I'm currently stuck on that little programm in C:
#include <stdio.h>
void print(int a[], int n) {
int index =0;
while (index <= n-1){
while (index != a[index]){
int temp;
temp = a[index];
a[index] = a[temp];
a[temp] = temp;
printf("%d\n", temp);
}
index = index+1;
}
int i;
for (i = 0; i<n; i++){
printf("%d\n", a[i]);
}
}
int main() {
int A[] = {0,6,5,1,3,20};
int n=6;
print(A,n);
return 0;
}
I have no idea why its not sorting and giving me that error...
Kind regard and thanks in advance
Related
The following code is trying to find the averages of a set of numbers in C, but the median and the mean both do not return anything. How do I make it so the mean and the median both return a float? Am I returning an invalid value or?
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int compare (const void * a, const void * b)
{
return ( *(int*)a - *(int*)b );
}
float mean(int arr[], int size){
int sum = 0;
for(int i = 0; i<size; i++){
sum += arr[i];
}
return ((float)sum/size);
}
int range(int arr[], int size){
int smallest = arr[0];
int largest = arr[0];
for(int i=0; i<size; i++){
if(smallest>arr[i]){
smallest = arr[i];
} if(largest<arr[i]){
largest = arr[i];
}
} int difference = largest - smallest;
return difference;
}
int mode(int arr[], int size){
int maxValue = 0;
int maxCount = 0;
for(int i = 0; i<size; i++){
int count = 0;
for(int j = 0; j<size; j++){
if(arr[j] == arr[i]){
count++;
}
} if(count > maxCount){
maxCount = count;
maxValue = arr[i];
}
} return maxValue;
}
float median(int arr[], int size){
qsort(arr, size, sizeof(int), compare);
float middleOfArray = size/2;
int roundedMiddleOfArray = rint(middleOfArray);
if(ceilf(middleOfArray) == middleOfArray){
return((float)arr[roundedMiddleOfArray]);
}
else{
return((float)arr[roundedMiddleOfArray] - arr[roundedMiddleOfArray-1]);
}
}
int main(){
int array[6] = {1,2,3,4,5,6};
int newMean = mean(array, 5);
int newRange = range(array, 5);
int newMode = mode(array,5);
int newMedian = median(array, 5);
printf("The mean is : %f \n", newMean);
printf("The range is : %d \n",newRange);
printf("The mode is : %d \n",newMode);
printf("The median is : %f \n", newMedian);
return 0;
}
Turns out the way you fix it is by declaring the mean and the median as floats not ints (in main).
can you help me with code which returns partial sum of 'X' numbers in array in c?
Complete :
int arr_sum( int arr[], int n )//Recursive sum of array
{
if (n < 0) //base case:
{
return arr[0];
}
else
{
return arr[n] + arr_sum(arr, n-1);
}
}
void sum_till_last (int *ar,int si )
{
**int sum,i;// the problem some where here
ar=(int*) malloc(si*sizeof(int));
for (i=0; i<si;i++)
{
sum=arr_sum(ar,i);
ar [i]=sum;
}
free (ar);**
}
void main ()
{
int i;
int a [5];
for (i = 0; i < 5; i++)
scanf_s("%d", &a[i]);
sum_till_last(a,5);
printf("%d\n",a[5]);
}
\i want to create new array with this this legality:
My input :
4
13
23
21
11
The output should be (without brackets or commas):
4
17
40
61
72
Now when we can see the full code, it's quite obvious that the problem is in the sum_till_last function where you overwrite the pointer you pass to the function with some new and uninitialized memory you allocate.
Drop the allocation (and the call to free of course). And fix the logical bug in arr_sum that causes you to get arr[0] + arr[0] when i is zero.
Here you go:
#include <stdio.h>
int main () {
int in_arr[5] = {4,13,23,21,11};
int out_arr[5];
int p_sum =0;
int i;
for ( i=0;i<5;i++){
out_arr[i] = in_arr[i]+p_sum;
p_sum=p_sum+in_arr[i];
}
for (i=0;i<5;i++){
printf("%d", out_arr[i] );
}
}
Fix according to your policy
#include <stdio.h>
#include <stdlib.h>
int arr_sum(int arr[], int n){
if (n == 0){//Change to this
return arr[0];
} else {
return arr[n] + arr_sum(arr, n-1);
}
}
void sum_till_last(int *ar, int si){
int sum,i;
int *temp = malloc(si * sizeof(int));//variable name ar is shadowing parameter name ar.
for(i = 0; i < si; i++){
temp[i] = arr_sum(ar, i);
if(i)
putchar(' ');
printf("%d", temp[i]);//need print out :D
}
free(temp);
}
int main(void){
int i, a[5];
for (i = 0; i < 5; i++)
scanf_s("%d", &a[i]);
sum_till_last(a, 5);
//printf("%d\n",a[5]);<-- this print only one element. and It's out of bounds element XD
}
I just made it simple so it´s easy to understand :)
I´m assuming "n" is always equal or less then array element number. Then you just print the SUM.
#include <stdio.h>
int arr_sum( int arr[], int n ){
int i=0,SUM=0;
for(; i < n;i++){
SUM= SUM+ arr[i];
printf("%d ",SUM);
}
}
int main(void) {
int array[] = {4, 13, 23, 21, 11};
arr_sum(array,5);
return 0;
}
I coded as below to print all the permutations of three number :1,2,3.
But the output is:
1,1,1
1,1,2
1,1,3
1,2,1
1,2,2
1,2,3
The code is as follows:
#include<stdio.h>
#include<conio.h>
void perm(int);
int a[10],l=2;
int main()
{
int k;
k=0;
perm(k);
getch();
return 0;
}
void perm(int k)
{
int i;
for(a[k]=1;a[k]<=3;a[k]++)
{
if(k==2)
{
for(i=0;i<3;i++)
{
printf("%d ",a[i]);
}
printf("\n");
}
else
{
k++;
perm(k);
}
}
}
Please give the correct code.
Why do you increment k? k should not change for a given call to perm().
Also it's a bit too bad to be stuck with 3 permutations, you can easily generalize this way:
#include<stdio.h>
#include<conio.h>
static void perm(int, int);
static void all_perm(int);
int a[10];
int main()
{
all_perm(3);
getch();
return 0;
}
void all_perm(int n)
{
perm(0, n);
}
void perm(int k, int n)
{
if (k == n)
{
for(int i = 0; i < n; i++)
{
printf("%d ", a[i]);
}
printf("\n");
}
else
{
for(a[k]=1; a[k] <= n; a[k]++)
{
perm(k + 1, n);
}
}
}
Edit: Well, what you name permutations are not permutations.
The logic that I have used is to some extent similar to yours.
I have included the entire code to make it clear.
#include <stdio.h>
void recn(int*,int,int);
void print_arr(int*,int);
void main()
{
int arr[3] = {1,2,3};
recn(arr,3,0);
}
void print_arr(int *arr, int n){
int i;
for(i = 0,printf("\n"); i < n; printf("%d",arr[i++]));
}
void recn(int *arr, int n, int l) {
int i, j, f, k, xx = 0;
static int tst[15], a[14]={0};
if (l == n) {
for (i = 0; i < n; i++) {
tst[i] = arr[a[i]];
}
print_arr(tst,n);
return;
}
for (i = 0; i < n; i++) {
f = 0;
for (j = 0; j < l; j++)
if (a[j] == i)
f = 1;
if (!f) {
a[l] = i;
recn(arr, n, l + 1);
}
}
}
I want to a fill a small 2 d array with arbitrary values before moving onto a bigger array. However, when I compile and run my program, I get some weird output. It is not perfectly square. if someone could point out what im doing wrong that would be great.
void startarray(char (*arr)[10], int y_length, int x_length);
void printarray(char (*arr)[10], int y_length);
int main()
{
char arr[10][10];
startarray(arr, 10,10);
printarray(arr, 10);
return 0;
}
void startarray(char (*arr)[10], int y_length, int x_length)
{
int i;
int j;
for(i = 0; i <= y_length; i++)
{
for(j = 0; j < x_length; j++)
{
arr[i][j] = 'a';
}//end for
arr[i][j] = '\0';
}//end for
}
void printarray(char (*arr)[10], int y_length)
{
int i = 0;
while(i < y_length)
{
printf("\n%s", arr[i]);
i++;
}//end while
}
I am trying to access a structure with a pointer to an integer , from main. But the program crashes. It needs to be built with "std=c99" option as it is the requirement in a test.
The code is as follows:
#include <stdio.h>
#include <malloc.h>
struct Results{
int *A;
int N;
};
struct Results solution(int A[], int N, int K) {
struct Results result;
// write your code in C99 (gcc 4.8.2)
int* T = (int*) malloc(N*sizeof(int));
result.A = A;
result.N = N;
int count = 0;
while(count < K)
{
for(int i = 0; i < N; i++)
{
if(i > 0)
{
T[i] = A[i-1];
}
else
{
T[0] = A[N-1];
}
}
count++;
for(int i = 0; i < N; i++)
{
A[i] = T[i];
}
};
for(int i = 0;i < N; i++)
{
A[i] = T[i];
}
return result;
}
struct Results solution(int A[], int N, int K);
void main()
{
int B[5] = {3,8,9,7,6};
struct Results st;
solution(B,sizeof(B),1);
}
The trouble is at line:
" solution(B,sizeof(B),1);"
What am I doing wrong?
Please help.
You see sizeof(B) would give the number of elements in B times the size of an int, use sizeof(B) / sizeof(B[0]) instead.