral type T simply discards
all but the n lowest order bits, where n is the number of bits used to represent
type T. In addition to a possible loss of information about the magnitude of
the numeric value, this may cause the sign of the resulting value to differ from the
sign of the input value.
A narrowing conversion of a char to an integral type T likewise simply discards
all but the n lowest order bits, where n is the number of bits used to represent
type T. In addition to a possible loss of information about the magnitude of
the numeric value, this may cause the resulting value to be a negative number,
even though chars represent 16-bit unsigned integer values.
A narrowing conversion of a floating-point number to an integral type T takes
two steps:
1. In the first step, the floating-point number is converted either to a long, if T is
long, or to an int, if T is byte, short, char, or int, as follows:
◆ If the floating-point number is NaN (§4.2.3), the result of the first step of
the conversion is an int or long 0.
◆ Otherwise, if the floating-point number is not an infinity, the floating-point
value is rounded to an integer value V, rounding toward zero using IEEE
754 round-toward-zero mode (§4.2.3). Then there are two cases:
❖ If T is long, and this integer value can be represented as a long, then the
result of the first step is the long value V.
❖ Otherwise, if this integer value can be represented as an int, then the
result of the first step is the int value V.
◆ Otherwise, one of the following two cases must be true:
CONVERSIONS AND PROMOTIONS Narrowing Primitive Conversions 5.1.3
83
❖ The value must be too small (a negative value of large magnitude or negative
infinity), and the result of the first step is the smallest representable
value of type int or long.
❖ The value must be too large (a positive value of large magnitude or positive
infinity), and the result of the first step is the largest representable
value of type int or long.
2. In the second step:
◆ If T is int or long,the result of the conversion is the result of the first step.
◆ If T is byte, char, or short, the result of the conversion is the result of a
narrowing conversion to type T (§5.1.3) of the result of the first step.
The example:
class Test {
public static void main(String[] args) {
float fmin = Float.NEGATIVE_INFINITY;
float fmax = Float.POSITIVE_INFINITY;
System.out.println("long: " + (long)fmin +
".." + (long)fmax);
System.out.println("int: " + (int)fmin +
".." + (int)fmax);
System.out.println("short: " + (short)fmin +
".." + (short)fmax);
System.out.println("char: " + (int)(char)fmin +
".." + (int)(char)fmax);
System.out.println("byte: " + (byte)fmin +
".." + (byte)fmax);
}
}
produces the output:
long: -9223372036854775808..9223372036854775807
int: -2147483648..2147483647
short: 0..-1
char: 0..65535
byte: 0..-1

The results for char, int, and long are unsurprising, producing the minimum
and maximum representable values of the type.
The results for byte and short lose information about the sign and magnitude
of the numeric values and also lose precision. The results can be understood
by examining the low order bits of the minimum and maximum int. The minimum
int is, in hexadecimal, 0x80000000, and the maximum int is 0x7fffffff.
This explains the short results, which are the low 16 bits of these values, namely,

0x0000 and 0xffff; it explains the char results, which also are the low 16 bits of
these values, namely, '\u0000' and '\uffff'; and it explains the byte results,
which are the low 8 bits of these values, namely, 0x00 and 0xff.
Despite the fact that overflow, underflow, or other loss of information may
occur, narrowing conversions among primitive types never result in a run-time
exception (§11).
Here is a small test program that demonstrates a number of narrowing conversions
that lose information:
class Test {
public static void main(String[] args) {
// A narrowing of int to short loses high bits:
System.out.println("(short)0x12345678==0x" +
Integer.toHexString((short)0x12345678));
// A int value not fitting in byte changes sign and magnitude:
System.out.println("(byte)255==" + (byte)255);
// A float value too big to fit gives largest int value:
System.out.println("(int)1e20f==" + (int)1e20f);
// A NaN converted to int yields zero:
System.out.println("(int)NaN==" + (int)Float.NaN);
// A double value too large for float yields infinity:
System.out.println("(float)-1e100==" + (float)-1e100);
// A double value too small for float underflows to zero:
System.out.println("(float)1e-50==" + (float)1e-50);
}
}
This test program produces the following output:
(short)0x12345678==0x5678
(byte)255==-1
(int)1e20f==2147483647
(int)NaN==0
(float)-1e100==-Infinity
(float)1e-50==0.0

char 타입으로 음의 byte 타입 변환은 불가능..

음 이 예제는 아무리 봐도 핵심이 중요한것 같다 -..-

핵심 : 프로그램을 보는 것만으로는 프로그램이 무엇을 하는지 알 수 없다면 프로그램은 아마도 여러분이 원하는

작업을 하지 않을 것이다.

Posted by 유쾌한순례자