I know that the most common method to test endianity programmatically is to cast to char* like this:
short temp = 0x1234;
char* tempChar = (char*)&temp;
But can it be done by casting to short* like this:
unsigned char test[2] = {1,0};
if ( *(short *)test == 1)
//Little-Endian
else
//Big-Endian
Am I right that the "test" buffer will be saved (on x86 platforms) in the memory using Little-Endian convention (from right-to-left: "0" at lower address, "1" at higher) just like in case with the "temp" var?
And more generally if I have a string:
char tab[] = "abcdef";
How would it be stored in the memory? Will it be reversed like: "fedcba"?
Thx. in advance:-)
PS.
Is there any way to see how exactly the data of a program look like in the system memory?.
I would like to see that byte-swap in Little-Endian in "real life".
Your code would probably work in practice (you could have just tried it!). However, technically, it invokes undefined behaviour; the standard doesn't allow you to access a char array through a pointer of another type.
And more generally if I have a string: char tab[] = "abcdef"; How would it be stored in the memory? Will it be reversed like: "fedcba"?
No. Otherwise tab[0] would give you f.
Your alternative method for checking endianness would work.
char tab[] = "abcdef" would be stored in that same order: abcdef
Endianness comes into play when you access multiple bytes (short, int, and so on). When you try to access tab[] as a short array using a little endian machine, you'd read it as ba, dc, fe (whatever their actual byte equivalents are, this is the order the chars are "evaluated" in the short).
It would be safer, i.e. standards-compliant, to use a union.
Both ways are not guaranteed to work, furthermore, latter invokes undefined behavior.
First fails if sizeof(char) == sizeof(short).
Second may fail for the same reason, and is also unsafe: result of the pointer cast may have wrong alignment for short, and accessing the (short) value invokes undefined behavior (3.10.15).
But yes, the char buffer is stored sequentially into memory so that &test[0] < &test[1],
and more generally, as others have already said, char tab[] = "abcdef" is not reversed or otherwise permuted regardless of endianness.
Related
Is there anything wrong with the casting a double pointer to a char pointer? Goal in the following code is to change the 1 element in three different ways.
double vec1[100];
double *vp = vec1;
char *yp = (char*) vp;
vp++;
vec1[1] = 19.0;
*vp = 12.0;
*((double*) (yp + (1*sizeof (vec1[0])))) = 34.0;
Casts of this type fall into the category of "OK if you know what you're doing but dangerous if you don't".
For example, in this case you already know the pointer value of "yp" (it was pointing to a double) so it is technically safe to increase its value by the size of a double and re-cast back to a double*.
A counter-example: suppose you didn't know where the char* came from...say, it was given to you as a function parameter. Now, your cast would be a big problem: since char* is technically 1-byte-aligned and a double is usually 8-byte-aligned, you can't be sure if you were given an 8-byte-aligned address. If it's aligned, your arithmetic would produce a valid double*; if not, it would crash when dereferenced.
This is just one example of how casts can go wrong. What you're doing (at first glance) looks like it will work but in general you really have to pay attention when you cast things.
With newer INTEL processors the main problem you can run into is alignment. Say you were to write something like this:
*((double*) (yp + 4)) = 34.0;
Then you are likely to have a runtime error because a double should be aligned on 8 bytes. This was also true on processors such as 68k, or MIPS.
This is similar to having a structure and doing casts on that structure. You are not unlikely to break things.
In most cases, if you can avoid such, your code will be a lot stronger. Personally, I do not even use such casts when reading a file. Instead, I get the data from the file and put it in a structure as required. Say I read 4 bytes in a buffer to convert to an integer, I'd write something like this:
unsigned char buf[4];
...
fread(buf, 1, 4, f);
my_struct.integer = buf[0] | (buf[1] << 8) | (buf[2] << 16) | (buf[3] << 24);
Now I did not do an ugly cast and I could control the endianess of the integer in the file whatever the endian of the processor you are running with.
What is the use of unsigned char pointers? I have seen it at many places that pointer is type cast to pointer to unsinged char Why do we do so?
We receive a pointer to int and then type cast it to unsigned char*. But if we try to print element in that array using cout it does not print anything. why? I do not understand. I am new to c++.
EDIT Sample Code Below
int Stash::add(void* element)
{
if(next >= quantity)
// Enough space left?
inflate(increment);
// Copy element into storage, starting at next empty space:
int startBytes = next * size;
unsigned char* e = (unsigned char*)element;
for(int i = 0; i < size; i++)
storage[startBytes + i] = e[i];
next++;
return(next - 1); // Index number
}
You are actually looking for pointer arithmetic:
unsigned char* bytes = (unsigned char*)ptr;
for(int i = 0; i < size; i++)
// work with bytes[i]
In this example, bytes[i] is equal to *(bytes + i) and it is used to access the memory on the address: bytes + (i* sizeof(*bytes)). In other words: If you have int* intPtr and you try to access intPtr[1], you are actually accessing the integer stored at bytes: 4 to 7:
0 1 2 3
4 5 6 7 <--
The size of type your pointer points to affects where it points after it is incremented / decremented. So if you want to iterate your data byte by byte, you need to have a pointer to type of size 1 byte (that's why unsigned char*).
unsigned char is usually used for holding binary data where 0 is valid value and still part of your data. While working with "naked" unsigned char* you'll probably have to hold the length of your buffer.
char is usually used for holding characters representing string and 0 is equal to '\0' (terminating character). If your buffer of characters is always terminated with '\0', you don't need to know it's length because terminating character exactly specifies the end of your data.
Note that in both of these cases it's better to use some object that hides the internal representation of your data and will take care of memory management for you (see RAII idiom). So it's much better idea to use either std::vector<unsigned char> (for binary data) or std::string (for string).
In C, unsigned char is the only type guaranteed to have no trapping values, and which guarantees copying will result in an exact bitwise image. (C++ extends this guarantee to char as well.) For this reason, it is traditionally used for "raw memory" (e.g. the semantics of memcpy are defined in terms of unsigned char).
In addition, unsigned integral types in general are used when bitwise operations (&, |, >> etc.) are going to be used. unsigned char is the smallest unsigned integral type, and may be used when manipulating arrays of small values on which bitwise operations are used. Occasionally, it's also used because one needs the modulo behavior in case of overflow, although this is more frequent with larger types (e.g. when calculating a hash value). Both of these reasons apply to unsigned types in general; unsigned char will normally only be used for them when there is a need to reduce memory use.
The unsinged char type is usually used as a representation of a single byte of binary data. Thus, and array is often used as a binary data buffer, where each element is a singe byte.
The unsigned char* construct will be a pointer to the binary data buffer (or its 1st element).
I am not 100% sure what does c++ standard precisely says about size of unsigned char, whether it is fixed to be 8 bit or not. Usually it is. I will try to find and post it.
After seeing your code
When you use something like void* input as a parameter of a function, you deliberately strip down information about inputs original type. This is very strong suggestion that the input will be treated in very general manner. I.e. as a arbitrary string of bytes. int* input on the other hand would suggest it will be treated as a "string" of singed integers.
void* is mostly used in cases when input gets encoded, or treated bit/byte wise for whatever reason, since you cannot draw conclusions about its contents.
Then In your function you seem to want to treat the input as a string of bytes. But to operate on objects, e.g. performing operator= (assignment) the compiler needs to know what to do. Since you declare input as void* assignment such as *input = something would have no sense because *input is of void type. To make compiler to treat input elements as the "smallest raw memory pieces" you cast it to the appropriate type which is unsigned int.
The cout probably did not work because of wrong or unintended type conversion. char* is considered a null terminated string and it is easy to confuse singed and unsigned versionin code. If you pass unsinged char* to ostream::operator<< as a char* it will treat and expect the byte input as normal ASCII characters, where 0 is meant to be end of string not an integer value of 0. When you want to print contents of memory it is best to explicitly cast pointers.
Also note that to print memory contents of a buffer you would need to use a loop, since other wise the printing function would not know when to stop.
Unsigned char pointers are useful when you want to access the data byte by byte. For example, a function that copies data from one area to another could need this:
void memcpy (unsigned char* dest, unsigned char* source, unsigned count)
{
for (unsigned i = 0; i < count; i++)
dest[i] = source[i];
}
It also has to do with the fact that the byte is the smallest addressable unit of memory. If you want to read anything smaller than a byte from memory, you need to get the byte that contains that information, and then select the information using bit operations.
You could very well copy the data in the above function using a int pointer, but that would copy chunks of 4 bytes, which may not be the correct behavior in some situations.
Why nothing appears on the screen when you try to use cout, the most likely explanation is that the data starts with a zero character, which in C++ marks the end of a string of characters.
I am trying to insert a uint16_t value into a uint8_t array using pointers. I would think below would work, but haven't been able to do it. Any clues as to what the problem is?
uint8_t myarray[10];
uint16_t value = 10000;
uint16_t * myptr = (uint16_t *)(myarray+2);
*myptr = value;
I know I can do it like so, but why doesn't above work?
uint8_t myarray[10];
uint16_t value = 10000;
uint8_t * myptr = (myarray+2);
uint8_t * myptr2 =(myarray+3);
*myptr = value>>8;
*myptr2 =value;
The second version writes the most significant byte (with value 39) to myarray[2], and the least significant (with value 16) to myarray[3].
The first version will write the two bytes in an order determined by the endianness of your computer. Most modern computers are little-endian, meaning that the least significant byte of a multi-byte integer value comes first in memory - so this version will write the two bytes in the opposite order to the other version.
I'm assuming that that's the problem you're seeing; if it's something else, then please be more specific than "haven't been able to do it".
Also, the first version technically has undefined behaviour, and might do something completely unexpected on a sufficiently exotic computer. I suggest that you stick to well-defined code like the second version; only use dubious optimisations if profiling reveals both that the well-defined code is too slow, and that the dodgy pointer-aliasing code is faster. I would also suggest using reinterpret_cast rather than the evil C-style cast; it wouldn't change the behaviour, but it would be easier to see that there's something wonky going on.
You can do it like this:
uint8_t * value_data = reinterpret_cast<uint8_t*>(&value); // cast to `(unsigned) char*` is allowed by standard
myarray[0] = value_data[0];
myarray[1] = value_data[1];
Lets us consider this snippet:
int s;
scanf("%c",&s);
Here I have used int, and not char, for variable s, now for using s for character conversion safely I have to make it char again because when scanf reads a character it only overwrites one byte of the variable it is assigning it to, and not all four that int has.
For conversion I could use s = (char)s; as the next line, but is it possible to implement the same by subtracting something from s ?
What you've done is technically undefined behaviour. The %c format calls for a char*, you've passed it an int* which will (roughly speaking) be reinterpreted. Even assuming that the pointer value is still good after reinterpreting, storing an arbitrary character to the first byte of an int and then reading it back as int is undefined behaviour. Even if it were defined, reading an int when 3 bytes of it are uninitialized, is undefined behaviour.
In practice it probably does something sensible on your machine, and you just get garbage in the top 3 bytes (assuming little-endian).
Writing s = (char)s converts the value from int to char and then back to int again. This is implementation-defined behaviour: converting an out-of-range value to a signed type. On different implementations it might clean up the top 3 bytes, it might return some other result, or it might raise a signal.
The proper way to use scanf is:
char c;
scanf("%c", &c);
And then either int s = c; or int s = (unsigned char)c;, according to whether you want negative-valued characters to result in a negative integer, or a positive integer (up to 255, assuming 8-bit char).
I can't think of any good reason for using scanf improperly. There are good reasons for not using scanf at all, though:
int s = getchar();
Are you trying to convert a digit to its decimal value? If so, then
char c = '8';
int n = c - '0';
n should 8 at this point.
That's probably not a good idea; GCC gives me a warning for that code:
main.c:10: warning: format ‘%c’ expects type ‘char *’, but
argument 2 has type ‘int *’
In this case you're ok since you're passing a pointer to more space than you need (for most systems), but what if you did it the other way around? Could be crash city. If you really want to do something like what you have there, just do the typecast or mask it - the mask will be endian-dependent.
As written this won't work reliably . The argument, &s, to scanf is a pointer to int and scanf is expecting a pointer to char. The two data type (int and char) have different sizes (at least on most architectures) so the data may get put in the wrong spot in memeory, and the other part of s may not get properly cleared.
The answers suggesting manipulation of the result after using a pointer to int rely on unspecified behavior (i.e. that scanf will put the character value it has in the least significant byte of the int you're pointing to), and are not safe.
Not but you could use the following:
s = s & 0xFF
That will blank out all of the data except the first byte. But in general all these ideas (and the ones above) are bad ideas, since not all systems store the lowest part of the integer in memory first. So if you ever have to port this code to a big endian system, you'll be screwed.
True, you may never have to port the code, but why write unportable code to begin with?
See this for more info:
http://en.wikipedia.org/wiki/Endianness
I have been working on a legacy C++ application and am definitely outside of my comfort-zone (a good thing). I was wondering if anyone out there would be so kind as to give me a few pointers (pun intended).
I need to cast 2 bytes in an unsigned char array to an unsigned short. The bytes are consecutive.
For an example of what I am trying to do:
I receive a string from a socket and place it in an unsigned char array. I can ignore the first byte and then the next 2 bytes should be converted to an unsigned char. This will be on windows only so there are no Big/Little Endian issues (that I am aware of).
Here is what I have now (not working obviously):
//packetBuffer is an unsigned char array containing the string "123456789" for testing
//I need to convert bytes 2 and 3 into the short, 2 being the most significant byte
//so I would expect to get 515 (2*256 + 3) instead all the code I have tried gives me
//either errors or 2 (only converting one byte
unsigned short myShort;
myShort = static_cast<unsigned_short>(packetBuffer[1])
Well, you are widening the char into a short value. What you want is to interpret two bytes as an short. static_cast cannot cast from unsigned char* to unsigned short*. You have to cast to void*, then to unsigned short*:
unsigned short *p = static_cast<unsigned short*>(static_cast<void*>(&packetBuffer[1]));
Now, you can dereference p and get the short value. But the problem with this approach is that you cast from unsigned char*, to void* and then to some different type. The Standard doesn't guarantee the address remains the same (and in addition, dereferencing that pointer would be undefined behavior). A better approach is to use bit-shifting, which will always work:
unsigned short p = (packetBuffer[1] << 8) | packetBuffer[2];
This is probably well below what you care about, but keep in mind that you could easily get an unaligned access doing this. x86 is forgiving and the abort that the unaligned access causes will be caught internally and will end up with a copy and return of the value so your app won't know any different (though it's significantly slower than an aligned access). If, however, this code will run on a non-x86 (you don't mention the target platform, so I'm assuming x86 desktop Windows), then doing this will cause a processor data abort and you'll have to manually copy the data to an aligned address before trying to cast it.
In short, if you're going to be doing this access a lot, you might look at making adjustments to the code so as not to have unaligned reads and you'll see a perfromance benefit.
unsigned short myShort = *(unsigned short *)&packetBuffer[1];
The bit shift above has a bug:
unsigned short p = (packetBuffer[1] << 8) | packetBuffer[2];
if packetBuffer is in bytes (8 bits wide) then the above shift can and will turn packetBuffer into a zero, leaving you with only packetBuffer[2];
Despite that this is still preferred to pointers. To avoid the above problem, I waste a few lines of code (other than quite-literal-zero-optimization) it results in the same machine code:
unsigned short p;
p = packetBuffer[1]; p <<= 8; p |= packetBuffer[2];
Or to save some clock cycles and not shift the bits off the end:
unsigned short p;
p = (((unsigned short)packetBuffer[1])<<8) | packetBuffer[2];
You have to be careful with pointers, the optimizer will bite you, as well as memory alignments and a long list of other problems. Yes, done right it is faster, done wrong the bug can linger for a long time and strike when least desired.
Say you were lazy and wanted to do some 16 bit math on an 8 bit array. (little endian)
unsigned short *s;
unsigned char b[10];
s=(unsigned short *)&b[0];
if(b[0]&7)
{
*s = *s+8;
*s &= ~7;
}
do_something_With(b);
*s=*s+8;
do_something_With(b);
*s=*s+8;
do_something_With(b);
There is no guarantee that a perfectly bug free compiler will create the code you expect. The byte array b sent to the do_something_with() function may never get modified by the *s operations. Nothing in the code above says that it should. If you don't optimize your code then you may never see this problem (until someone does optimize or changes compilers or compiler versions). If you use a debugger you may never see this problem (until it is too late).
The compiler doesn't see the connection between s and b, they are two completely separate items. The optimizer may choose not to write *s back to memory because it sees that *s has a number of operations so it can keep that value in a register and only save it to memory at the end (if ever).
There are three basic ways to fix the pointer problem above:
Declare s as volatile.
Use a union.
Use a function or functions whenever changing types.
You should not cast a unsigned char pointer into an unsigned short pointer (for that matter cast from a pointer of smaller data type to a larger data type). This is because it is assumed that the address will be aligned correctly. A better approach is to shift the bytes into a real unsigned short object, or memcpy to a unsigned short array.
No doubt, you can adjust the compiler settings to get around this limitation, but this is a very subtle thing that will break in the future if the code gets passed around and reused.
Maybe this is a very late solution but i just want to share with you. When you want to convert primitives or other types you can use union. See below:
union CharToStruct {
char charArray[2];
unsigned short value;
};
short toShort(char* value){
CharToStruct cs;
cs.charArray[0] = value[1]; // most significant bit of short is not first bit of char array
cs.charArray[1] = value[0];
return cs.value;
}
When you create an array with below hex values and call toShort function, you will get a short value with 3.
char array[2];
array[0] = 0x00;
array[1] = 0x03;
short i = toShort(array);
cout << i << endl; // or printf("%h", i);
static cast has a different syntax, plus you need to work with pointers, what you want to do is:
unsigned short *myShort = static_cast<unsigned short*>(&packetBuffer[1]);
Did nobody see the input was a string!
/* If it is a string as explicitly stated in the question.
*/
int byte1 = packetBuffer[1] - '0'; // convert 1st byte from char to number.
int byte2 = packetBuffer[2] - '0';
unsigned short result = (byte1 * 256) + byte2;
/* Alternatively if is an array of bytes.
*/
int byte1 = packetBuffer[1];
int byte2 = packetBuffer[2];
unsigned short result = (byte1 * 256) + byte2;
This also avoids the problems with alignment that most of the other solutions may have on certain platforms. Note A short is at least two bytes. Most systems will give you a memory error if you try and de-reference a short pointer that is not 2 byte aligned (or whatever the sizeof(short) on your system is)!
char packetBuffer[] = {1, 2, 3};
unsigned short myShort = * reinterpret_cast<unsigned short*>(&packetBuffer[1]);
I (had to) do this all the time. big endian is an obvious problem. What really will get you is incorrect data when the machine dislike misaligned reads! (and write).
you may want to write a test cast and an assert to see if it reads properly. So when ran on a big endian machine or more importantly a machine that dislikes misaligned reads an assert error will occur instead of a weird hard to trace 'bug' ;)
On windows you can use:
unsigned short i = MAKEWORD(lowbyte,hibyte);
I realize this is an old thread, and I can't say that I tried every suggestion made here. I'm just making my self comfortable with mfc, and I was looking for a way to convert a uint to two bytes, and back again at the other end of a socket.
There are alot of bit shifting examples you can find on the net, but none of them seemed to actually work. Alot of the examples seem overly complicated; I mean we're just talking about grabbing 2 bytes out of a uint, sending them over the wire, and plugging them back into a uint at the other end, right?
This is the solution I finally came up with:
class ByteConverter
{
public:
static void uIntToBytes(unsigned int theUint, char* bytes)
{
unsigned int tInt = theUint;
void *uintConverter = &tInt;
char *theBytes = (char*)uintConverter;
bytes[0] = theBytes[0];
bytes[1] = theBytes[1];
}
static unsigned int bytesToUint(char *bytes)
{
unsigned theUint = 0;
void *uintConverter = &theUint;
char *thebytes = (char*)uintConverter;
thebytes[0] = bytes[0];
thebytes[1] = bytes[1];
return theUint;
}
};
Used like this:
unsigned int theUint;
char bytes[2];
CString msg;
ByteConverter::uIntToBytes(65000,bytes);
theUint = ByteConverter::bytesToUint(bytes);
msg.Format(_T("theUint = %d"), theUint);
AfxMessageBox(msg, MB_ICONINFORMATION | MB_OK);
Hope this helps someone out.