英文:
Why results of bitwise left in Java and in C are different?
问题
在C语言中，当将128以上的数值转换成无符号字符(unsigned char)时，会导致截断（wrap around）现象，即结果会回到0。而在Java中，char类型是16位的，所以没有这种截断现象。如果你想在Java中获得类似C语言的结果，可以使用以下方法：
char x = (char) 128;
char y;
y = (char)(x << 1);
int ke = y & 0xFF; // 使用位掩码将高位截断，保留低8位
System.out.println(ke);
这将在Java中产生与C语言相似的输出结果。
英文:
I have a code in c :
unsigned char x = (char)128;
unsigned char y;
y = (x << 1);
int ye = (int) y;
printf("%d" , ye);
output = 0
I have a code in Java :
char x = (char) 128;
char y;
y = (char)(x << 1);
int ke = (char) y;
System.out.println(ke);
output = 256
if x = 1-127 in c and java are the same, but when 128 and above the results are different,
is there anything I can do for results in java like in C?
答案1
得分: 1
在C中，诸如'char'等基本类型的'width'和属性未定义：这取决于编译器和架构。
在Java中，它们都有明确定义。
在Java中，'char'是一个16位的无符号整数。大多数接受字符的方法将解释这个数字作为Unicode中的代码点（例如'System.out.println()'：'System.out.println((char) 65);'打印的是'A'，而不是65。
显然，在您的C代码中，'char'被定义为一个8位宽度的数据类型；在一个8位系统中，128左移1位等于0（它是256... 这是100 000 000 - 9位；第9位是溢出并被丢弃，得到0）。
Java没有任何表示无符号8位的数据类型。
您可以使用'int'并自己处理溢出规则。您还可以使用'byte'，并处理在Java中，字节是有符号的（操作它们的方法倾向于以有符号方式解释位；位本身就是它们是什么，不关心我们人类可能认为它们意味着什么。它们既不是有符号的也不是无符号的）。
英文:
The 'width' and properties of basic types such as 'char' are undefined in C: It's up to the compiler and architecture.
In java they are all exactly defined.
In java, a char is a 16-bit unsigned whole number. Most methods that take chars will interpret this number as a codepoint in unicode (such as System.out.println(): `System.out.println((char) 65); prints 'A', not 65.
Evidently, in your C code, char is defined as an 8-bit width data type; 128 shifted left by 1, in an 8 bit system, is 0. (it's 256.. which is 100 000 000 - 9 bits; the 9th bit is overflow and is discarded, giving you 0).
Java does not have any data type that represents unsigned 8-bit.
You can use int and take care of the overflow rules yourself. You can also use byte, and deal with the fact that in java, bytes are signed (methods that operate on them tend to interpret the bits in a signed fashion; bits themselves are what they are and care not for what us humans imagine they might mean. They are neither signed nor unsigned).
答案2
得分: 0
C 和 Java 中的 char 类型大小不同
- 在 C 中为 8 位（检查
sizeof(char)） - 在 Java 中为 16 位（https://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html）
因此，移位后的位会消失，因此
- 在 C 中，
x = 128的二进制表示为10000000，x << 1的结果为00000000，因此x << 1为0 - 在 Java 中，
x = 128的二进制表示为0000000010000000，x << 1的结果为0000000100000000，因此x << 1为256
英文:
The size of char is different in C and Java
- 8 bits in C (check
sizeof(char)) - 16 bits in Java (https://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html)
The shift out bit would disappears, thus
- In C,
x = 128is10000000in binary,x << 1results00000000, thusx << 1is0 - In Java,
x = 128is0000000010000000in binary,x << 1results0000000100000000, thusx << 1is256
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