code like this:
#include <stdio.h>
int main(){
struct{
unsigned char a:4;
unsigned char b:4;
}i;
struct{
unsigned char a:4;
unsigned char b:4;
unsigned char c:4;
}j;
i.a = 1;
i.b = 1;
j.a = 1;
j.b = 1;
j.c = 1;
printf("size of i is: %d, size of j is: %d", sizeof(i), sizeof(j));
return 0;
}
why the output is 1 2? means size of i possess 1 byte, j possess 2 bytes. we know unsigned char have 1 byte, so why i not equal 2? i am sorry for my english.
All variables in C++ are padded upto next byte.
In struct i, both a and b are of 4 bit summing up to 1 byte.
In j, variables sum up to 12 bits, but size is 2 byte due to padding.
Reference: http://www.cplusplus.com/forum/general/51911/
Related
Using bitwise operations, is it possible to package and read back the following set of values into an uint16_t variable? I think yes but I am trying to figure out how using a sample program.
Lets say following is the set of values I want to package into an uint16_t.
unsigned int iVal1 = 165 // which is 8 bits
unsigned int iVal2 = 28 // which is 5 bits
unsigned int iVal3 = 3 // which is 2 bits
bool bVal = true; // which can stored in 1 bit if we use 0 for true and 1 for false
Following is my program which aims to write in the values and needs to read back. How can write and read the values back using C++ 11?
#include <iostream>
uint16_t Write(unsigned int iVal1, unsigned int iVal2, unsigned int iVal3, bool bVal) {
// Is this technique correct to package the 3 values into an uint16_t?
return static_cast<uint16_t>(iVal1) + static_cast<uint16_t>(iVal2) + static_cast<uint16_t>(iVal3) + static_cast<uint16_t>(bVal);
}
unsigned int ReadVal1(const uint16_t theNumber) {
// How to read back iVal1
}
unsigned int ReadVal2(const uint16_t theNumber) {
// How to read back iVal2
}
unsigned int ReadVal3(const uint16_t theNumber) {
// How to read back iVal3
}
bool ReadVal4(const uint16_t theNumber) {
// How to read back bVal
}
int main() {
unsigned int iVal1 = 165; // which is 8 bits
unsigned int iVal2 = 28; // which is 5 bits
unsigned int iVal3 = 3; // which is 2 bits
bool bVal = true; // which can stored in 1 bit if we use 0 for true and 1 for false
const uint16_t theNumber = Write(iVal1, iVal2, iVal3, bVal);
std::cout << "The first 8 bits contain the number: " << ReadVal1(theNumber) << std::endl;
std::cout << "Then after 8 bits contain the number: " << ReadVal2(theNumber) << std::endl;
std::cout << "Then after 2 bits contain the number: " << ReadVal3(theNumber) << std::endl;
std::cout << "Then after 1 bit contains the number: " << ReadVal4(theNumber) << std::endl;
}
For this, you need to play on bit shifts.
uint16_t Write(unsigned int iVal1, unsigned int iVal2, unsigned int iVal3, bool bVal) {
// this will encode ival1 on the 8 first bits, ival2 on bits 4 to 8,
// ival3 on bits 2 and 3, and bval on last bit
return (static_cast<uint16_t>(iVal1)<<8) + (static_cast<uint16_t>(iVal2)<<3) + (static_cast<uint16_t>(iVal3)<<1) + static_cast<uint16_t>(bVal);
}
Then your uint16_t will hold all the value you need.
To read back, let us say ival2, you need to shift back and to use the and operator:
unsigned int ReadVal1(const uint16_t theNumber) {
// ival1 is the first 8 bits from place 9 to 16
uint16_t check1 = 255; // in bits 0000000011111111
return (theNumber>>8)&check1;
}
unsigned int ReadVal2(const uint16_t theNumber) {
// ival2 is the 5 bits from place 3 to place 8
uint16_t check2 = 31; // in bits 0000000000011111
return (theNumber>>3)&check2;
}
unsigned int ReadVal3(const uint16_t theNumber) {
// ival3 is the 2 bits on places 2 and 3
uint16_t check3 = 3; // in bits 0000000000000011
return (theNumber>>1)&check3;
}
bool ReadVal4(const uint16_t theNumber) {
// ival4 is the last bit
uint16_t check4 = 1; // in bits 0000000000000001
return theNumber&check4;
}
NB : here true is equal to 1 and false to 0.
I am trying to put the values from the vector into the int.
Given vector :'1','0','1','1','1','0','1','1','1','0','1','1','1','0','1','1' :
Expected output (binary representation for the variable out):
00000000000000001011101110111011.
However, I am getting the following output:
10111011101110110000000000000000
Notice: the insertion begun at the 16bit from right end instead of beginning from the leftmost bit
#include<vector>
#include<iostream>
int main() {
std::vector<unsigned char> test = {'1','0','1','1','1','0','1','1','1','0','1','1','1','0','1','1'};
std::vector<unsigned int> out(1);
int j = 0;
for (int i =0; i < test.size(); i++) {
out[j] = out[j] << 1;
if (test[i] == '1') {out[j] |=0x1;}
}
j++;
for (int p = 0; p < j; p++) {
for (int k = 0; k<32; k++ ) {
std::cout << !!((out[p]<<k)&0x8000);
}
std::cout << std::endl;
}
std::cout << "Size Of:" << sizeof(int);
return 0;
}
The reason why this happens is that you are using a wrong constant for the mask: 0x8000 has its 16-bit set, while you probably meant to use 0x80000000 with the 32-nd bit set. To avoid mistakes like that it's best to construct masks with shifts, for example
(1 << 31)
This expression is evaluated at compile time, so the result is the same as if you computed the constant yourself.
Note that both 0x8000 and 0x80000000 constants are system-dependent. Moreover, 0x80000000 assumes 32-bit int, which is not guaranteed.
A better approach would be shifting the number right instead of left, and masking with 1.
The block of code creating out[j] works just fine.
Your problem is in the output block, due to use of 0x8000. Whenever k >= 16, the low 16 bits will be zero, guaranteeing that 0x8000 is zero.
Your code seems overly complicated to me. Here's my version of a C program that transforms a string of 1's and 0's into an int and one going from int to string.
#include <stdlib.h>
#include <stdio.h>
int main(int argc, char **argv);
int main (int argc, char **argv) {
char str [] = "1010101010101010";
int x;
int out;
for (x=0;x<16;x++) {
if (str[x] == '1') {
out |= (1 << x);
}
}
printf("%d", out) ;
}
and
#include <stdlib.h>
#include <stdio.h>
int main(int argc, char **argv);
int main (int argc, char **argv) {
char str [] = "1010101010101010";
int in = 21845;
char out[17] = {0};
for (x=0;x<16;x++) {
if (in & (1<<x)) {
out[x] = '1';
}
else {
out[x] = '0';
}
}
printf("%s", out) ;
}
I need to convert rgba8 to rgba5551 manually. I found some helpful code from another post and want to modify it to convert from rgba8 to rgba5551. I don't really have experience with bitewise stuff and haven't had any luck messing with the code myself.
void* rgba8888_to_rgba4444( void* src, int src_bytes)
{
// compute the actual number of pixel elements in the buffer.
int num_pixels = src_bytes / 4;
unsigned long* psrc = (unsigned long*)src;
unsigned short* pdst = (unsigned short*)src;
// convert every pixel
for(int i = 0; i < num_pixels; i++){
// read a source pixel
unsigned px = psrc[i];
// unpack the source data as 8 bit values
unsigned r = (px << 8) & 0xf000;
unsigned g = (px >> 4) & 0x0f00;
unsigned b = (px >> 16) & 0x00f0;
unsigned a = (px >> 28) & 0x000f;
// and store
pdst[i] = r | g | b | a;
}
return pdst;
}
The value of RGBA5551 is that it has color info condensed into 16 bits - or two bytes, with only one bit for the alpha channel (on or off). RGBA8888, on the other hand, uses a byte for each channel. (If you don't need an alpha channel, I hear RGB565 is better - as humans are more sensitive to green). Now, with 5 bits, you get the numbers 0 through 31, so r, g, and b each need to be converted to some number between 0 and 31, and since they are originally a byte each (0-255), we multiply each by 31/255. Here is a function that takes RGBA bytes as input and outputs RGBA5551 as a short:
short int RGBA8888_to_RGBA5551(unsigned char r, unsigned char g, unsigned char b, unsigned char a){
unsigned char r5 = r*31/255; // All arithmetic is integer arithmetic, and so floating points are truncated. If you want to round to the nearest integer, adjust this code accordingly.
unsigned char g5 = g*31/255;
unsigned char b5 = b*31/255;
unsigned char a1 = (a > 0) ? 1 : 0; // 1 if a is positive, 0 else. You must decide what is sensible.
// Now that we have our 5 bit r, g, and b and our 1 bit a, we need to shift them into place before combining.
short int rShift = (short int)r5 << 11; // (short int)r5 looks like 00000000000vwxyz - 11 zeroes. I'm not sure if you need (short int), but I've wasted time tracking down bugs where I didn't typecast properly before shifting.
short int gShift = (short int)g5 << 6;
short int bShift = (short int)b5 << 1;
// Combine and return
return rShift | gShift | bShift | a1;
}
You can, of course condense this code.
I'm working in C on a PC, trying to leverage as little C++ as possible, working with binary data stored in unsigned char format, although other formats are certainly possible if worthwhile. The goal is subtracting two signed integer values (which can be ints, signed ints, longs, signed longs, signed shorts, etc.) in binary without converting to other data formats. The raw data is just prepackaged as unsigned char, though, with the user basically knowing which of the signed integer formats should be used for reading (i.e. we know how many bytes to read at once). Even though data is stored as an unsigned char array, data are meant to be read signed as two's-complement integers.
One common way we're often taught in school is adding the negative. Negation, in turn, is often taught to be performed as flipping bits and adding 1 (0x1), resulting in two additions (perhaps a bad thing?); or, as other posts point out, flipping bits past the first zero starting from the MSB. I'm wondering if there is a more efficient way, that may not be easily described as a pen-and-paper operation, but works because of the way data is stored in bit format. Here are some prototypes I've written, which may not be the most efficient way, but which summarizes my progress so far based on textbook methodology.
The addends are passed by reference in case I have to manually extend them to balance their length. Any and all feedback will be appreciated! Thanks in advance for considering.
void SubtractByte(unsigned char* & a, unsigned int & aBytes,
unsigned char* & b, unsigned int & bBytes,
unsigned char* & diff, unsigned int & nBytes)
{
NegateByte(b, bBytes);
// a - b == a + (-b)
AddByte(a, aBytes, b, bBytes, diff, nBytes);
// Restore b to its original state so input remains intact
NegateByte(b, bBytes);
}
void AddByte(unsigned char* & a, unsigned int & aBytes,
unsigned char* & b, unsigned int & bBytes,
unsigned char* & sum, unsigned int & nBytes)
{
// Ensure that both of our addends have the same length in memory:
BalanceNumBytes(a, aBytes, b, bBytes, nBytes);
bool aSign = !((a[aBytes-1] >> 7) & 0x1);
bool bSign = !((b[bBytes-1] >> 7) & 0x1);
// Add bit-by-bit to keep track of carry bit:
unsigned int nBits = nBytes * BITS_PER_BYTE;
unsigned char carry = 0x0;
unsigned char result = 0x0;
unsigned char a1, b1;
// init sum
for (unsigned int j = 0; j < nBytes; ++j) {
for (unsigned int i = 0; i < BITS_PER_BYTE; ++i) {
a1 = ((a[j] >> i) & 0x1);
b1 = ((b[j] >> i) & 0x1);
AddBit(&a1, &b1, &carry, &result);
SetBit(sum, j, i, result==0x1);
}
}
// MSB and carry determine if we need to extend:
if (((aSign && bSign) && (carry != 0x0 || result != 0x0)) ||
((!aSign && !bSign) && (result == 0x0))) {
++nBytes;
sum = (unsigned char*)realloc(sum, nBytes);
sum[nBytes-1] = (carry == 0x0 ? 0x0 : 0xFF); //init
}
}
void FlipByte (unsigned char* n, unsigned int nBytes)
{
for (unsigned int i = 0; i < nBytes; ++i) {
n[i] = ~n[i];
}
}
void NegateByte (unsigned char* n, unsigned int nBytes)
{
// Flip each bit:
FlipByte(n, nBytes);
unsigned char* one = (unsigned char*)malloc(nBytes);
unsigned char* orig = (unsigned char*)malloc(nBytes);
one[0] = 0x1;
orig[0] = n[0];
for (unsigned int i = 1; i < nBytes; ++i) {
one[i] = 0x0;
orig[i] = n[i];
}
// Add binary representation of 1
AddByte(orig, nBytes, one, nBytes, n, nBytes);
free(one);
free(orig);
}
void AddBit(unsigned char* a, unsigned char* b, unsigned char* c,
unsigned char* result) {
*result = ((*a + *b + *c) & 0x1);
*c = (((*a + *b + *c) >> 1) & 0x1);
}
void SetBit(unsigned char* bytes, unsigned int byte, unsigned int bit,
bool val)
{
// shift desired bit into LSB position, and AND with 00000001
if (val) {
// OR with 00001000
bytes[byte] |= (0x01 << bit);
}
else{ // (!val), meaning we want to set to 0
// AND with 11110111
bytes[byte] &= ~(0x01 << bit);
}
}
void BalanceNumBytes (unsigned char* & a, unsigned int & aBytes,
unsigned char* & b, unsigned int & bBytes,
unsigned int & nBytes)
{
if (aBytes > bBytes) {
nBytes = aBytes;
b = (unsigned char*)realloc(b, nBytes);
bBytes = nBytes;
b[nBytes-1] = ((b[0] >> 7) & 0x1) ? 0xFF : 0x00;
} else if (bBytes > aBytes) {
nBytes = bBytes;
a = (unsigned char*)realloc(a, nBytes);
aBytes = nBytes;
a[nBytes-1] = ((a[0] >> 7) & 0x1) ? 0xFF : 0x00;
} else {
nBytes = aBytes;
}
}
The first thing to notice is that signed vs. unsigned doesn't matter to the generated bit pattern in two's complement. All that changes is the interpretation of the result.
The second thing to notice is that an addition has carried if the result is less than either input when done with unsigned arithmetic.
void AddByte(unsigned char* & a, unsigned int & aBytes,
unsigned char* & b, unsigned int & bBytes,
unsigned char* & sum, unsigned int & nBytes)
{
// Ensure that both of our addends have the same length in memory:
BalanceNumBytes(a, aBytes, b, bBytes, nBytes);
unsigned char carry = 0;
for (int j = 0; j < nbytes; ++j) { // need to reverse the loop for big-endian
result[j] = a[j] + b[j];
unsigned char newcarry = (result[j] < a[j] || (unsigned char)(result[j]+carry) < a[j]);
result[j] += carry;
carry = newcarry;
}
}
i have an unsigned char test_SHA1[20] of 20 bytes which is a return value from a hash function. With the following code, I get this output
unsigned char test_SHA1[20];
char hex_output[41];
for(int di = 0; di < 20; di++)
{
sprintf(hex_output + di*2, "%02x", test_SHA1[di]);
}
printf("SHA1 = %s\n", hex_output);
50b9e78177f37e3c747f67abcc8af36a44f218f5
The last 9 bits of this number is 0x0f5 (= 245 in decimal) which I would get by taking a mod 512 of test_SHA1. To take mod 512 of test_SHA1, I do
int x = (unsigned int)test_SHA1 % 512;
printf("x = %d\n", x);
But x turns out to be 158 instead of 245.
I suggest doing a bit-wise and with 0x1ff instead of using the % operator.