我有一个像这样的短数组a
:
a = [ 16748,
26979,
25888,
30561,
115
] //in decimal
a = [ 0100000101101100,
0110100101100011,
0110010100100000,
0111011101100001,
0000000001110011
] //in binary
我想获得另一个由表示每个短路的每对位组成的短路数组b
。
(解释起来很困难,但通过示例很容易理解)。
因此,对于数组a
,我将获得数组b
:
b = [ 01, 00, 00, 01, 01, 10, 11, 00,
01, 10, 10, 01, 01, 10, 00, 11,
01, 10, 01, 01, 00, 10, 00, 00,
01, 11, 01, 11, 01, 10, 00, 01,
00, 00, 00, 00, 01, 11, 00, 11
]
在伪代码中我想到这样做:
int lenght = (16/2) * a.length; //16*2 because I had short (16 bit) and I want sequences of 2 bit
short[] b = new short[length]; //I create the new array of short
int j = 0; //counter of b array
foreach n in a { //foreach short in array a
for(int i = 16 - 2; i > 0; i-2) { //shift of 2 positions to right
b[j] = ( (n >> i) & ((2^2)-1) ); //shift and &
j++;
}
}
我尝试将这个伪代码(假设它是正确的)翻译为 Java:
public static short[] createSequencesOf2Bit(short[] a) {
int length = (16/2) * a.length;
short[] b = new short[length];
for(int i = 0; i < a.length; i++) {
int j = 0;
for(short c = 16 - 2; c > 0; c -= 2) {
short shift = (short)(a[i] >> c);
b[j] = (short)(shift & 0x11);
j++;
}
}
return b;
}
但是如果我打印 b[]
我就得不到我想要的东西。
例如,仅考虑 a
(16748 = 0100000101101100)
中的第一个短路。
我得到:
[1, 0, 16, 1, 1, 16, 17]
这是完全错误的。事实上我应该得到:
b = [ 01, 00, 00, 01, 01, 10, 11, 00,
...
] //in binary
b = [ 1, 0, 0, 1, 1, 2, 3, 0,
...
] //in decimal
有人可以帮助我吗? 非常感谢。
<小时/>这很奇怪。如果我只考虑 a 中的第一个短路并打印 b 我得到:
public static short[] createSequencesOf2Bit(short[] a) {
int length = (16/2) * a.length;
short[] b = new short[length];
//for(int i = 0; i < a.length; i++) {
int j = 0;
for(short c = (16 - 2); c >= 0; c -= 2) {
short shift = (short)(a[0] >> c);
b[j] = (short)(shift & 0x3);
j++;
}
//}
for(int i = 0; i < b.length; i++) {
System.out.println("b[" + i + "]: " + b[i]);
}
return b;
}
b = [1 0 0 1 1 2 3 0 0 0 0 0 ... 0]
但是如果我打印这个:
public static short[] createSequencesOf2Bit(short[] a) {
int length = (16/2) * a.length;
short[] b = new short[length];
for(int i = 0; i < a.length; i++) {
int j = 0;
for(short c = (16 - 2); c >= 0; c -= 2) {
short shift = (short)(a[i] >> c);
b[j] = (short)(shift & 0x3);
j++;
}
}
for(int i = 0; i < b.length; i++) {
System.out.println("b[" + i + "]: " + b[i]);
}
return b;
}
b = [0 0 0 0 1 3 0 3 0 0 0 0 ... 0]
最佳答案
我怀疑主要问题是您需要 & 0x3
而不是 & 0x11
。请记住,0x
表示十六进制,因此 0x11
为您提供数字 17,而不是 3。或者您可以编写 0b11
来获取 11
二进制。
此外,正如评论中指出的,循环条件应该是 c >= 0
,而不是 c > 0
。
关于java - 从短裤中获得 8 条短裤,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/35805084/