I have an array that holds 28 ints which are all 1's and 0's. However, I need to print this information as 4 characters so how do I get each 7 bytes of data to become one bit in order to print.
Not sure this makes sense so I will illustrate what I need to:
Right now my array (in order) is this: 0101101111011101011000100010
But I need to somehow take those first 7 numbers (0101101) and print that out as Z and do that with the next 7, the next 7...
Thanks for your help!
I think this might be something along the lines you are looking for.
int to_int(int *bits) {
int power = 2;
int digit = 1;
int value = 0;
int i=0;
for(i=0; i <= 6; i++) {
if(bits[i] == 1) {
value += digit;
}
digit *= power;
}
return value;
}
int main() {
int myArray[28] = {0, 1, 0, 1, 1, 0, 1,
1, 1, 1, 0, 1, 1, 1,
0, 1, 0, 1, 1, 0, 0,
0, 1, 0, 0 ,0, 1, 0};
char theChars[5];
theChars[0] = to_char(&myArray[0]);
theChars[1] = to_char(&myArray[7]);
theChars[2] = to_char(&myArray[14]);
theChars[3] = to_char(&myArray[21]);
theChars[4] = '\0';
printf("%s\n",&theChars[0]);
}
Also, I don't think your expected output is correct.
Well, there is always the stupid way:
Cycle through each 7 blocks.
int bytes=7;
for(int i=0; i++;i<4){
double ASCII = 0;
for(int j=0; i++;j<bytes){
ASCII+=Math.pow(2, bytes-j-1)*array[i*bytes + j]
}
char c = (char) ASCII // you'll have some trouble with types here
}
Assuming your input array is called inputBits[] Try something like this:
const int input_bit_count = 28;
char output[input_bit_count / 7];
int outIdx = 0;
// step through the bit stream converting bits to 7-bit characters
for( int inIdx = 0; inIdx < input_bit_count; ){
// shift over and add the next bit to this character
output[outIdx] <<= 1;
if( inputBits[inIdx] != 0 ){
output[outIdx] |= 1;
}
inIdx++;
if( inIdx % 7 == 0)
// after each 7 bits, increment to next output character
outIdx++;
}
// done processing, now print it out
for( int chIdx = 0; chIdx < input_bit_count / 7; chIdx++ ){
printf( "%c", output[chIdx] );
}
Related
Declare an array containing these number and print the evens numbers and odd numbers
Now I initialized an array that containing 11 integers.
Here is my code
#include <stdio.h>
int main(void) {
int nums[11] = {11,3,9,7,6,10,13,17,2,8,3}; // create an variables that store integers
int evens[11] = {0}; // initialized an array to store even numbers
int odds[11] = {0}; // initialized an array to store even numbers
int length = sizeof(nums) / sizeof(nums[0]); // get the length of nums
int nums_index = 0;
int evens_index = 0;
int odds_index = 0;
for (nums_index; nums_index < length;nums_index++) {
if (nums[nums_index] % 2 == 0) {
evens[evens_index] = nums[nums_index];
evens_index++;
}
else if(nums[nums_index] % 2 != 0) {
odds[odds_index] = nums[nums_index];
odds_index++;
}
printf("%d\n",evens[evens_index]);
printf("%d\n",odds[odds_index]);
}
return 0;
}
The major question is whether the output has problems when I compile my code.
The output is ：0 11 0 3 0 9 0 7 6 0 10 0 0 13 0 17 2 0 8 0 0 3
Why it could happened?
Thank you all.
You need separate indexing for each array, advancing the index for evens and odds only when nums[i] value is one of the two.
Otherwise you would get sort of a copy of nums with zeroes in place of those numbers of the opposite type (odd/even).
For instance:
int j = 0;
int k = 0;
for (int i = 0; i < length; i++) {
if (nums[i] % 2 == 0) {
evens[j] = nums[i];
j++;
}
else if(nums[i] % 2 != 0) {
odds[k] = nums[i];
k++;
}
printf("%d\n",evens[i]);
printf("%d\n",odds[i]);
}
This will compose the arrays like:
11 3 9 7 13 17 3 0 0 0 0 --- for odds
6 10 2 8 0 0 0 0 0 0 0 0 --- for evens
The second problem is that you are printing inside the loop, firstly a value from evens and immediately after a value for odds.
So if you want to display them nice and separate, you can move both printf outside the first loop, then looping again on each result array for displaying it completely, before proceding to the other.
#include <stdio.h>
void PrintNumbers(int*, int);
int main(void) {
int nums[11] = {11,3,9,7,6,10,13,17,2,8,3}; // create an variables that store integers
int evens[11] = {0}; // initialized an array to store even numbers
int odds[11] = {0}; // initialized an array to store even numbers
int length = sizeof(nums) / sizeof(nums[0]); // get the length of nums
int nums_index = 0;
int evens_index = 0;
int odds_index = 0;
for (nums_index; nums_index < length; nums_index++)
{
if (nums[nums_index] % 2 == 0)
{
evens[evens_index] = nums[nums_index];
evens_index++;
}
else if(nums[nums_index] % 2 != 0)
{
odds[odds_index] = nums[nums_index];
odds_index++;
}
}
printf("Original List: ");
PrintNumbers(nums, length);
printf("Even numbers: ");
PrintNumbers(evens, length);
printf("Odd numbers: ");
PrintNumbers(odds, length);
return 0;
}
void PrintNumbers(int* numbers, int n)
{
for (int i = 0; i < n; i++)
{
printf("%d, ", numbers[i]);
}
printf("\n");
}
Output:
Original List: 11, 3, 9, 7, 6, 10, 13, 17, 2, 8, 3,
Even numbers: 6, 10, 2, 8, 0, 0, 0, 0, 0, 0, 0,
Odd numbers: 11, 3, 9, 7, 13, 17, 3, 0, 0, 0, 0,
So given an array of:
input[3] = {0, 0, 0}
this outputs :
output[3] = {3, 0 ,0}
code:
void create_hist(double input[], int num_of_inputs, int output[])
{
int num_to_check = input[0];
int counter = 0;
for (int i = 0; i < num_of_inputs; i++)
{
int j = output[i];
if ((int)input[i] == num_to_check)
{
counter++; /* it was found */
}
output[j] = counter;
}
return;
}
but if I have a floating point array
input[5] = {0.0000, 1.0000, 2.0000, 3.0000, 4.000}
and I want to truncate the values to int, and count how many times each integer in the range 0 - 10 appears in the input array then output it to:
output[5] = {1, 1, 1, 1, 1}
output[0] = {1} //indicates how many times 0 appeared in the array
or
input[10] = {1.000, 4.000, 5.0000, 2.000, 4.000, 7.000, 9.000, 6.000, 0.000, 0.000}
and output
output[10] = {2, 1, 1, 0, 2, 1, 1, 1, 0, 1}
output[0] = {2} // this indicates how many times 0 appeared in the first array
Can anyone tell me how to do this?
You shouldn't use output[i] as an array index. It's a counter, not the value whose count you want. You should use (int)input[i] as the index.
You first need to initialize all elements of output to 0, then you increment the elements corresponding to the integer part of each input.
memset(output, 0, sizeof(output[0]) * MAX_INPUT_VALUE);
for (int i = 0; i < num_of_inputs; i++) {
output[(int)input[i]]++;
}
I am trying to make a program that will count the number of even numbers in the provided arrays. When I run the program now, it will return the amount of numbers in the array, but not the amount of even numbers. For some reason my count_even function doesn't work. Can anyone help?
#include <stdio.h>
int main()
{
int data_array_1[] = { 1, 3, 5, 7, 9, 11 };
int data_array_2[] = { 2, -4, 6, -8, 10, -12, 14, -16 };
int data_array_3[] = { 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0 };
int data_array_4[] = { 6, 2, 4, 5, 1, -9 };
int data_array_5[] = { 1, 3, 9, 23, 5, -2, 4 };
int result_1 = count_even(data_array_1, 6);
printf("data_array_1 has %d even numbers.\n", result_1);
int result_2 = count_even(data_array_2, 8);
printf("data_array_2 has %d even numbers.\n", result_2);
int result_3 = count_even(data_array_3, 11);
printf("data_array_3 has %d even numbers.\n", result_3);
int result_4 = count_even(data_array_4, 6);
printf("data_array_4 has %d even numbers.\n", result_4);
int result_5 = count_even(data_array_5, 7);
printf("data_array_5 has %d even numbers.\n", result_5);
return 0;
}
int count_even(int* data_array, int size)
{
int even_num = 0;
for (int i = 0; i == size; i++)
{
if (data_array[size] % 2 == 0)
{
even_num++;
}
}
return even_num;
}
The condition in your for loop is wrong.
The correct condition should say "as long as the index is smaller than size", but yours say "as long as the index equal to to size".
The condition should be i < size.
As for the result, it seems like it should return 0 (for the non-working code), not size.
Also, you are using size as an index, when you should use i.
In your count_even function, you are using the size attribute as the array index, when it should be i
int count_even(int* data_array, int size)
{
int even_num = 0
for(int i = 0; i <= size, ++i)
{
if(data_array[i] % 2 == 0)
{
even_num++;
}
}
return even_num;
}
these two lines are the root of the problems in the code:
for (int i = 0; i == size; i++)
{
if (data_array[size] % 2 == 0)
the for() statement, should be:
for (int i = 0; i < size; i++)
so the loop exits when reaching the end of the array
the if() statement is always looking at the same entry beyond the end of the array, This is undefined behaviour
The if() statement should be:
if (data_array[i] % 2 == 0)
However, the modulo operator & is not a good choice for negative numbers
a better choice would be:
if ( !(data_array[i] & 1 ) )
Given a sequence of digits, a valley is defined as the region in the sequence that is surrounded (to the left and right) by higher values. The task is to find the number of valleys in the sequence.
For example,
{9,8,7,7,8,9} has one valley at {7,7}
{9,8,7,7,8,6,9} has two valleys at {7,7} and {6}
{7,8,9,8,7} has no valleys
The code I have to compute the number of valleys is as follows:
#include <stdio.h>
#define SIZE 40
int main()
{
int input;
int store[SIZE];
int i = 0;
int j;
int valley = 0;
int count = 0;
printf("Enter sequence: ");
scanf("%d", &input);
while(input != -1)
{
store[i] = input;
i++;
scanf("%d", &input);
}
count = count + i;
for(i = 1; i < count; i++)
{
for(j = i; j < i + 1; j++)
{
if((store[j-1] > store[j]) && (store[j] < store[j+1]))
{
valley = valley + 1;
break;
}
}
}
printf("Number of valleys: %d", valley);
return 0;
}
I am able to display the correct answer if the input is "3 2 1 2 3". However, if in between the number is equal to another and they are side by side (for example, "3 1 1 2"), the program will compute the wrong answer.
How do I go about writing the program so that I am able to display the correct number of valleys?
Look for slope changes from down to up.
Rather than a double nested for loop, march along looking for slope changes from down to up. Consider any slope of 0 to be the same as the previous slope.
size_t Valley(const int *store, size_t count) {
size_t valley = 0;
int slope = -1;
size_t i;
// Find first down slope
for (i = 1; i < count; i++) {
if (store[i] < store[i - 1]) {
break;
}
}
for (; i < count; i++) {
int newslope = (store[i] > store[i - 1]) - (store[i] < store[i - 1]);
// Loop for slope changes
if (newslope == -slope) {
if (newslope > 0)
valley++;
slope = newslope;
}
}
return valley;
}
Test code.
void Vtest(const int *store, size_t count) {
size_t n = Valley(store, count);
printf("%zu %zu\n", count, n);
}
void Vtests(void) {
int a1[] = { 9, 8, 7, 7, 8, 9 };
Vtest(a1, sizeof a1 / sizeof a1[0]);
int a2[] = { 9, 8, 7, 7, 8, 6, 9 };
Vtest(a2, sizeof a2 / sizeof a2[0]);
int a3[] = { 7, 8, 9, 8, 7 };
Vtest(a3, sizeof a3 / sizeof a3[0]);
int a4[] = { 3, 2, 1, 2, 3 };
Vtest(a4, sizeof a4 / sizeof a4[0]);
int a5[] = { 8, 7, 7, 8, 6 };
Vtest(a5, sizeof a5 / sizeof a5[0]);
}
int main(void) {
Vtests();
return 0;
}
Output
6 1
7 2
5 0
5 1
5 1
The problem is here:
if((store[j-1] > store[j] )&&(store[j] < store[j+1]))
In both comparations you are using index j, so this program finds only valleys with length 1. Try this modification:
if((store[i-1] > store[i] )&&(store[j] < store[j+1]))
Also I am not sure, that it is right to break; in this situation. But it is not clear now, which answer is correct in case 3 1 2 3 - one (1) or two (1 and 1 2). From your first example we can see, that right answer is one, but it is not obvious from the definition.
Depending on whether you define valley as a higher value to the IMMEDIATE left/right of a given point you may need to adjust the Valley function provided by chux as follows:
size_t Valley (const int *store, size_t count) {
...
i++;
for (; i < count; i++) {
int newslope = (store[i] > store[i - 1]) - (store[i] < store[i - 1]);
if (newslope == -slope) {
if (newslope > 0)
valley++;
}
slope = newslope;
}
...
}
output:
$ ./bin/valleyt
6 0
7 1
5 0
5 1
5 0
This is a supplement to the answer provided by chux, and the input data is as he provided in his answer. This code just limits the definition of a valley to being created by 3 adjacent points. (a special case of the general answer of a change from negative to positive slope with intervening equivalent points)
I have a C array fftArray[64] that contains values that I want averaged and placed into another array frequencyBar[8]. Getting the average of the entire array would be easy enough using a for statement.
int average, sum = 0;
for (i = 0; i < 64; i++)
{
sum += fftArray[i];
}
average = sum/64;
But I just can't seem to figure out how to get the average from fftArray[0] through fftArray[8] and store this in frequencyBar[0], the average of fftArray[9] through fftArray[16] and store this in frequencyBar[1], etc. Can anyone help me out with this? Thanks
This looks like a homework assignment, so, rather than give you the outright answer, I'd rather just point you in the right direction...
use a nested loop (one inside the other). One loop cycles 0-7, the other one 0 - 63. Use the smaller one to populate your sliced averages.
or better yet use the % operator to see when you've gone through 8 elements and do an average of your total, then reset the total for the next set. Then you'll have learned how to use the % operator too! :)
[EDIT]
ok, if not homework then something like this... I haven't written C in 5 years, so treat this as pseudo code:
//assuming you have a fftArray[64] with data, as per your question
int i,sum,avCounter,total;
int averages[8];
for(i=0 , avCounter=0, total=0 ; i<64; ){
total += fftArray[i];
if(++i % 8 == 0){ //%gives you the remainder which will be 0 every 8th cycle
averages[avCounter++] = total / 8
total = 0; //reset for next cycle
}
}
I think this will work better than a nested loop... but I'm not sure since % is division which is more processor heavy than addition... however... I doubt anyone would notice :)
int i, j;
for (i = 0; i < 8; i++) {
int sum = 0;
for (j = 0; j < 8; j++) {
sum += fftArray[ 8*i + j ];
}
frequencyBar[i] = sum / 8;
}
Bonus exercise: Optimize this code for speed on your chosen platform.
TF,
DISCLAIMER: This code is just off the top of my head... it hasn't even been compiled, let alone tested.
// returns the average of array[first..last] inclusive.
int average(int[] array, int first, int last) {
int sum = 0;
for (i = first; i <= last; i++)
sum += array[i];
return sum / (last - first + 1); // not sure about the +1
}
Then what you'd do is loop through the indexes of your frequencyBar array [0..7], setting frequencyBar[i] = average(array, first, last);... the tricky bit is calculating the first and last indexes... try i*8 and (i+1)*8 respectively... that may not be exactly right, but it'll be close ;-)
Cheers. Keith.
EDIT: Bored... waiting for my test results to come back. No news is good news, right? ;-)
It turns out that passing the length is a fair bit simpler than passing the last index.
#include <stdio.h>
int sum(int array[], int first, int length) {
int sum = 0;
for (int i = first; i < first+length; i++)
sum += array[i];
return sum;
}
double average(int array[], int first, int length) {
double total = sum(array, first, length);
#ifdef DEBUG
printf("DEBUG: [%2d..%2d] %d", first, first+length-1, array[first]);
for (int i = first+1; i < first+length; i++)
printf(" + %d", array[i]);
printf(" = %d / %d = %f\n", (int)total, length, total/length);
#endif
return total / length;
}
int main(int argc, char* argv[]) {
int array[] = { // average
1, 2, 3, 4, 5, 1, 2, 3, // 2.625
4, 5, 1, 2, 3, 4, 5, 1, // 3.125
2, 3, 4, 5, 1, 2, 3, 4, // 3
5, 1, 2, 3, 4, 5, 1, 2, // 2.875
3, 4, 5, 1, 2, 3, 4, 5, // 3.375
1, 2, 3, 4, 5, 1, 2, 3, // 2.625
4, 5, 1, 2, 3, 4, 5, 1, // 3.125
2, 3, 4, 5, 1, 2, 3, 4 // 3
};
double frequency[8];
for (int i = 0; i < 8; i++)
frequency[i] = average(array, i*8, 8);
for (int i = 0; i < 8; i++)
printf("%f ", frequency[i]);
printf("\n");
}
Watch your sum doesn't wrap around if fftArray has large value in!