I have an array defined as
int data[k];
where k is the size of the array. Each element of the array is either 0 or 1. I want to save the binary data in another array defined as
uint8_t new_data[k/8];
(k is usually a multiple of 8).
How can I do this in C?
Thanks in advance
Assuming k is a multiple of 8, assuming that by "each element is binary" you mean "each int is either 0 or 1", also assuming the bits in data are packed from most significant to least significant and the bytes of new_data are packed as big-endian (all reasonable assumptions), then this is how you do it:
for (int i = 0; i < k/8; ++i)
{
new_data[i] = (data[8*i ] << 7) | (data[8*i+1] << 6)
| (data[8*i+2] << 5) | (data[8*i+3] << 4)
| (data[8*i+4] << 3) | (data[8*i+5] << 2)
| (data[8*i+6] << 1) | data[8*i+7];
}
Assuming new_data starts initialized at 0, data[i] contains only zeroes and ones and that you want to fill lowest bits first:
for(unsigned i = 0; i < k; ++i) {
new_data[i/8] |= data[i]<<(i%8);
}
A possibly faster implementation1 may be:
for(int i = 0; i < k/8; ++i) {
uint8_t o = 0;
for(int j = 0; j < 8; ++j) {
o |= data[i*8]<<j;
}
new_data[i] = o;
}
(notice that this essentially assumes that k is multiple of 8)
It's generally easier to optimize, as the inner loop has small, known boundaries and it writes on a variable with just that small scope; this is easier for optimizers to handle, and you can see for example that with gcc the inner loop gets completely unrolled.
I am looking for a more faster way of dealing with the following C code. I have an image of 640x480 and I want to decimate it by a factor of 2 by removing every other rows and columns in the image. I have attached the code in the following. Is there any better way to optimize the code.
#define INPUT_NUM_ROW 480
#define INPUT_NUM_COL 640
#define OUTPUT_NUM_ROW 240
#define OUTPUT_NUM_COL 320
unsigned char inputBuf[INPUT_NUM_ROW* INPUT_NUM_COL];
unsigned char outputBuf[OUTPUT_NUM_ROW* OUTPUT_NUM_COL];
void imageDecimate(unsigned char *outputImage , unsigned char *inputImage)
{
/* Fill in your code here */
for (int p = 0; p< OUTPUT_NUM_ROW; p++) {
for (int q = 0; q < OUTPUT_NUM_COL; q++) {
outputImage[p*OUTPUT_NUM_COL + q] = inputImage[(p*INPUT_NUM_COL+q)*2];
// cout << "The pixel at " << p*OUTPUT_NUM_COL+q << " is " << outputImage[p*OUTPUT_NUM_COL+q] << endl;
}
}
}
Rather than doing the math every time in the inner loop, you could do this:
int outputIndex;
int inputIndex;
for (int p = 0; p< OUTPUT_NUM_ROW; p++) {
inputIndex = p * INPUT_NUM_COL * 2;
outputIndex = p * OUTPUT_NUM_COL;
for (int q = 0; q < OUTPUT_NUM_COL; q++) {
outputImage[outputIndex] = inputImage[inputIndex];
inputIndex += 2;
outputIndex++;
// cout << "The pixel at " << p*OUTPUT_NUM_COL+q << " is " << outputImage[p*OUTPUT_NUM_COL+q] << endl;
}
}
}
You could do the incrementing inline with the copying assignment too, and you could also only assign inputIndex and outputIndex the first time, but it wouldn't get you as much of a performance boost as moving the calculation out of the inner loop. I assume that bulk copying functions don't have this kind of incrementing flexibility, but if they do and they use hardware acceleration that is available on all of your target platforms, then that would be a better choice.
I am also assuming that array access like this compiles down to the most optimized pointer arithmetic that you could use.
I'm having trouble with dynamic array. The code I wrote is suppose to input the # of coins and check if 1 is included. If it is not include in the arrays include 1 to the array. But the array size is "fixed" so i can't change the size of array while keeping the other numbers inputted. How can I do this without messing up with my arrays?
#include <iostream>
using namespace std;
int main()
{
int N,coin;
cout << "Enter the value N to produce: " << endl;
cin >> N;
cout << "Enter number of different coins: " << endl;
cin >> coin;
int *S = new int[coin];
cout << "Enter the denominations to use with a space after it" << endl;
cout << "(1 will be added if necessary): " << endl;
for(int i = 0; i < coin; i++)
{
cin >> S[i];
if(S[i] != 1)
S[coin] = 1; // confused at this part of how to set the last element to 1
cout << S[i] << endl;
}
//system("PAUSE");
return 0;
}
here is pseudo code/comments
bool hasOne;
for(int i = 0; i < coin; i++) {
cin >> S[i];
if(S[i] == 1) hasOne = true;
}
if(!hasOne) {
// create a new array size one more than S
// copy elements from S to the new array
// set the last element to 1 in the new array
// assign the new array to S
}
Do you need to append 1 to the end of the array? If so, I would use std::vector object instead of an array
#include <vector>
vector<int> S;
for(int i = 0; i < coin; i++)
{
cin >> S[i];
if(S[i] != 1)
S.push_back(1);
cout << S[i] << endl;
}
If you must use arrays you need to use malloc() to dynamically allocate memory. But since you are using c++ std::vector is the way to go.
This question already has answers here:
Pointer subtraction confusion
(8 answers)
Closed 9 years ago.
I have a pointer to array, why it gives me the following output?
int main() {
int b[] = {1, 2};
cout << "size of int = " << sizeof(int) << endl;
int *pt = b;
int i = 0;
while( i++ < 2) {
cout << "pt = " << pt << ", b = " << b << endl;
cout << pt - b << endl;
(pt)++;
}
return 0;
}
code output:
size of int = 4
pt = 0x7fff576f0c2c, b = 0x7fff576f0c2c
0
pt = 0x7fff576f0c30, b = 0x7fff576f0c2c
1
pt is a pointer to the start of array b initially, why pt-b gives me the index of the array that pt points to rather than the index of the array times the size of one element.
it's because your array is really an address, and int *pt = b; essentially makes pt the exact same as b
your output is simply printing the number of times you've incremented pt since you set it to b
Remember, in the end, pointers are plain old ints, which represent an address in your logical space. Here the address b is pointing to remains constant while pt gets shifted by one.
I want to extract bits of a decimal number.
For example, 7 is binary 0111, and I want to get 0 1 1 1 all bits stored in bool. How can I do so?
OK, a loop is not a good option, can I do something else for this?
If you want the k-th bit of n, then do
(n & ( 1 << k )) >> k
Here we create a mask, apply the mask to n, and then right shift the masked value to get just the bit we want. We could write it out more fully as:
int mask = 1 << k;
int masked_n = n & mask;
int thebit = masked_n >> k;
You can read more about bit-masking here.
Here is a program:
#include <stdio.h>
#include <stdlib.h>
int *get_bits(int n, int bitswanted){
int *bits = malloc(sizeof(int) * bitswanted);
int k;
for(k=0; k<bitswanted; k++){
int mask = 1 << k;
int masked_n = n & mask;
int thebit = masked_n >> k;
bits[k] = thebit;
}
return bits;
}
int main(){
int n=7;
int bitswanted = 5;
int *bits = get_bits(n, bitswanted);
printf("%d = ", n);
int i;
for(i=bitswanted-1; i>=0;i--){
printf("%d ", bits[i]);
}
printf("\n");
}
As requested, I decided to extend my comment on forefinger's answer to a full-fledged answer. Although his answer is correct, it is needlessly complex. Furthermore all current answers use signed ints to represent the values. This is dangerous, as right-shifting of negative values is implementation-defined (i.e. not portable) and left-shifting can lead to undefined behavior (see this question).
By right-shifting the desired bit into the least significant bit position, masking can be done with 1. No need to compute a new mask value for each bit.
(n >> k) & 1
As a complete program, computing (and subsequently printing) an array of single bit values:
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char** argv)
{
unsigned
input = 0b0111u,
n_bits = 4u,
*bits = (unsigned*)malloc(sizeof(unsigned) * n_bits),
bit = 0;
for(bit = 0; bit < n_bits; ++bit)
bits[bit] = (input >> bit) & 1;
for(bit = n_bits; bit--;)
printf("%u", bits[bit]);
printf("\n");
free(bits);
}
Assuming that you want to calculate all bits as in this case, and not a specific one, the loop can be further changed to
for(bit = 0; bit < n_bits; ++bit, input >>= 1)
bits[bit] = input & 1;
This modifies input in place and thereby allows the use of a constant width, single-bit shift, which may be more efficient on some architectures.
Here's one way to do it—there are many others:
bool b[4];
int v = 7; // number to dissect
for (int j = 0; j < 4; ++j)
b [j] = 0 != (v & (1 << j));
It is hard to understand why use of a loop is not desired, but it is easy enough to unroll the loop:
bool b[4];
int v = 7; // number to dissect
b [0] = 0 != (v & (1 << 0));
b [1] = 0 != (v & (1 << 1));
b [2] = 0 != (v & (1 << 2));
b [3] = 0 != (v & (1 << 3));
Or evaluating constant expressions in the last four statements:
b [0] = 0 != (v & 1);
b [1] = 0 != (v & 2);
b [2] = 0 != (v & 4);
b [3] = 0 != (v & 8);
Here's a very simple way to do it;
int main()
{
int s=7,l=1;
vector <bool> v;
v.clear();
while (l <= 4)
{
v.push_back(s%2);
s /= 2;
l++;
}
for (l=(v.size()-1); l >= 0; l--)
{
cout<<v[l]<<" ";
}
return 0;
}
Using std::bitset
int value = 123;
std::bitset<sizeof(int)> bits(value);
std::cout <<bits.to_string();
#prateek thank you for your help. I rewrote the function with comments for use in a program. Increase 8 for more bits (up to 32 for an integer).
std::vector <bool> bits_from_int (int integer) // discern which bits of PLC codes are true
{
std::vector <bool> bool_bits;
// continously divide the integer by 2, if there is no remainder, the bit is 1, else it's 0
for (int i = 0; i < 8; i++)
{
bool_bits.push_back (integer%2); // remainder of dividing by 2
integer /= 2; // integer equals itself divided by 2
}
return bool_bits;
}
#include <stdio.h>
int main(void)
{
int number = 7; /* signed */
int vbool[8 * sizeof(int)];
int i;
for (i = 0; i < 8 * sizeof(int); i++)
{
vbool[i] = number<<i < 0;
printf("%d", vbool[i]);
}
return 0;
}
If you don't want any loops, you'll have to write it out:
#include <stdio.h>
#include <stdbool.h>
int main(void)
{
int num = 7;
#if 0
bool arr[4] = { (num&1) ?true: false, (num&2) ?true: false, (num&4) ?true: false, (num&8) ?true: false };
#else
#define BTB(v,i) ((v) & (1u << (i))) ? true : false
bool arr[4] = { BTB(num,0), BTB(num,1), BTB(num,2), BTB(num,3)};
#undef BTB
#endif
printf("%d %d %d %d\n", arr[3], arr[2], arr[1], arr[0]);
return 0;
}
As demonstrated here, this also works in an initializer.