Sorting an array - C language [closed] - c

Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 5 years ago.
Improve this question
I have tried to write a sorting function sort(int *buffer, int array[], int size) which works in a way similar to the insertion sort - it takes the first element from the array, sets it as the first element of the buffer and then checks whether or not the next value showing up in the array is greater than the last value stored in the buffer. If yes, it keeps swapping the two elements until everything is in its place. This is my minimal working example:
#include <stdio.h>
void sort(int *buffer, int array[], int size) {
for(int i = 0; i < size; i++) {
buffer[i] = array[i];
while(i >= 1 && buffer[i] < buffer[i-1]) {
int tmp = buffer[i-1];
buffer[i-1] = buffer[i];
buffer[i] = tmp;
printf("i = %d i: %d, i -1 : %d \n",i, buffer[i], buffer[i-1]);
i--;
}
}
}
int main(void) {
int array[3] = {4,3,2};
int buffer[3];
sort(buffer, array, 3);
for(int i = 0; i < 3; i++) {
printf("%d", buffer[i]);
}
}
However, the output of this program is 222
To be honest, I don't see how it's even possible that three identical elements got placed in the buffer.
What can have gone wrong?

You are using the same variable for the inner while cycle and for the outer for loop. Use a different variable and copy the value of i to it in each iteration of the for.

Related

C simple question of getting average value from array each components addition [closed]

Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 2 years ago.
Improve this question
This is a c code for getting the average value of the addition of array components.
But once I run this which is not outputting anything.
Can anyone help me out where I got the code wrong?
#include <stdio.h>
#include <stdbool.h>
#include <stdlib.h>
double solution(int arr[], size_t arr_len);
int main()
{
int array[10] = { 1,2,3,4,5,6,7,8,9,10 };
int length = sizeof(array[10]);
double out = solution(array, length);
printf("solution is %f\n", out);
return 0;
}
double solution(int arr[], size_t arr_len) {
double answer = 0;
int total = 0;
for (int i = 0; i < arr_len;){
total += arr[i];
}
answer = total / arr_len;
return answer;
}
You are not incrementing the loop counter in solution so its stuck in an infinite loop.
for (int i = 0; i < arr_len;){
needs to be
for (int i = 0; i < arr_len; i++) {
Edit:
sizeof is also wrong. It returns the total memory used by the array. So you need to do
int length = sizeof(array) / sizeof(array[0])
which divides the total memory by the size of one element to give you the total number of elements.

Not getting the correct output elements from an array in C [closed]

Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 3 years ago.
Improve this question
I'm trying to write a code that inputs a number each time into an array and than print out the result from the arrays elements but for some reason i either get an infintie loop or it prints out the same number.
void main() {
char arr[SIZE];
int k = 1;
int i = 0;
while (k != 0) {
scanf("%d", &k);
arr[i] = k;
i++;
}
arr[i] = '\0';
int b = 0;
while (b < i) {
printf("elements are %d\n", arr[i]);
b++;
}
You want to print arr[b] and not arr[i].
Thus, you want: printf("elements are %d\n", arr[b]);
As you [currently] have it, printing arr[i] will always print the same element and it's UB because at that point i is one beyond the end of the arr array, so the value will be unknown/undefined.

Converting "c99" loop to regular stuff [closed]

Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
Closed 3 years ago.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Improve this question
Story: I tried to convert a c99 script to regular gcc.
Problem: The output is empty.
Expected output: 3,2,1
length is the number of elements in the array.
Update: the script is designed to sort the elements of the array in a descending order.
The code:
#include <stdio.h>
int main() {
int arr[] = { 1,2,3 };
int temp = 0;
int length = sizeof(arr) / sizeof(arr[0]);
int i = 0;
int j = i + 1;
for (i < length; i++;) {
for (j < length; j++;) {
if (arr[i] < arr[j]) {
temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
}
int y = 0;
for (y < length; y++;) {
printf("%d ", arr[y]);
}
return 0;
}
Your syntax for for loops is the issue.
Here is the correct way to write your loops.
int i, j;
for (i = 0; i < length; ++i) // for (initialisation; test condition; operation)
{
for (j = i + 1; j < length; ++j) // note that j is initialized with i + 1 on each iteration of
// the outer loop. That's what makes the bubble sort work.
{
/* test and swap if needed */
}
}
for (i = 0; i < length; ++i) // note that i is reset to zero, so we can scan the array from
// a known position (the top) to bottom.
{
/* printout */
}
Your semicolon is in the wrong place, move it to the far left just inside the parentheses.
Loop syntax is:
for (intializer; break condition; iterator)

rand() doesn't give me random numbers [closed]

Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 3 years ago.
Improve this question
I want to print 1000 random numbers saved in a array. Numbers have to be between 1 and 10000.
I put srand(time(NULL)) in my main function and the array have to be filled with random numbers in my init function. The ausgabe function is for formatted output.
But rand fills my array with numbers all in row.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define ARR_SIZE 1000
void init(int * ptr, int size){
for (int i = 0; i < size; i++){
*(ptr + i) = (rand() %10000+1);
}
}
void ausgabe(int * ptr, int size){
for (int i = 0; i < size; i++){
printf("%5i\t", * ptr + i);
if ((i + 1) %10 == 0){
printf("\n");
}
}
printf("\n\n");
}
int main(){
int zufall[ARR_SIZE];
srand(time(NULL));
init(zufall, ARR_SIZE);
printf("\n\t\t\t---unsortierte Ausgabe---\n\n");
ausgabe(zufall, ARR_SIZE);
return 0;
}
* ptr + i is (*ptr)+i, not *(ptr+i). You need to be more careful with operator precedence. (And to learn to use your debugger: 30 seconds in your debugger would have clearly revealed that the problem was the printing, not the initialization.)

C Function: Count the occurrences of a digit in an integer [closed]

Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 5 years ago.
Improve this question
I'm receiving Output: 1. I should count the number of times a digit appear in an integer, for example, for number 1222345 and n = 2 Should appear 3 times.
int countOccurrences(int n, int num)
{
int i,k;
i=0;
while(num!=0)
{
k=num%10;
num=num/10;
if(k==n)
{
i++;
}
}
}
// Main
void main()
{
int num= 1222345;
int n = 2;
printf("Occurance of a number: %d", countOccurrences(n,num));
}
You have undefined behavior in the code. The function is supposed to return an int and it didn't.
Solution is to add return i in the end of other function. This will give you correct result. In the countOccurrences() function
...
if(k==n)
{
i++;
}
}
return i;
}
I was skipping the discussion of error check and all that. As chux mentioned for n<=0 case you might want to add a different way of handling it but you didn't add it. Atleast consider those case and put an error message on whatever input you need.
Some corner cases are
n=0,m=0.
Negative value of n or m.
Put a return on your countOccurrences function please
int countOccurrences (int n, int num) {
int i, k;
i = 0;
while (num! = 0)
{
k = num% 10;
num = num / 10;
if (k == n)
{
i ++;
}
}
return i; }
As other have pointed out, there are important issues with your code.
Here is a recursive solution that you may find interesting:
int countOccurrences(int n, int num)
{
int count = ((num % 10) == n);
return (num < 10) ? count : count + countOccurrences(n, num / 10);
}
Few general remarks about your code:
When using printf(), you should #include <stdio.h>.
main() should return int.
Place spaces around operators and format your code consistently. This k = num % 10; is more readable than k=num%10;. (There's more to code formatting than a matter of taste; without spaces you create areas full of characters which are more difficult to parse for our visual system.)

Resources