转换字节数组中的字节到整数位阵列和存储 [英] Convert the bytes of byte array to bits and store in Integer array
问题描述
我有一个字节数组。我需要每个字节的位值存储在一个整数数组。
例如，
字节数组是
字节HexToBin [] = {（字节）为0x9A（字节）为0xFF，（字节）0×05，（字节）0x16};
那么整数数组应该有
A = [10011010111111110000010100010110]
我曾尝试以下code，在那里我能够打印每个字节（S2）的二进制值，但我couldnot在整数数组存储ALLBITS
字节hexToBin [] = {（字节）为0x9A（字节）为0xFF，（字节）0×05，（字节）0x16};
INT [] = ALLBITS新INT [32];
INT A = 0;的for（int i = 0; I＆LT; hexToBin.length;我++）
{
字节eachByte = hexToBin [I]
字符串s2 =的String.format（％8S，Integer.toBinaryString（（eachByte）及为0xFF））。REPLACE（''，'0'）;
的System.out.println（S2）;
的char [] = totalCharArr s2.toCharArray（）;
对于（INT K = 0; K＆LT; 8; k ++）
{
ALLBITS [K +一] = totalCharArr [K];
}
A = A + 8;
}对于（INT B = 0; B＆LT; 32; b ++）
{
System.out.print（ALLBITS并[b]）;
}
以上code的输出
10011010
11111111
00000101
00010110
4948484949484948494949494949494948484848484948494848484948494948
整数数组不具备二进制值。
////////////////////////////////////////////////////////////////////////////////////////////////////
感谢您的帮助。
校正code是
字节hexToBin [] = {（字节）为0x9A（字节）为0xBF，（字节）0×05，（字节）0x16};
INT [] = ALLBITS新INT [32]; //没有位被许可code确定 对于（INT N = 0; N＆LT; hexToBin.length; N ++）
{
//使用整数，以避免任何可能的混乱，由于符号字节值
INT sourceByte = 0xFF的及（INT）hexToBin [N]; //转换字节无符号整型
INT面膜= 0x80的;
的for（int i = 0; I＆LT; 8;我++）
{
INT maskResult = sourceByte＆安培;面具; //提取单个位
如果（maskResult大于0）{
ALLBITS [8 * N + I] = 1;
}
其他{
ALLBITS [8 * N + I] = 0; //因为数组不必要的启动零，但良好的文档
}
面膜面膜=＆GT;＆GT; 1;
}
}
对于（INT K = 0; K＆LT; 32; k ++）
{
System.out.print（ALLBITS [K]）;
}
假定为一个循环内 N = 0至3
//使用整数，以避免任何可能的混乱，由于符号字节值
INT sourceByte = 0xFF的＆安培; （中间体）（hexToBin [N]）; //将字节无符号整型
INT面膜= 0x80的;
的for（int i = 0; I＆LT; 8;我++）{
INT maskResult = sourceByte＆安培;面具; //提取单个位
如果（maskResult！= 0）{
ALLBITS [8 * N + I] = 1;
}
其他{
ALLBITS [8 * N + 1] = 0; //因为数组不必要的inited为零，但良好的机制的文档
}
面膜面膜=＆GT;＆GT; 1;
}
I have bytes in a byte array. I need to store the bit value of each byte in an integer array .
For example ,
the byte array is
byte HexToBin[] = {(byte)0x9A, (byte)0xFF,(byte) 0x05,(byte) 0x16};
then the integer array should have
a = [10011010111111110000010100010110]
I have tried the following code, where i was able to print the binary value of each byte (s2) but i couldnot store in integer array allBits.
byte hexToBin[] = {(byte)0x9A, (byte)0xFF,(byte) 0x05,(byte) 0x16};
int[] allBits = new int[32];
int a =0;
for (int i =0; i < hexToBin.length ; i++)
{
byte eachByte = hexToBin[i];
String s2 = String.format("%8s", Integer.toBinaryString((eachByte)& 0xFF)).replace(' ', '0');
System.out.println(s2);
char [] totalCharArr = s2.toCharArray();
for (int k=0; k <8; k++)
{
allBits[k+a]= totalCharArr[k];
}
a= a+8;
}
for (int b=0; b<32;b++)
{
System.out.print(allBits[b]);
}
The output of above code is
10011010
11111111
00000101
00010110
4948484949484948494949494949494948484848484948494848484948494948
The integer array does not have the binary value.
////////////////////////////////////////////////////////////////////////////////////////////////////
Thank you for the help
The Corrected code is
byte hexToBin[] = {(byte)0x9A, (byte)0xBF,(byte) 0x05,(byte) 0x16};
int[] allBits = new int[32]; // no of bits is determined by the license code
for (int n =0; n<hexToBin.length; n++)
{
//Use ints to avoid any possible confusion due to signed byte values
int sourceByte = 0xFF &(int)hexToBin[n];//convert byte to unsigned int
int mask = 0x80;
for (int i=0; i<8; i++)
{
int maskResult = sourceByte & mask; // Extract the single bit
if (maskResult>0) {
allBits[8*n + i] = 1;
}
else {
allBits[8*n + i] = 0; // Unnecessary since array is initiated to zero but good documentation
}
mask = mask >> 1;
}
}
for (int k= 0; k<32; k++)
{
System.out.print(allBits[k]);
}
Assumed to be inside a loop of n = 0 to 3
// Use ints to avoid any possible confusion due to signed byte values
int sourceByte = 0xFF & (int)(hexToBin[n]); // Convert byte to unsigned int
int mask = 0x80;
for (int i = 0; i < 8; i++) {
int maskResult = sourceByte & mask; // Extract the single bit
if (maskResult != 0) {
allBits[8*n + i] = 1;
}
else {
allBits[8*n + 1] = 0; // Unnecessary since array is inited to zero but good documention
}
mask = mask >> 1;
}
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