I am trying to take a given unsigned char and store the 8 bit value in an unsigned char array of size 8 (1 bit per array index).
So given the unsigned char A
Id like to create an unsigned char array containing 0 1 0 0 0 0 0 1 (one number per index)
What would be the most effective way to achieve this? Happy Thanksgiving btw!!
The fastest (not sure if that's what you menat by "effective") way of doing this is probably something like
void char2bits1(unsigned char c, unsigned char * bits) {
int i;
for(i=sizeof(unsigned char)*8; i; c>>=1) bits[--i] = c&1;
}
The function takes the char to convert as the first argument and fills the array bits with the corresponding bit pattern. It runs in 2.6 ns on my laptop. It assumes 8-bit bytes, but not how many bytes long a char is, and does not require the input array to be zero-initialized beforehand.
I didn't expect this to be the fastest approach. My first attempt looked like this:
void char2bits2(unsigned char c, unsigned char * bits) {
for(;c;++bits,c>>=1) *bits = c&1;
}
I thought this would be faster by avoiding array lookups, by looping in the natural order (at the cost of producing the bits in the opposite order of what was requested), and by stopping as soon as c is zero (so the bits array would need to be zero-initialized before calling the function). But to my surprise, this version had a running time of 5.2 ns, double that of the version above.
Investigating the corresponding assembly revealed that the difference was loop unrolling, which was being performed in the former case but not the latter. So this is an illustration of how modern compilers and modern CPUs often have surprising performance characteristics.
Edit: If you actually want the unsigned chars in the result to be the chars '0' and '1', use this modified version:
void char2bits3(unsigned char c, unsigned char * bits) {
int i;
for(i=sizeof(unsigned char)*8; i; c>>=1) bits[--i] = '0'+(c&1);
}
You could use bit operators as recommended.
#include <stdio.h>
main() {
unsigned char input_data = 8;
unsigned char array[8] = {0};
int idx = sizeof(array) - 1;
while (input_data > 0) {
array[idx--] = input_data & 1;
input_data /= 2; // or input_data >>= 1;
}
for (unsigned long i = 0; i < sizeof(array); i++) {
printf("%d, ", array[i]);
}
}
Take the value, right shift it and mask it to keep only the lower bit. Add the value of the lower bit to the character '0' so that you get either '0' or '1' and write it into the array:
unsigned char val = 65;
unsigned char valArr[8+1] = {};
for (int loop=0; loop<8; loop++)
valArr[7-loop] = '0' + ((val>>loop)&1);
printf ("val = %s", valArr);
Related
size of short int is 2 bytes(16 bits) on my 64 bit processor and mingw compiler but when I convert short int variable to a binary string using itoa function
it returns string of 32 bits
#include<stdio.h>
int main(){
char buffer [50];
short int a=-2;
itoa(a,buffer,2); //converting a to binnary
printf("%s %d",buffer,sizeof(a));
}
Output
11111111111111111111111111111110 2
The answer is in understanding C's promotion of short datatypes (and char's, too!) to int's when those values are used as parameters passed to a function and understanding the consequences of sign extension.
This may be more understandable with a very simple example:
#include <stdio.h>
int main() {
printf( "%08X %08X\n", (unsigned)(-2), (unsigned short)(-2));
// Both are cast to 'unsigned' to avoid UB
return 0;
}
/* Prints:
FFFFFFFE 0000FFFE
*/
Both parameters to printf() were, as usual, promoted to 32 bit int's. The left hand value is -2 (decimal) in 32bit notation. By using the cast to specify the other parameter should not be subjected to sign extension, the printed value shows that it was treated as a 32 bit representation of the original 16 bit short.
itoa() is not available in my compiler for testing, but this should give the expected results
itoa( (unsigned short)a, buffer, 2 );
your problem is so simple , refer to itoa() manual , you will notice its prototype which is
char * itoa(int n, char * buffer, int radix);
so it takes an int that to be converted and you are passing a short int so it's converted from 2 byte width to 4 byte width , that's why it's printing a 32 bits.
to solve this problem :
you can simply shift left the array by 16 position by the following simple for loop :
for (int i = 0; i < 17; ++i) {
buffer[i] = buffer[i+16];
}
and it shall give the same result , here is edited version of your code:
#include<stdio.h>
#include <stdlib.h>
int main(){
char buffer [50];
short int a= -2;
itoa(a,buffer,2);
for (int i = 0; i < 17; ++i) {
buffer[i] = buffer[i+16];
}
printf("%s %d",buffer,sizeof(a));
}
and this is the output:
1111111111111110 2
I have experience with Java and Python, but this is my first time really using C, for my first assignment also haha.
I'm having trouble figuring out how to convert an unsigned char to a bit, so I would be able to get/set/swap some bit values.
I'm not looking for someone to do my assignment of course, I just need help accessing the bit. I came across this Access bits in a char in C
But it seems like that method only showed how to get the last two bits.
Any help or guidance is much appreciated. I tried Googling to see if there was some sort of documentation on this, but couldn't find any. Thanks in advance!
Edit: Made changes in accordance with Chux's comment. Also introduced rotl function which rotates bits. Originally reset function was wrong (should have used the rotation instead of shift tmp = tmp << n;
unsigned char setNthBit(unsigned char c, unsigned char n) //set nth bit from right
{
unsigned char tmp=1<<n;
return c | tmp;
}
unsigned char getNthBit(unsigned char c, unsigned char n)
{
unsigned char tmp=1<<n;
return (c & tmp)>>n;
}
//rotates left the bits in value by n positions
unsigned char rotl(unsigned char value, unsigned char shift)
{
return (value << shift) | (value >> (sizeof(value) * 8 - shift));
}
unsigned char reset(unsigned char c, unsigned char n) //set nth bit from right to 0
{
unsigned char tmp=254; //set all bits to 1 except the right=most one
//tmp = tmp << n; <- wrong, sets to zero n least signifacant bits
//use rotl instead
tmp = rotl(tmp,n);
return c & tmp;
}
//Combine the two for swapping of the bits ;)
char swap(unsigned char c, unsigned char n, unsigned char m)
{
unsigned char tmp1=getNthBit(c,n), tmp2=getNthBit(c,m);
char tmp11=tmp2<<n, tmp22=tmp1<<m;
c=reset(c,n); c=reset(c,m);
return c | tmp11 | tmp22;
}
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.
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.
Hello I have an unsigned char * that looks (after printf) like this (it's a SHA-1 hash):
n\374\363\327=\3103\231\361P'o]Db\251\360\316\203
I need to convert this unsigned char * to an unsigned int, what do you think it would be the best way to do it ? I have some ideas, but I'm not a C expert so wanted to see someone else ideas before trying my own stuff.
Why would you need a conversion? It's a 160 bit long digest. Digests are used only in two ways:
You print a digest with something like
for (i = 0; i < 20; ++i) {
printf("%2x", digest[i]);
}
and compare against another digest with something like
for (i = 0, equals = 1; i < 20; ++i) {
if (a[i] != b[i]) {
equals = 0;
}
}
It works just fine the way it is as a 20-byte long array of bytes. You don't have to worry about endianness, word length, nothing.
Well, that's more than 4 bytes, so if your system uses 32 bits for an unsigned int you can't do it without potentially losing information. IOW, it will have to be a hash of some kind.
That's 160 bits, so would be hard to fit in a single unsigned int. However, it'd certainly be possible to fit it into an array of unsigned ints.
Something like this (ugly, makes a couple of assumptions about machine architecture, should probably use CHAR_BITS and a couple of other things compile-time to have the right constants, but should be enough as a proof-of-concept):
unsigned int (*convert)(unsigned char *original)
{
unsigned int *rv = malloc(5*sizeof(unsigned int));
char *tp = original;
for (rvix=0;rvix<5;rvix++) {
rv[rvix] = *(tp++)<<24;
rv[rvix] |= *(tp++)<<16;
rv[rvix] |= *(tp++)<<8;
rv[rvix] |= *(tp++);
}
return rv;
}