I'm new on C programming and I'm testing some code where I receive and process an UDP packet formatted as follow:
UINT16 port1
UINT16 port2
The corresponding values on this test are:
6005
5555
If I print the whole packet buffer I get something like this:
u^W³^U><9e>^D
So I thought that I would just have to break it and process as an unsigned int of 16 bytes. So I've tried something like this:
int l = 0;
unsigned int *primaryPort = *(unsigned int) &buffer[l];
AddToLog(logInfo, "PrimaryPort: %u\n", primaryPort);
l += sizeof(primaryPort);
unsigned int *secondaryPort = *(unsigned int) &buffer[l];
AddToLog(logInfo, "SecondaryPort: %u\n", secondaryPort);
l += sizeof(secondaryPort);
But I get wrong numbers with 8 digits.
I even tried another approach like follow, but also get the wrong number as well.
int l = 0;
unsigned char primaryPort[16];
snprintf(primaryPort, sizeof(primaryPort), "%u", &buffer[l]);
AddToLog(logInfo, "PrimaryPort: %d\n", primaryPort);
l += sizeof(primaryPort);
unsigned char secondaryPort[16];
snprintf(secondaryPort, sizeof(secondaryPort), "%u", &buffer[l]);
AddToLog(logInfo, "SecondaryPort: %d\n", secondaryPort);
l += sizeof(secondaryPort);
What I'm doing wrong? Also, why I have to free on a char string variables, but I don't need to free on int variables?
You are passing to AddToLog and snprintf pointers to the integers. So what you're seeing are the addresses of the integers, not the integers themselves.
You need to dereference your pointers -- for example, put an asterisk (*) in front of primaryPort in your calls to AddToLog in your first approach.
As #rileyberton suggests, most likely unsigned int is 4 bytes on your system, which is the C99 type uint32_t. For a 16-bit integer, use uint16_t. These are defined in stdint.h. These are traditionally called "short integers" or "half integers" and require the %hu qualifier in printf or similar functions, rather than just %u (which stands for unsigned int, whose size depends on the target machine.)
Also, as #igor-tandetnik suggests, you may need to switch the byte order of the integers in your packet, if for example the packet is using network order (big-endian) format and your target machine is using little-endian format.
unsigned int on your system is likely 4 bytes (uint32_t). You can use unsigned int here if you mask out the values in the correct endianess, or simply use a short.
int l = 0;
unsigned short *primaryPort = *(unsigned short) &buffer[l];
AddToLog(logInfo, "PrimaryPort: %u\n", primaryPort);
l += sizeof(*primaryPort);
unsigned short *secondaryPort = *(unsigned short) &buffer[l];
AddToLog(logInfo, "SecondaryPort: %u\n", secondaryPort);
l += sizeof(*secondaryPort);
You declared primaryPort and secondaryPort to be pointers to unsigned short.
But when you assign them values from a section of buffer, you already de-referenced the pointer. You don't need pointers-to-unsigned-short. You just need an unsigned short.
Change it to:
unsigned short primaryPort = *((unsigned short*) &buffer[l]);
unsigned short secondaryPort = *((unsigned short *) &buffer[l]);
Note the removal of a * in the variable declarations.
If you're still having problems, you'll need to examine buffer byte-by-byte, looking for the value you expect. You can expect that 6005 will show up as either hex 17 75 or 75 17, depending on your platform's endianness.
Related
What will be the output of the following C code. Assuming it runs on Little endian machine, where short int takes 2 Bytes and char takes 1 Byte.
#include<stdio.h>
int main() {
short int c[5];
int i = 0;
for(i = 0; i < 5; i++)
c[i] = 400 + i;
char *b = (char *)c;
printf("%d", *(b+8));
return 0;
}
In my machine it gave
-108
I don't know if my machine is Little endian or big endian. I found somewhere that it should give
148
as the output. Because low order 8 bits of 404(i.e. element c[4]) is 148. But I think that due to "%d", it should read 2 Bytes from memory starting from the address of c[4].
The code gives different outputs on different computers because on some platforms the char type is signed by default and on others it's unsigned by default. That has nothing to do with endianness. Try this:
char *b = (char *)c;
printf("%d\n", (unsigned char)*(b+8)); // always prints 148
printf("%d\n", (signed char)*(b+8)); // always prints -108 (=-256 +148)
The default value is dependent on the platform and compiler settings. You can control the default behavior with GCC options -fsigned-char and -funsigned-char.
c[4] stores 404. In a two-byte little-endian representation, that means two bytes of 0x94 0x01, or (in decimal) 148 1.
b+8 addresses the memory of c[4]. b is a pointer to char, so the 8 means adding 8 bytes (which is 4 two-byte shorts). In other words, b+8 points to the first byte of c[4], which contains 148.
*(b+8) (which could also be written as b[8]) dereferences the pointer and thus gives you the value 148 as a char. What this does is implementation-defined: On many common platforms char is a signed type (with a range of -128 .. 127), so it can't actually be 148. But if it is an unsigned type (with a range of 0 .. 255), then 148 is fine.
The bit pattern for 148 in binary is 10010100. Interpreting this as a two's complement number gives you -108.
This char value (of either 148 or -108) is then automatically converted to int because it appears in the argument list of a variable-argument function (printf). This doesn't change the value.
Finally, "%d" tells printf to take the int argument and format it as a decimal number.
So, to recap: Assuming you have a machine where
a byte is 8 bits
negative numbers use two's complement
short int is 2 bytes
... then this program will output either -108 (if char is a signed type) or 148 (if char is an unsigned type).
To see what sizes types have in your system:
printf("char = %u\n", sizeof(char));
printf("short = %u\n", sizeof(short));
printf("int = %u\n", sizeof(int));
printf("long = %u\n", sizeof(long));
printf("long long = %u\n", sizeof(long long));
Change the lines in your program
unsigned char *b = (unsigned char *)c;
printf("%d\n", *(b + 8));
And simple test (I know that it is not guaranteed but all C compilers I know do it this way and I do not care about old CDC or UNISYS machines which had different addresses and pointers to different types of data
printf(" endianes test: %s\n", (*b + (unsigned)*(b + 1) * 0x100) == 400? "little" : "big");
Another remark: it is only because in your program c[0] == 400
This code snippet is excerpted from a linux book.
If this is not appropriate to post the code snippet here, please let me know.
I will delete it. Thanks.
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
char buf[30];
char *p;
int i;
unsigned int index = 0;
//unsigned long index = 0;
printf("index-1 = %lx (sizeof %d)\n", index-1, sizeof(index-1));
for(i = 'A'; i <= 'Z'; i++)
buf[i - 'A'] = i;
p = &buf[1];
printf("%c: buf=%p p=%p p[-1]=%p\n", p[index-1], buf, p, &p[index-1]);
return 0;
}
On 32-bit OS environment:
This program works fine no matter the data type of index is unsigned int or unsigned long.
On 64-bit OS environment:
The same program will run into "core dump" if index is declared as unsigned int.
However, if I only change the data type of index from unsigned int to a) unsigned long or b) unsigned short,
this program works fine too.
The reason from the book only tells me that 64-bit will cause the core-dump due to non-negative number. But I have no idea exactly about the reason why unsigned long and unsigned short work but unsigned int.
What I am confused is that
p + (0u -1) == p + UINT_MAX when index is unsigned int.
BUT,
p + (0ul - 1) == p[-1] when index is unsigned long.
I get stuck at here.
If anyone can help to elaborate the details, it is highly appreciated!
Thank you.
Here comes some result on my 32 bit(RHEL5.10/gcc version 4.1.2 20080704)
and 64 bit machine (RHEL6.3/gcc version 4.4.6 20120305)
I am not sure if gcc version makes any difference here.
So, I paste the information as well.
On 32 bit:
I tried two changes:
1) Modify unsigned int index = 0 to unsigned short index = 0.
2) Modify unsigned int index = 0 to unsigned char index = 0.
The program can run without problem.
index-1 = ffffffff (sizeof 4)
A: buf=0xbfbdd5da p=0xbfbdd5db p[-1]=0xbfbdd5da
It seems that the data type of index will be promoted to 4 bytes due to -1.
On 64 bit:
I tried three changes:
1) Modify unsigned int index = 0 to unsigned char index = 0.
It works!
index-1 = ffffffff (sizeof 4)
A: buf=0x7fffef304ae0 p=0x7fffef304ae1 p[-1]=0x7fffef304ae0
2) Modify unsigned int index = 0 to unsigned short index = 0.
It works!
index-1 = ffffffff (sizeof 4)
A: buf=0x7fff48233170 p=0x7fff48233171 p[-1]=0x7fff48233170
3) Modify unsigned int index = 0 to unsigned long index = 0.
It works!
index-1 = ffffffff (sizeof 8)
A: buf=0x7fffb81d6c20 p=0x7fffb81d6c21 p[-1]=0x7fffb81d6c20
BUT, only
unsigned int index = 0 runs into the core dump at the last printf.
index-1 = ffffffff (sizeof 4)
Segmentation fault (core dumped)
Do not lie to the compiler!
Passing printf an int where it expects a long (%ld) is undefined behavior.
(Creating a pointer pointing outside any valid object (and not just behind one) is UB too...)
Correct the format specifiers and the pointer arithmetic (that includes indexing as a special case) and everything will work.
UB includes "It works as expected" as well as "Catastrophic failure".
BTW: If you politely ask your compiler for all warnings, it would warn you. Use -Wall -Wextra -pedantic or similar.
One other problem is code has is in your printf():
printf("index-1 = %lx (sizeof %d)\n", index-1, sizeof(index-1));
Lets simplify:
int i = 100;
print("%lx", i-1);
You are telling printf here is a long but in reality you are sending an int. clang does tell you the corrent warning (I think gcc should also spit the correct waring). See:
test1.c:6:19: warning: format specifies type 'unsigned long' but the argument has type 'int' [-Wformat]
printf("%lx", i - 100);
~~~ ^~~~~~~
%x
1 warning generated.
Solution is simple: you need to pass a long to printf or tell printf to print an int:
printf("%lx", (long)(i-100) );
printf("%x", i-100);
You got luck on 32bit and your app did not crash. Porting it to 64bit revealed a bug in your code and you can now fix it.
Arithmetic on unsigned values is always defined, in terms of wrap-around. E.g. (unsigned)-1 is the same as UINT_MAX. So an expression like
p + (0u-1)
is equivalent to
p + UINT_MAX
(&p[0u-1] is equivalent to &*(p + (0u-1)) and p + (0u-1)).
Maybe this is easier to understand if we replace the pointers with unsigned integer types. Consider:
uint32_t p32; // say, this is a 32-bit "pointer"
uint64_t p64; // a 64-bit "pointer"
Assuming 16, 32, and 64 bit for short, int, and long, respectively (entries on the same line equal):
p32 + (unsigned short)-1 p32 + USHRT_MAX p32 + (UINT_MAX>>16)
p32 + (0u-1) p32 + UINT_MAX p32 - 1
p32 + (0ul-1) p32 + ULONG_MAX p32 + UINT_MAX p32 - 1
p64 + (0u-1) p64 + UINT_MAX
p64 + (0ul-1) p64 + ULONG_MAX p64 - 1
You can always replace operands of addition, subtraction and multiplication on unsigned types by something congruent modulo the maximum value + 1. For example,
-1 ☰ ffffffffhex mod 232
(ffffffffhex is 232-1 or UINT_MAX), and also
ffffffffffffffffhex ☰ ffffffffhex mod 232
(for a 32-bit unsigned type you can always truncate to the least-significant 8 hex-digits).
Your examples:
32-bit
unsigned short index = 0;
In index - 1, index is promoted to int. The result has type int and value -1 (which is negative). Same for unsigned char.
64-bit
unsigned char index = 0;
unsigned short index = 0;
Same as for 32-bit. index is promoted to int, index - 1 is negative.
unsigned long index = 0;
The output
index-1 = ffffffff (sizeof 8)
is weird, it’s your only correct use of %lx but looks like you’ve printed it with %x (expecting 4 bytes); on my 64-bit computer (with 64-bit long) and with %lx I get:
index-1 = ffffffffffffffff (sizeof 8)
ffffffffffffffffhex is -1 modulo 264.
unsigned index = 0;
An int cannot hold any value unsigned int can, so in index - 1 nothing is promoted to int, the result has type unsigned int and value -1 (which is positive, being the same as UINT_MAX or ffffffffhex, since the type is unsigned). For 32-bit-addresses, adding this value is the same as subtracting one:
bfbdd5db bfbdd5db
+ ffffffff - 1
= 1bfbdd5da
= bfbdd5da = bfbdd5da
(Note the wrap-around/truncation.) For 64-bit addresses, however:
00007fff b81d6c21
+ ffffffff
= 00008000 b81d6c20
with no wrap-around. This is trying to access an invalid address, so you get a segfault.
Maybe have a look at 2’s complement on Wikipedia.
Under my 64-bit Linux, using a specifier expecting a 32-bit value while passing a 64-bit type (and the other way round) seems to “work”, only the 32 least-significant bits are read. But use the correct ones. lx expects an unsigned long, unmodified x an unsigned int, hx an unsigned short (an unsigned short is promoted to int when passed to printf (it’s passed as a variable argument), due to default argument promotions). The length modifier for size_t is z, as in %zu:
printf("index-1 = %lx (sizeof %zu)\n", (unsigned long)(index-1), sizeof(index-1));
(The conversion to unsigned long doesn’t change the value of an unsigned int, unsigned short, or unsigned char expression.)
sizeof(index-1) could also have been written as sizeof(+index), the only effect on the size of the expression are the usual arithmetic conversions, which are also triggered by unary +.
A char is 1 byte and an integer is 4 bytes. I want to copy byte-by-byte from a char[4] into an integer. I thought of different methods but I'm getting different answers.
char str[4]="abc";
unsigned int a = *(unsigned int*)str;
unsigned int b = str[0]<<24 | str[1]<<16 | str[2]<<8 | str[3];
unsigned int c;
memcpy(&c, str, 4);
printf("%u %u %u\n", a, b, c);
Output is
6513249 1633837824 6513249
Which one is correct? What is going wrong?
It's an endianness issue. When you interpret the char* as an int* the first byte of the string becomes the least significant byte of the integer (because you ran this code on x86 which is little endian), while with the manual conversion the first byte becomes the most significant.
To put this into pictures, this is the source array:
a b c \0
+------+------+------+------+
| 0x61 | 0x62 | 0x63 | 0x00 | <---- bytes in memory
+------+------+------+------+
When these bytes are interpreted as an integer in a little endian architecture the result is 0x00636261, which is decimal 6513249. On the other hand, placing each byte manually yields 0x61626300 -- decimal 1633837824.
Of course treating a char* as an int* is undefined behavior, so the difference is not important in practice because you are not really allowed to use the first conversion. There is however a way to achieve the same result, which is called type punning:
union {
char str[4];
unsigned int ui;
} u;
strcpy(u.str, "abc");
printf("%u\n", u.ui);
Neither of the first two is correct.
The first violates aliasing rules and may fail because the address of str is not properly aligned for an unsigned int. To reinterpret the bytes of a string as an unsigned int with the host system byte order, you may copy it with memcpy:
unsigned int a; memcpy(&a, &str, sizeof a);
(Presuming the size of an unsigned int and the size of str are the same.)
The second may fail with integer overflow because str[0] is promoted to an int, so str[0]<<24 has type int, but the value required by the shift may be larger than is representable in an int. To remedy this, use:
unsigned int b = (unsigned int) str[0] << 24 | …;
This second method interprets the bytes from str in big-endian order, regardless of the order of bytes in an unsigned int in the host system.
unsigned int a = *(unsigned int*)str;
This initialization is not correct and invokes undefined behavior. It violates C aliasing rules an potentially violates processor alignment.
You said you want to copy byte-by-byte.
That means the the line unsigned int a = *(unsigned int*)str; is not allowed. However, what you're doing is a fairly common way of reading an array as a different type (such as when you're reading a stream from disk.
It just needs some tweaking:
char * str ="abc";
int i;
unsigned a;
char * c = (char * )&a;
for(i = 0; i < sizeof(unsigned); i++){
c[i] = str[i];
}
printf("%d\n", a);
Bear in mind, the data you're reading may not share the same endianness as the machine you're reading from. This might help:
void
changeEndian32(void * data)
{
uint8_t * cp = (uint8_t *) data;
union
{
uint32_t word;
uint8_t bytes[4];
}temp;
temp.bytes[0] = cp[3];
temp.bytes[1] = cp[2];
temp.bytes[2] = cp[1];
temp.bytes[3] = cp[0];
*((uint32_t *)data) = temp.word;
}
Both are correct in a way:
Your first solution copies in native byte order (i.e. the byte order the CPU uses) and thus may give different results depending on the type of CPU.
Your second solution copies in big endian byte order (i.e. most significant byte at lowest address) no matter what the CPU uses. It will yield the same value on all types of CPUs.
What is correct depends on how the original data (array of char) is meant to be interpreted.
E.g. Java code (class files) always use big endian byte order (no matter what the CPU is using). So if you want to read ints from a Java class file you have to use the second way. In other cases you might want to use the CPU dependent way (I think Matlab writes ints in native byte order into files, c.f. this question).
If your using CVI (National Instruments) compiler you can use the function Scan to do this:
unsigned int a;
For big endian:
Scan(str,"%1i[b4uzi1o3210]>%i",&a);
For little endian:
Scan(str,"%1i[b4uzi1o0123]>%i",&a);
The o modifier specifies the byte order.
i inside the square brackets indicates where to start in the str array.
I have an unsigned int number (2 byte) and I want to convert it to unsigned char type. From my search, I find that most people recommend to do the following:
unsigned int x;
...
unsigned char ch = (unsigned char)x;
Is the right approach? I ask because unsigned char is 1 byte and we casted from 2 byte data to 1 byte.
To prevent any data loss, I want to create an array of unsigned char[] and save the individual bytes into the array. I am stuck at the following:
unsigned char ch[2];
unsigned int num = 272;
for(i=0; i<2; i++){
// how should the individual bytes from num be saved in ch[0] and ch[1] ??
}
Also, how would we convert the unsigned char[2] back to unsigned int.
Thanks a lot.
You can use memcpy in that case:
memcpy(ch, (char*)&num, 2); /* although sizeof(int) would be better */
Also, how would be convert the unsigned char[2] back to unsigned int.
The same way, just reverse the arguments of memcpy.
How about:
ch[0] = num & 0xFF;
ch[1] = (num >> 8) & 0xFF;
The converse operation is left as an exercise.
How about using a union?
union {
unsigned int num;
unsigned char ch[2];
} theValue;
theValue.num = 272;
printf("The two bytes: %d and %d\n", theValue.ch[0], theValue.ch[1]);
It really depends on your goal: why do you want to convert this to an unsigned char? Depending on the answer to that there are a few different ways to do this:
Truncate: This is what was recomended. If you are just trying to squeeze data into a function which requires an unsigned char, simply cast uchar ch = (uchar)x (but, of course, beware of what happens if your int is too big).
Specific endian: Use this when your destination requires a specific format. Usually networking code likes everything converted to big endian arrays of chars:
int n = sizeof x;
for(int y=0; n-->0; y++)
ch[y] = (x>>(n*8))&0xff;
will does that.
Machine endian. Use this when there is no endianness requirement, and the data will only occur on one machine. The order of the array will change across different architectures. People usually take care of this with unions:
union {int x; char ch[sizeof (int)];} u;
u.x = 0xf00
//use u.ch
with memcpy:
uchar ch[sizeof(int)];
memcpy(&ch, &x, sizeof x);
or with the ever-dangerous simple casting (which is undefined behavior, and crashes on numerous systems):
char *ch = (unsigned char *)&x;
Of course, array of chars large enough to contain a larger value has to be exactly as big as this value itself.
So you can simply pretend that this larger value already is an array of chars:
unsigned int x = 12345678;//well, it should be just 1234.
unsigned char* pChars;
pChars = (unsigned char*) &x;
pChars[0];//one byte is here
pChars[1];//another byte here
(Once you understand what's going on, it can be done without any variables, all just casting)
You just need to extract those bytes using bitwise & operator. OxFF is a hexadecimal mask to extract one byte. Please look at various bit operations here - http://www.catonmat.net/blog/low-level-bit-hacks-you-absolutely-must-know/
An example program is as follows:
#include <stdio.h>
int main()
{
unsigned int i = 0x1122;
unsigned char c[2];
c[0] = i & 0xFF;
c[1] = (i>>8) & 0xFF;
printf("c[0] = %x \n", c[0]);
printf("c[1] = %x \n", c[1]);
printf("i = %x \n", i);
return 0;
}
Output:
$ gcc 1.c
$ ./a.out
c[0] = 22
c[1] = 11
i = 1122
$
Endorsing #abelenky suggestion, using an union would be a more fail proof way of doing this.
union unsigned_number {
unsigned int value; // An int is 4 bytes long
unsigned char index[4]; // A char is 1 byte long
};
The characteristics of this type is that the compiler will allocate memory only for the biggest member of our data structure unsigned_number, which in this case is going to be 4 bytes - since both members (value and index) have the same size. Had you defined it as a struct instead, we would have 8 bytes allocated on memory, since the compiler does its allocation for all the members of a struct.
Additionally, and here is where your problem is solved, the members of an union data structure all share the same memory location, which means they all refer to same data - think of that like a hard link on GNU/Linux systems.
So we would have:
union unsigned_number my_number;
// Assigning decimal value 202050300 to my_number
// which is represented as 0xC0B0AFC in hex format
my_number.value = 0xC0B0AFC; // Representation: Binary - Decimal
// Byte 3: 00001100 - 12
// Byte 2: 00001011 - 11
// Byte 1: 00001010 - 10
// Byte 0: 11111100 - 252
// Printing out my_number one byte at time
for (int i = 0; i < (sizeof(my_number.value)); i++)
{
printf("index[%d]: %u, 0x%x\n", \
i, my_number.index[i], my_number.index[i]);
}
// Printing out my_number as an unsigned integer
printf("my_number.value: %u, 0x%x", my_number.value, my_number.value);
And the output is going to be:
index[0]: 252, 0xfc
index[1]: 10, 0xa
index[2]: 11, 0xb
index[3]: 12, 0xc
my_number.value: 202050300, 0xc0b0afc
And as for your final question, we wouldn't have to convert from unsigned char back to unsigned int since the values are already there. You just have to choose by which way you want to access it
Note 1: I am using an integer of 4 bytes in order to ease the understanding of the concept. For the problem you presented you must use:
union unsigned_number {
unsigned short int value; // A short int is 2 bytes long
unsigned char index[2]; // A char is 1 byte long
};
Note 2: I have assigned byte 0 to 252 in order to point out the unsigned characteristic of our index field. Was it declared as a signed char, we would have index[0]: -4, 0xfc as output.
I have a char array that is really used as a byte array and not for storing text. In the array, there are two specific bytes that represent a numeric value that I need to store into an unsigned int value. The code below explains the setup.
char* bytes = bytes[2];
bytes[0] = 0x0C; // For the sake of this example, I'm
bytes[1] = 0x88; // assigning random values to the char array.
unsigned int val = ???; // This needs to be the actual numeric
// value of the two bytes in the char array.
// In other words, the value should equal 0x0C88;
I can not figure out how to do this. I would assume it would involve some casting and recasting of the pointers, but I can not get this to work. How can I accomplish my end goal?
UPDATE
Thank you Martin B for the quick response, however this doesn't work. Specifically, in my case the two bytes are 0x00 and 0xbc. Obviously what I want is 0x000000bc. But what I'm getting in my unsigned int is 0xffffffbc.
The code that was posted by Martin was my actual, original code and works fine so long as all of the bytes are less than 128 (.i.e. positive signed char values.)
unsigned int val = (unsigned char)bytes[0] << CHAR_BIT | (unsigned char)bytes[1];
This if sizeof(unsigned int) >= 2 * sizeof(unsigned char) (not something guaranteed by the C standard)
Now... The interesting things here is surely the order of operators (in many years still I can remember only +, -, * and /... Shame on me :-), so I always put as many brackets I can). [] is king. Second is the (cast). Third is the << and fourth is the | (if you use the + instead of the |, remember that + is more importan than << so you'll need brakets)
We don't need to upcast to (unsigned integer) the two (unsigned char) because there is the integral promotion that will do it for us for one, and for the other it should be an automatic Arithmetic Conversion.
I'll add that if you want less headaches:
unsigned int val = (unsigned char)bytes[0] << CHAR_BIT;
val |= (unsigned char)bytes[1];
unsigned int val = (unsigned char) bytes[0]<<8 | (unsigned char) bytes[1];
The byte ordering depends on the endianness of your processor. You can do this, which will work on big or little endian machines. (without ntohs it will work on big-endian):
unsigned int val = ntohs(*(uint16_t*)bytes)
unsigned int val = bytes[0] << 8 + bytes[1];
I think this is a better way to go about it than relying on pointer aliasing:
union {unsigned asInt; char asChars[2];} conversion;
conversion.asInt = 0;
conversion.asChars[0] = 0x0C;
conversion.asChars[1] = 0x88;
unsigned val = conversion.asInt;