java运算符总结 - 编程入门网
作者 佚名技术
来源 NET编程
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发布时间 2012-07-04
// Compound assignment:
x += y;
x -= y;
x *= y;
x /= y;
x %= y;
x <<= 1;
x >>= 1;
x >>>= 1;
x &= y;
x ^= y;
x |= y;
// Casting:
//! boolean b = (boolean)x;
char c = (char)x;
byte B = (byte)x;
short s = (short)x;
int i = (int)x;
float f = (float)x;
double d = (double)x;
}
void floatTest(float x, float y) {
// Arithmetic operators:
x = x * y;
x = x / y;
x = x % y;
x = x + y;
x = x - y;
x++;
x--;
x = +y;
x = -y;
// Relational and logical:
f(x > y);
f(x >= y);
f(x < y);
f(x <= y);
f(x == y);
f(x != y);
//! f(!x);
//! f(x && y);
//! f(x || y);
// Bitwise operators:
//! x = ~y;
//! x = x & y;
//! x = x | y;
//! x = x ^ y;
//! x = x << 1;
//! x = x >> 1;
//! x = x >>> 1;
// Compound assignment:
x += y;
x -= y;
x *= y;
x /= y;
x %= y;
//! x <<= 1;
//! x >>= 1;
//! x >>>= 1;
//! x &= y;
//! x ^= y;
//! x |= y;
// Casting:
//! boolean b = (boolean)x;
char c = (char)x;
byte B = (byte)x;
short s = (short)x;
int i = (int)x;
long l = (long)x;
double d = (double)x;
}
void doubleTest(double x, double y) {
// Arithmetic operators:
x = x * y;
x = x / y;
x = x % y;
x = x + y;
x = x - y;
x++;
x--;
x = +y;
x = -y;
// Relational and logical:
f(x > y);
f(x >= y);
f(x < y);
f(x <= y);
f(x == y);
f(x != y);
//! f(!x);
//! f(x && y);
//! f(x || y);
// Bitwise operators:
//! x = ~y;
//! x = x & y;
//! x = x | y;
//! x = x ^ y;
//! x = x << 1;
//! x = x >> 1;
//! x = x >>> 1;
// Compound assignment:
x += y;
x -= y;
x *= y;
x /= y;
x %= y;
//! x <<= 1;
//! x >>= 1;
//! x >>>= 1;
//! x &= y;
//! x ^= y;
//! x |= y;
// Casting:
//! boolean b = (boolean)x;
char c = (char)x;
byte B = (byte)x;
short s = (short)x;
int i = (int)x;
long l = (long)x;
float f = (float)x;
}
} ///:~
注意布尔值（boolean）的能力非常有限。我们只能为其赋予true和false值。而且可测试它为真还是为假，但不可为它们再添加布尔值，或进行其他其他任何类型运算。在char，byte和short中，我们可看到算术运算符的“转型”效果。对这些类型的任何一个进行算术运算，都会获得一个int结果。必须将其明确“造型”回原来的类型（缩小转换会造成信息的丢失），以便将值赋回那个类型。但对于int值，却不必进行造型处理，因为所有数据都已经属于int类型。然而，不要放松警惕，认为一切事情都是安全的。如果对两个足够大的int值执行乘法运算，结果值就会溢出。下面这个例子向大家展示了这一点：
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