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;
Related
I need to store a large number, but due to limitations in an old game engine, I am restricted to working with signed short (I can, however, use as many of these as I want).
I need to split an unsigned long (0 to 4,294,967,295) into multiple signed short (-32,768 to 32,767). Then I need to recombine the multiple signed short into a new unsigned long later.
For example, take the number 4,000,000,000. This should be split into multiple signed short and then recombined into unsigned long.
Is this possible in C? Thanks.
In addition to dbush's answer you can also use a union, e.g.:
union
{
unsigned long longvalue;
signed short shortvalues[2];
}
value;
The array of two shorts overlays the single long value.
I assume your problem is finding a place to store these large values. There are options we haven't yet explored which don't involve splitting the values up and recombining them:
Write them to a file, and read them back later. This might seem silly at first, but considering the bigger picture, if the values end up in a file later on then this might seem like the most attractive option.
Declare your unsigned long to have static storage duration e.g. outside of any blocks of code A.K.A globally (I hate that term) or using the static keyword inside a block of code.
None of the other answers so far are strictly portable, not that it seems like it should matter to you. You seem to be describing a twos complement 16-bit signed short representation and a 32-bit unsigned long representation (you should put assertions in place to ensure this is the case), which has implications that restrict the options for the implementation (that is, the C compiler, the OS, the CPU, etc)... so the portability issues associated with them are unlikely to occur. In case you're curious, however, I'll discuss those issues anyway.
The portability issues associated are that one type or the other might have padding bits causing the sizes to mismatch, and that there might be trap representations for short.
Changing the type but not the representation is by far much cleaner and easier to get right, though not portable; this includes the union hack, you could also avoid the union by casting an unsigned long * to a short *. These solutions are the cleanest solutions, which makes Ken Clement's answer my favourite so far, despite the non-portability.
Binary shifts (the >> and << operators), and (the & operator), or (|) operators introduce additional portability issues when you use them on signed types; they're also bulky and clumsy leading to more code to debug and a higher chance that mistakes are made.
You need to consider that while ULONG_MAX is guaranteed to be at least 4,294,967,295, SHORT_MIN is not guaranteed by the C standard to be -32,768; it might be -32,767 (which is quite uncommon indeed, though still possible)... There might be a negative zero or trap representation in place of that -32,768 value.
This means you can't portably rely upon a pair of signed shorts being able to represent all of the values of an unsigned long; even when the sizes match up you need another bit to account for the two missing values.
With this in mind, you could use a third signed char... The implementation-defined and undefined behaviours of the shift approaches could be avoided that way.
signed short x = (value ) & 0xFFF,
y = (value >> 12) & 0xFFF,
z = (value >> 24) & 0xFFF;
value = (unsigned long) x
+ ((unsigned long) y << 12)
+ ((unsigned long) z << 24);
You can do it like this (I used fixed size types to properly illustrate how it works):
#include<stdio.h>
#include<stdint.h>
int main()
{
uint32_t val1;
int16_t val2a, val2b;
uint32_t val3;
val1 = 0x11223344;
printf("val1=%08x\n", val1);
// to short
val2a = val1 >> 16;
val2b = val1 & 0xFFFF;
printf("val2a=%04x\n", val2a);
printf("val2b=%04x\n", val2b);
// to long
val3 = (uint32_t)val2a << 16;
val3 |= (uint32_t)val2b;
printf("val3=%08x\n", val3);
return 0;
}
Output:
val1=11223344
val2a=1122
val2b=3344
val3=11223344
There are any number of ways to do it. One thing to consider is that unsigned long may not have the same size on different hardware/operating systems. You can use exact length types found in stdint.h to avoid ambiguity (e.g. uint8_t, uint16_t, etc.). One implementation incorporating exact types (and cheezy hex values) would be:
#include <stdio.h>
#include <stdint.h>
#include <inttypes.h>
#include <limits.h>
int main (void) {
uint64_t a = 0xfacedeadbeefcafe, b = 0;
uint16_t s[4] = {0};
uint32_t i = 0, n = 0;
printf ("\n a : 0x%16"PRIx64"\n\n", a);
/* separate uint64_t into 4 uint16_t */
for (i = 0; i < sizeof a; i += 2, n++)
printf (" s[%"PRIu32"] : 0x%04"PRIx16"\n", n,
(s[n] = (a >> (i * CHAR_BIT))));
/* combine 4 uint16_t into uint64_t */
for (n = i = 0; i < sizeof b; i += 2, n++)
b |= (uint64_t)s[n] << i * CHAR_BIT;
printf ("\n b : 0x%16"PRIx64"\n\n", b);
return 0;
}
Output
$ ./bin/uint64_16
a : 0xfacedeadbeefcafe
s[0] : 0xcafe
s[1] : 0xbeef
s[2] : 0xdead
s[3] : 0xface
b : 0xfacedeadbeefcafe
This is one possible solution (which assumes ulong is 32-bits, and sshort is 16-bits):
unsigned long L1, L2;
signed short S1, S2;
L1 = 0x12345678; /* Initial ulong to store away into two sshort */
S1 = L1 & 0xFFFF; /* Store component 1 */
S2 = L1 >> 16; /* Store component 2*/
L2 = S1 | (S2<<16); /* Retrive ulong from two sshort */
/* Print results */
printf("Initial value: 0x%08lx\n",L1);
printf("Stored component 1: 0x%04hx\n",S1);
printf("Stored component 2: 0x%04hx\n",S2);
printf("Retrieved value: 0x%08lx\n",L2);
I want to convert an unsigned int and break it into 2 chars. For example: If the integer is 1, its binary representation would be 0000 0001. I want the 0000 part in one char variable and the 0001 part in another binary variable. How do I achieve this in C?
If you insist that you have a sizeof(int)==2 then:
unsigned int x = (unsigned int)2; //or any other value it happens to be
unsigned char high = (unsigned char)(x>>8);
unsigned char low = x & 0xff;
If you have eight bits total (one byte) and you are breaking it into two 4-bit values:
unsigned char x=2;// or whatever
unsigned char high = (x>>4);
unsigned char low = x & 0xf;
Shift and mask off the part of the number you want. Unsigned ints are probably four bytes, and if you wanted all four bytes, you'd just shift by 16 and 24 for the higher order bytes.
unsigned char low = myuint & 0xff;
unsigned char high = (myuint >> 8) & 0xff;
This is assuming 16 bit ints check with sizeof!! On my platform ints are 32bit so I will use a short in this code example. Mine wins the award for most disgusting in terms of pulling apart the pointer - but it also is the clearest for me to understand.
unsigned short number = 1;
unsigned char a;
a = *((unsigned char*)(&number)); // Grab char from first byte of the pointer to the int
unsigned char b;
b = *((unsigned char*)(&number) + 1); // Offset one byte from the pointer and grab second char
One method that works is as follows:
typedef union
{
unsigned char c[sizeof(int)];
int i;
} intchar__t;
intchar__t x;
x.i = 2;
Now x.c[] (an array) will reference the integer as a series of characters, although you will have byte endian issues. Those can be addressed with appropriate #define values for the platform you are programming on. This is similar to the answer that Justin Meiners provided, but a bit cleaner.
unsigned short s = 0xFFEE;
unsigned char b1 = (s >> 8)&0xFF;
unsigned char b2 = (((s << 8)>> 8) & 0xFF);
Simplest I could think of.
int i = 1 // 2 Byte integer value 0x0001
unsigned char byteLow = (i & 0x00FF);
unsinged char byteHigh = ((i & 0xFF00) >> 8);
value in byteLow is 0x01 and value in byteHigh is 0x00
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.
These are my two variables with which I want to do an xor operation (in C).
unsigned long int z=0xB51A06CD;
unsigned char array[] = {0xF0,0xCC,0xAA,0xF0};
desired output= 0X45D6AC3D
I know I cannot do a simple z ^ array, because it's a character array and not a single character. Will I have to do XOR of one byte at a time or is there a function for it in C?
I am trying all kinds of crazy things to get it done, but failing miserably all the time. If anyone can help me out with a small code snippet or at least a rough idea, I would be extremely grateful.
Cast the array, which is treat as a pointer to the first element in an expression like this one, to a long pointer instead of char pointer , and dereference it.
unsigned long result = z ^ *(unsigned long *)array;
Just make an unsigned long int out of your array (warning, it depends on the machine endianness!):
unsigned long int z=0xB51A06CD;
unsigned char array[] = {0xF0,0xCC,0xAA,0xF0};
unsigned long int w = 0;
w |= array[0] << 0;
w |= array[1] << 8;
w |= array[2] << 16;
w |= array[3] << 24;
unsigned long output = z ^ w;
unsigned long int z=0xB51A06CD;
unsigned char array[] = {0xF0,0xCC,0xAA,0xF0};
unsigned long tmp;
memcpy(&tmp, array, sizeof tmp);
... z ^ tmp ...
Note that this still makes a number of non-portable assumptions: that unsigned long is 4 bytes, and that the system's endianness is what you expect it to be.
As others mentioned, you have to worry about endianness and size of long. Here's how to make it safe:
1) instead of unsigned long, use uint32_t (from inttypes.h), to be sure you get a 4 byte type.
2) use htonl() as a platform-independent way to ensure you interpret the array as a big-endian value
z ^ htonl(*(uint32_t *)array);
As part of my CS course I've been given some functions to use. One of these functions takes a pointer to unsigned chars to write some data to a file (I have to use this function, so I can't just make my own purpose built function that works differently BTW). I need to write an array of integers whose values can be up to 4095 using this function (that only takes unsigned chars).
However am I right in thinking that an unsigned char can only have a max value of 256 because it is 1 byte long? I therefore need to use 4 unsigned chars for every integer? But casting doesn't seem to work with larger values for the integer. Does anyone have any idea how best to convert an array of integers to unsigned chars?
Usually an unsigned char holds 8 bits, with a max value of 255. If you want to know this for your particular compiler, print out CHAR_BIT and UCHAR_MAX from <limits.h> You could extract the individual bytes of a 32 bit int,
#include <stdint.h>
void
pack32(uint32_t val,uint8_t *dest)
{
dest[0] = (val & 0xff000000) >> 24;
dest[1] = (val & 0x00ff0000) >> 16;
dest[2] = (val & 0x0000ff00) >> 8;
dest[3] = (val & 0x000000ff) ;
}
uint32_t
unpack32(uint8_t *src)
{
uint32_t val;
val = src[0] << 24;
val |= src[1] << 16;
val |= src[2] << 8;
val |= src[3] ;
return val;
}
Unsigned char generally has a value of 1 byte, therefore you can decompose any other type to an array of unsigned chars (eg. for a 4 byte int you can use an array of 4 unsigned chars). Your exercise is probably about generics. You should write the file as a binary file using the fwrite() function, and just write byte after byte in the file.
The following example should write a number (of any data type) to the file. I am not sure if it works since you are forcing the cast to unsigned char * instead of void *.
int homework(unsigned char *foo, size_t size)
{
int i;
// open file for binary writing
FILE *f = fopen("work.txt", "wb");
if(f == NULL)
return 1;
// should write byte by byte the data to the file
fwrite(foo+i, sizeof(char), size, f);
fclose(f);
return 0;
}
I hope the given example at least gives you a starting point.
Yes, you're right; a char/byte only allows up to 8 distinct bits, so that is 2^8 distinct numbers, which is zero to 2^8 - 1, or zero to 255. Do something like this to get the bytes:
int x = 0;
char* p = (char*)&x;
for (int i = 0; i < sizeof(x); i++)
{
//Do something with p[i]
}
(This isn't officially C because of the order of declaration but whatever... it's more readable. :) )
Do note that this code may not be portable, since it depends on the processor's internal storage of an int.
If you have to write an array of integers then just convert the array into a pointer to char then run through the array.
int main()
{
int data[] = { 1, 2, 3, 4 ,5 };
size_t size = sizeof(data)/sizeof(data[0]); // Number of integers.
unsigned char* out = (unsigned char*)data;
for(size_t loop =0; loop < (size * sizeof(int)); ++loop)
{
MyProfSuperWrite(out + loop); // Write 1 unsigned char
}
}
Now people have mentioned that 4096 will fit in less bits than a normal integer. Probably true. Thus you can save space and not write out the top bits of each integer. Personally I think this is not worth the effort. The extra code to write the value and processes the incoming data is not worth the savings you would get (Maybe if the data was the size of the library of congress). Rule one do as little work as possible (its easier to maintain). Rule two optimize if asked (but ask why first). You may save space but it will cost in processing time and maintenance costs.
The part of the assignment of: integers whose values can be up to 4095 using this function (that only takes unsigned chars should be giving you a huge hint. 4095 unsigned is 12 bits.
You can store the 12 bits in a 16 bit short, but that is somewhat wasteful of space -- you are only using 12 of 16 bits of the short. Since you are dealing with more than 1 byte in the conversion of characters, you may need to deal with endianess of the result. Easiest.
You could also do a bit field or some packed binary structure if you are concerned about space. More work.
It sounds like what you really want to do is call sprintf to get a string representation of your integers. This is a standard way to convert from a numeric type to its string representation. Something like the following might get you started:
char num[5]; // Room for 4095
// Array is the array of integers, and arrayLen is its length
for (i = 0; i < arrayLen; i++)
{
sprintf (num, "%d", array[i]);
// Call your function that expects a pointer to chars
printfunc (num);
}
Without information on the function you are directed to use regarding its arguments, return value and semantics (i.e. the definition of its behaviour) it is hard to answer. One possibility is:
Given:
void theFunction(unsigned char* data, int size);
then
int array[SIZE_OF_ARRAY];
theFunction((insigned char*)array, sizeof(array));
or
theFunction((insigned char*)array, SIZE_OF_ARRAY * sizeof(*array));
or
theFunction((insigned char*)array, SIZE_OF_ARRAY * sizeof(int));
All of which will pass all of the data to theFunction(), but whether than makes any sense will depend on what theFunction() does.