here is a part of my program code:
int test;
for(uint i = 0; i < 1700; i++) {
test++;
}
the whole program takes 0.5 seconds to finish, but when I change it to:
int test[1];
for(uint i = 0; i < 1700; i++) {
test[0]++;
}
it will takes 3.5 seconds! and when I change the int to double, it will gets very worse:
double test;
for(uint i = 0; i < 1700; i++) {
test++;
}
it will takes about 18 seconds to finish !!!
I have to increase an int array element and a double variable in my real for loop, and it will takes about 30 seconds!
What's happening here?! Why should it takes that much time for just an increment?!
I know a floating point data type like double has different structure from a fixed point data type like int, but is it the only cause for such a big different time? and what about the second example which is also an int array element?!
Thanks
You have answered your question yourself.
float (double) operations are different from integer ones. Even if you just add 1.0f.
Your second example takes longer than the first one just because you added some pointer refernces. An array in C is -bottom down- not much different from a pointer to the first element. Accessing any element, even the first one, would cause the machine code to load the starting address of the array multiply the index (0 in this case) with the length of each member (4 or whatever bytes int has) and add that (0) to the pointer. Then it has to dereference the pointer, meaning to acutally load the value at that very address. Add one and write back the result.
A smart modern compiler should optimize this a bit. When you want to avoid this optimization, then modify the code a bit and don`t use a constant for the index.
I never tried that with a modern objective-c compiler. But I guess that this code would take much loger than 3.5s to run:
int test[2];
int index = 0;
for(uint i = 0; i < 1700; i++) {
test[index]++;
}
If that does not make much of a change then try this:
-(void)foo:(int)index {
int test[2];
for(uint i = 0; i < 1700; i++) {
test[index]++;
}
}
and then call foo:0;
Give it a try and let us know :)
Related
It is my first attempt to implement recursion with CUDA. The goal is to extract all the combinations from a set of chars "12345" using the power of CUDA to parallelize dynamically the task. Here is my kernel:
__device__ char route[31] = { "_________________________"};
__device__ char init[6] = { "12345" };
__global__ void Recursive(int depth) {
// up to depth 6
if (depth == 5) return;
// newroute = route - idx
int x = depth * 6;
printf("%s\n", route);
int o = 0;
int newlen = 0;
for (int i = 0; i<6; ++i)
{
if (i != threadIdx.x)
{
route[i+x-o] = init[i];
newlen++;
}
else
{
o = 1;
}
}
Recursive<<<1,newlen>>>(depth + 1);
}
__global__ void RecursiveCount() {
Recursive <<<1,5>>>(0);
}
The idea is to exclude 1 item (the item corresponding to the threadIdx) in each different thread. In each recursive call, using the variable depth, it works over a different base (variable x) on the route device variable.
I expect the kernel prompts something like:
2345_____________________
1345_____________________
1245_____________________
1234_____________________
2345_345_________________
2345_245_________________
2345_234_________________
2345_345__45_____________
2345_345__35_____________
2345_345__34_____________
..
2345_245__45_____________
..
But it prompts ...
·_____________
·_____________
·_____________
·_____________
·_____________
·2345
·2345
·2345
·2345
...
What I´m doing wrong?
What I´m doing wrong?
I may not articulate every problem with your code, but these items should get you a lot closer.
I recommend providing a complete example. In my view it is basically required by Stack Overflow, see item 1 here, note use of the word "must". Your example is missing any host code, including the original kernel call. It's only a few extra lines of code, why not include it? Sure, in this case, I can deduce what the call must have been, but why not just include it? Anyway, based on the output you indicated, it seems fairly evident the launch configuration of the host launch would have to be <<<1,1>>>.
This doesn't seem to be logical to me:
I expect the kernel prompts something like:
2345_____________________
The very first thing your kernel does is print out the route variable, before making any changes to it, so I would expect _____________________. However we can "fix" this by moving the printout to the end of the kernel.
You may be confused about what a __device__ variable is. It is a global variable, and there is only one copy of it. Therefore, when you modify it in your kernel code, every thread, in every kernel, is attempting to modify the same global variable, at the same time. That cannot possibly have orderly results, in any thread-parallel environment. I chose to "fix" this by making a local copy for each thread to work on.
You have an off-by-1 error, as well as an extent error in this loop:
for (int i = 0; i<6; ++i)
The off-by-1 error is due to the fact that you are iterating over 6 possible items (that is, i can reach a value of 5) but there are only 5 items in your init variable (the 6th item being a null terminator. The correct indexing starts out over 0-4 (with one of those being skipped). On subsequent iteration depths, its necessary to reduce this indexing extent by 1. Note that I've chosen to fix the first error here by increasing the length of init. There are other ways to fix, of course. My method inserts an extra _ between depths in the result.
You assume that at each iteration depth, the correct choice of items is the same, and in the same order, i.e. init. However this is not the case. At each depth, the choices of items must be selected not from the unchanging init variable, but from the choices passed from previous depth. Therefore we need a local, per-thread copy of init also.
A few other comments about CUDA Dynamic Parallelism (CDP). When passing pointers to data from one kernel scope to a child scope, local space pointers cannot be used. Therefore I allocate for the local copy of route from the heap, so it can be passed to child kernels. init can be deduced from route, so we can use an ordinary local variable for myinit.
You're going to quickly hit some dynamic parallelism (and perhaps memory) limits here if you continue this. I believe the total number of kernel launches for this is 5^5, which is 3125 (I'm doing this quickly, I may be mistaken). CDP has a pending launch limit of 2000 kernels by default. We're not hitting this here according to what I see, but you'll run into that sooner or later if you increase the depth or width of this operation. Furthermore, in-kernel allocations from the device heap are by default limited to 8KB. I don't seem to be hitting that limit, but probably I am, so my design should probably be modified to fix that.
Finally, in-kernel printf output is limited to the size of a particular buffer. If this technique is not already hitting that limit, it will soon if you increase the width or depth.
Here is a worked example, attempting to address the various items above. I'm not claiming it is defect free, but I think the output is closer to your expectations. Note that due to character limits on SO answers, I've truncated/excerpted some of the output.
$ cat t1639.cu
#include <stdio.h>
__device__ char route[31] = { "_________________________"};
__device__ char init[7] = { "12345_" };
__global__ void Recursive(int depth, const char *oroute) {
char *nroute = (char *)malloc(31);
char myinit[7];
if (depth == 0) memcpy(myinit, init, 6);
else memcpy(myinit, oroute+(depth-1)*6, 6);
myinit[6] = 0;
if (nroute == NULL) {printf("oops\n"); return;}
memcpy(nroute, oroute, 30);
nroute[30] = 0;
// up to depth 6
if (depth == 5) return;
// newroute = route - idx
int x = depth * 6;
//printf("%s\n", nroute);
int o = 0;
int newlen = 0;
for (int i = 0; i<(6-depth); ++i)
{
if (i != threadIdx.x)
{
nroute[i+x-o] = myinit[i];
newlen++;
}
else
{
o = 1;
}
}
printf("%s\n", nroute);
Recursive<<<1,newlen>>>(depth + 1, nroute);
}
__global__ void RecursiveCount() {
Recursive <<<1,5>>>(0, route);
}
int main(){
RecursiveCount<<<1,1>>>();
cudaDeviceSynchronize();
}
$ nvcc -o t1639 t1639.cu -rdc=true -lcudadevrt -arch=sm_70
$ cuda-memcheck ./t1639
========= CUDA-MEMCHECK
2345_____________________
1345_____________________
1245_____________________
1235_____________________
1234_____________________
2345__345________________
2345__245________________
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2345__234________________
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2345__235___35___________
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2345__235___23____3______
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2345__235___23____2______
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2345__235___23____23____3
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2345__235___235___25____2
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2345__234___34___________
2345__234___24___________
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2345__234___234__________
2345__234___34____4______
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2345__234___34____3______
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2345__234___34____34____4
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2345__234___24____24____4
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2345__234___23____2______
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2345__2345__345__________
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2345__2345__345___45_____
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2345__2345__345___45____5
2345__2345__345___45____4
2345__2345__345___35____5
2345__2345__345___35____3
2345__2345__345___34____4
2345__2345__345___34____3
2345__2345__245___45_____
2345__2345__245___25_____
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2345__2345__245___45____5
2345__2345__245___45____4
2345__2345__245___25____5
2345__2345__245___25____2
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2345__2345__235___35_____
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2345__2345__234___34_____
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2345__2345__234___24____2
2345__2345__234___23____3
2345__2345__234___23____2
1345__345________________
1345__145________________
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1345__1345_______________
1345__345___45___________
1345__345___35___________
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1345__345___45____4______
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1345__345___45____45____5
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1345__345___35____5______
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1345__345___35____35____5
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1345__145___15____1_____1
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1245__245________________
1245__145________________
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1245__145___45____4______
...
1235__1235__235___25_____
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1235__1235__235___35____5
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1235__1235__135___35_____
1235__1235__135___15_____
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========= ERROR SUMMARY: 0 errors
$
The answer given by Robert Crovella is correct at the 5th point, the mistake was in the using of init in every recursive call, but I want to clarify something that can be useful for other beginners with CUDA.
I used this variable because when I tried to launch a child kernel passing a local variable I always got the exception: Error: a pointer to local memory cannot be passed to a launch as an argument.
As I´m C# expert developer I´m not used to using pointers (Ref does the low-level-work for that) so I thought there was no way to do it in CUDA/c programming.
As Robert shows in its code it is possible copying the pointer with memalloc for using it as a referable argument.
Here is a kernel simplified as an example of deep recursion.
__device__ char init[6] = { "12345" };
__global__ void Recursive(int depth, const char* route) {
// up to depth 6
if (depth == 5) return;
//declaration for a referable argument (point 6)
char* newroute = (char*)malloc(6);
memcpy(newroute, route, 5);
int o = 0;
int newlen = 0;
for (int i = 0; i < (6 - depth); ++i)
{
if (i != threadIdx.x)
{
newroute[i - o] = route[i];
newlen++;
}
else
{
o = 1;
}
}
printf("%s\n", newroute);
Recursive <<<1, newlen>>>(depth + 1, newroute);
}
__global__ void RecursiveCount() {
Recursive <<<1, 5>>>(0, init);
}
I don't add the main call because I´m using ManagedCUDA for C# but as Robert says it can be figured-out how the call RecursiveCount is.
About ending arrays of char with /0 ... sorry but I don't know exactly what is the benefit; this code works fine without them.
void KeyExpansion(unsigned char key[N_KEYS], unsigned int* w)
{
unsigned int temp;
for(int i=0; i< N_KEYS; i++)
{
w[i] = (key[N_KEYS*i]<<24) + (key[N_KEYS*i+1]<<16) + (key[N_KEYS*i+2]<<8) + key[N_KEYS*i+3];
}
for(int i = 4; i< EXPANDED_KEY_COUNT; i++)
{
temp = w[i-1];
if(i % 4 == 0)
temp = SubWord(RotWord(temp)) ^ Rcon[i/4];
w[i] = temp ^ w[i-4] ;
}
}
Big-O helps us do analysis based on the input. The issue with your question is that there seems to be several inputs, which may or may not relate with each other.
Input variables look like N_KEYS, and EXPANDED_KEY_COUNT. We also don't know what SubWord() or RotWord() do based on what is provided.
Since SubWord() and RotWord() aren't provided, lets assume they are constant for easy calculations.
You have basic loops and iterate over each value, so its pretty straight forward. This means you have O(N_KEYS) + O(EXPANDED_KEY_COUNT). So the overall time complexity depends on two inputs, and would be bound by the larger.
If SubWord() or RotWord() do anything special that aren't constant time, then that would affect the time complexity of O(EXPANDED_KEY_COUNT) portion of code. You could adjust the time complexity by multiplied against it. But by the names of the methods, it sounds like their time complexity would be based on the length of the string, would would be yet another different input variable.
So this isn't a clear answer, because the question isn't fully clear, but I tried to break things down for you as best as I could.
-(void)InitWithPwd:(char *)pPwd
{
char szResult[17];
//generate md5 checksum
CC_MD5(pPwd, strlen(pPwd),&szResult[0]);
szResult[16] = 0;
m_csPasswordHash[0]=0;
for(int i = 0;i < 16;i++)
{
char sz[3] = {'\0'};
//crash in blow row. The first pass is ok. The third pass crash.
//I can't understand.
sprintf(&sz[0],"%2.2x",szResult[i]);
strcat(m_csPasswordHash,sz);
}
m_csPasswordHash[32] = 0;
printf("pass:%s\n",m_csPasswordHash);
m_ucPacketType = 1;
}
I want to get the md5 of the password. But above code crash again and again. I can't understand why.
Your buffer (sz) is too small, causing sprintf() to generate a buffer overflow which leads to undefined behavior, in your case a crash.
Note that szResult[1] might be a negative value when viewed as an int (which happens when passing a char-type value to sprintf()), which can cause sprintf() to disregard your field width and precision directives in order to format the full value.
Here is an example showing this problem. The example code is written in C, but that shouldn't matter for this case.
This solves the problem by making sure the incoming data is considered unsigned:
sprintf(sz, "%02x", (unsigned char) szResult[i]);
The code below attempts to save a data stream to a file using fwrite. The first example using malloc works but with the second example the data stream is %70 corrupted. Can someone explain to me why the second example is corrupted and how I can remedy it?
short int fwBuffer[1000000];
// short int *fwBuffer[1000000];
unsigned long fwSize[1000000];
// Not Working *********
if (dataFlow) {
size = sizeof(short int)*length*inchannels;
short int tmpbuffer[length*inchannels];
int count = 0;
for (count = 0; count < length*inchannels; count++)
{
tmpbuffer[count] = (short int) (inbuffer[count]);
}
memcpy(&fwBuffer[saveBufferCount], tmpbuffer, sizeof(tmpbuffer));
fwSize[saveBufferCount] = size;
saveBufferCount++;
totalSize += size;
}
// Working ***********
if (dataFlow) {
size = sizeof(short int)*length*inchannels;
short int *tmpbuffer = (short int*)malloc(size);
int count = 0;
for (count = 0; count < length*inchannels; count++)
{
tmpbuffer[count] = (short int) (inbuffer[count]);
}
fwBuffer[saveBufferCount] = tmpbuffer;
fwSize[saveBufferCount] = size;
saveBufferCount++;
totalSize += size;
}
// Write to file ***********
for (int i = 0; i < saveBufferCount; i++) {
if (isRecording && outFile != NULL) {
// fwrite(fwBuffer[i], 1, fwSize[i],outFile);
fwrite(&fwBuffer[i], 1, fwSize[i],outFile);
if (fwBuffer[i] != NULL) {
// free(fwBuffer[i]);
}
fwBuffer[i] = NULL;
}
}
You initialize your size as
size = sizeof(short int) * length * inchannels;
then you declare an array of size
short int tmpbuffer[size];
This is already highly suspect. Why did you include sizeof(short int) into the size and then declare an array of short int elements with that size? The byte size of your array in this case is
sizeof(short int) * sizeof(short int) * length * inchannels
i.e. the sizeof(short int) is factored in twice.
Later you initialize only length * inchannels elements of the array, which is not entire array, for the reasons described above. But the memcpy that follows still copies the entire array
memcpy(&fwBuffer[saveBufferCount], &tmpbuffer, sizeof (tmpbuffer));
(Tail portion of the copied data is garbage). I'd suspect that you are copying sizeof(short int) times more data than was intended. The recipient memory overflows and gets corrupted.
The version based on malloc does not suffer from this problem since malloc-ed memory size is specified in bytes, not in short int-s.
If you want to simulate the malloc behavior in the upper version of the code, you need to declare your tmpbuffer as an array of char elements, not of short int elements.
This has very good chances to crash
short int tmpbuffer[(short int)(size)];
first size could be too big, but then truncating it and having whatever size results of that is probably not what you want.
Edit: Try to write the whole code without a single cast. Only then the compiler has a chance to tell you if there is something wrong.
I was recently asked to complete a task for a c++ role, however as the application was decided not to be progressed any further I thought that I would post here for some feedback / advice / improvements / reminder of concepts I've forgotten.
The task was:
The following data is a time series of integer values
int timeseries[32] = {67497, 67376, 67173, 67235, 67057, 67031, 66951,
66974, 67042, 67025, 66897, 67077, 67082, 67033, 67019, 67149, 67044,
67012, 67220, 67239, 66893, 66984, 66866, 66693, 66770, 66722, 66620,
66579, 66596, 66713, 66852, 66715};
The series might be, for example, the closing price of a stock each day
over a 32 day period.
As stored above, the data will occupy 32 x sizeof(int) bytes = 128 bytes
assuming 4 byte ints.
Using delta encoding , write a function to compress, and a function to
uncompress data like the above.
Ok, so before this point I had never looked at compression so my solution is far from perfect. The manner in which I approached the problem is by compressing the array of integers into a array of bytes. When representing the integer as a byte I keep the calculate most
significant byte (msb) and keep everything up to this point, whilst throwing the rest away. This is then added to the byte array. For negative values I increment the msb by 1 so that we can
differentiate between positive and negative bytes when decoding by keeping the leading
1 bit values.
When decoding I parse this jagged byte array and simply reverse my
previous actions performed when compressing. As mentioned I have never looked at compression prior to this task so I did come up with my own method to compress the data. I was looking at C++/Cli recently, had not really used it previously so just decided to write it in this language, no particular reason. Below is the class, and a unit test at the very bottom. Any advice / improvements / enhancements will be much appreciated.
Thanks.
array<array<Byte>^>^ CDeltaEncoding::CompressArray(array<int>^ data)
{
int temp = 0;
int original;
int size = 0;
array<int>^ tempData = gcnew array<int>(data->Length);
data->CopyTo(tempData, 0);
array<array<Byte>^>^ byteArray = gcnew array<array<Byte>^>(tempData->Length);
for (int i = 0; i < tempData->Length; ++i)
{
original = tempData[i];
tempData[i] -= temp;
temp = original;
int msb = GetMostSignificantByte(tempData[i]);
byteArray[i] = gcnew array<Byte>(msb);
System::Buffer::BlockCopy(BitConverter::GetBytes(tempData[i]), 0, byteArray[i], 0, msb );
size += byteArray[i]->Length;
}
return byteArray;
}
array<int>^ CDeltaEncoding::DecompressArray(array<array<Byte>^>^ buffer)
{
System::Collections::Generic::List<int>^ decodedArray = gcnew System::Collections::Generic::List<int>();
int temp = 0;
for (int i = 0; i < buffer->Length; ++i)
{
int retrievedVal = GetValueAsInteger(buffer[i]);
decodedArray->Add(retrievedVal);
decodedArray[i] += temp;
temp = decodedArray[i];
}
return decodedArray->ToArray();
}
int CDeltaEncoding::GetMostSignificantByte(int value)
{
array<Byte>^ tempBuf = BitConverter::GetBytes(Math::Abs(value));
int msb = tempBuf->Length;
for (int i = tempBuf->Length -1; i >= 0; --i)
{
if (tempBuf[i] != 0)
{
msb = i + 1;
break;
}
}
if (!IsPositiveInteger(value))
{
//We need an extra byte to differentiate the negative integers
msb++;
}
return msb;
}
bool CDeltaEncoding::IsPositiveInteger(int value)
{
return value / Math::Abs(value) == 1;
}
int CDeltaEncoding::GetValueAsInteger(array<Byte>^ buffer)
{
array<Byte>^ tempBuf;
if(buffer->Length % 2 == 0)
{
//With even integers there is no need to allocate a new byte array
tempBuf = buffer;
}
else
{
tempBuf = gcnew array<Byte>(4);
System::Buffer::BlockCopy(buffer, 0, tempBuf, 0, buffer->Length );
unsigned int val = buffer[buffer->Length-1] &= 0xFF;
if ( val == 0xFF )
{
//We have negative integer compressed into 3 bytes
//Copy over the this last byte as well so we keep the negative pattern
System::Buffer::BlockCopy(buffer, buffer->Length-1, tempBuf, buffer->Length, 1 );
}
}
switch(tempBuf->Length)
{
case sizeof(short):
return BitConverter::ToInt16(tempBuf,0);
case sizeof(int):
default:
return BitConverter::ToInt32(tempBuf,0);
}
}
And then in a test class I had:
void CTestDeltaEncoding::TestCompression()
{
array<array<Byte>^>^ byteArray = CDeltaEncoding::CompressArray(m_testdata);
array<int>^ decompressedArray = CDeltaEncoding::DecompressArray(byteArray);
int totalBytes = 0;
for (int i = 0; i<byteArray->Length; i++)
{
totalBytes += byteArray[i]->Length;
}
Assert::IsTrue(m_testdata->Length * sizeof(m_testdata) > totalBytes, "Expected the total bytes to be less than the original array!!");
//Expected totalBytes = 53
}
This smells a lot like homework to me. The crucial phrase is: "Using delta encoding."
Delta encoding means you encode the delta (difference) between each number and the next:
67497, 67376, 67173, 67235, 67057, 67031, 66951, 66974, 67042, 67025, 66897, 67077, 67082, 67033, 67019, 67149, 67044, 67012, 67220, 67239, 66893, 66984, 66866, 66693, 66770, 66722, 66620, 66579, 66596, 66713, 66852, 66715
would turn into:
[Base: 67497]: -121, -203, +62
and so on. Assuming 8-bit bytes, the original numbers require 3 bytes apiece (and given the number of compilers with 3-byte integer types, you're normally going to end up with 4 bytes apiece). From the looks of things, the differences will fit quite easily in 2 bytes apiece, and if you can ignore one (or possibly two) of the least significant bits, you can fit them in one byte apiece.
Delta encoding is most often used for things like sound encoding where you can "fudge" the accuracy at times without major problems. For example, if you have a change from one sample to the next that's larger than you've left space to encode, you can encode a maximum change in the current difference, and add the difference to the next delta (and if you don't mind some back-tracking, you can distribute some to the previous delta as well). This will act as a low-pass filter, limiting the gradient between samples.
For example, in the series you gave, a simple delta encoding requires ten bits to represent all the differences. By dropping the LSB, however, nearly all the samples (all but one, in fact) can be encoded in 8 bits. That one has a difference (right shifted one bit) of -173, so if we represent it as -128, we have 45 left. We can distribute that error evenly between the preceding and following sample. In that case, the output won't be an exact match for the input, but if we're talking about something like sound, the difference probably won't be particularly obvious.
I did mention that it was an exercise that I had to complete and the solution that I received was deemed not good enough, so I wanted some constructive feedback seeing as actual companies never decide to tell you what you did wrong.
When the array is compressed I store the differences and not the original values except the first as this was my understanding. If you had looked at my code I have provided a full solution but my question was how bad was it?