Passing pointer to an array as a parameter to a function - c

I tried to build a heap and finally print the elements in the form of an array.
Here it is the code (I know this doesn't really make sense but I just wanted to test my knowlwdge of heap and dynamic arrays):
#include <stdio.h>
#include <stdlib.h>
void heapiify(int *arr,int n, int i)
{
int largest=i;
int l=2*i+1; // left node
int r= 2*i+2; // right node
if(l<=n && *arr[l]>=*arr[i])
largest=l;
if (r <=n && *arr[r]<=*arr[i])
largest= r;
if(largest !=i)
{
int temp=*arr[i];
*arr[i]=*arr[largest];
*arr[largest]=temp;
}
heapify(*arr,n,largest);
}
void buildh(int *arr,int n,int r,int c)
{
int i;
for(i=n/2-1;i>=0;i--)
heapify(*arr,n,i);
output(*arr,r,c);
}
void output(int *arr,int r,int c)
{
int i,j;
for(i=0;i<r;i++)
{
for(j=0;j<c;j++)
{
printf("%d",*arr[i*c+j]);
}
printf("\n");
}
}
int main()
{
int i,j,r,c;
printf("enter the number of rows");
scanf("%d",&r);
printf("enter the number of columns");
scanf("%d",&c);
int n=r*c;
int *arr=malloc(n*sizeof(int));
for(i=0;i<r;i++)
{
for(j=0;j<c;j++)
scanf("%d",&arr[i*c+j]);
}
buildh(*arr,n,r,c);
}
I'm getting 9 errors which are all the same
invalid argument type of unary '*'( have int)

Your arr variable is of type pointer to int:
int *arr=malloc(n*sizeof(int));
So when you call buildh, which takes the same type, you have to pass it as-is:
buildh(arr,n,r,c);
Same for the other cases.

The problem is the dereference of arr, across your funtions in multiple places, and the passing of dereferenced *arr in your functions to int * parameters, you should pass arr, try:
//...
void heapify(int *arr, int n, int i)
{
int largest = i;
int l = 2 * i + 1; // left node
int r = 2 * i + 2; // right node
if (l <= n && arr[l] >= arr[i]) //here
largest = l;
if (r <= n && arr[r] <= arr[i]) //here
largest = r;
if (largest != i)
{
int temp = arr[i]; //here
arr[i] = arr[largest]; //here
arr[largest] = temp; //here
}
heapify(arr, n, largest); //here
}
void buildh(int *arr, int n, int r, int c)
{
int i;
for (i = n / 2 - 1; i >= 0; i--)
heapify(arr, n, i); //here
output(arr, r, c); //here
}
void output(int *arr, int r, int c)
{
int i, j;
for (i = 0; i < r; i++)
{
for (j = 0; j < c; j++)
{
printf("%d", arr[i * c + j]); //here
}
printf("\n");
}
}
int main()
{
//...
buildh(arr, n, r, c); //here
}

Related

[Solved by rewrite]Problem with passing pointer and recursive function

I'm really newbie to c programing, and I have no idea how can I make it work.
I want to make sort function that sorting array of integers by using 2 argument.
I want to use recursive on sort function. and sorting from end of array to first. by ascending order.
void ft_sorting(int *arr, int size);
but I don't know what went wrong. It totally out of my understanding with pointer and array. right now I really have no idea. Can some one pointing what I did wrong. And what I need to learn and fix. Thank you!
void ft_recure(int *a, int *b, int j, int k)
{
if (--j >= 0)
{
if (a[k] < b[j])
{
a[k] = b[j];
}
else
{
ft_recure(a[k], b[j], j, k);
}
}
else
return a[k];
}
void ft_sort(int *tab, int size)
{
int i;
int h;
while (size > 0)
{
i = size;
h = i;
tab[size] = ft_recure(tab, tab, i, h);
size--;
}
}
and also I try this.
int ft_recurs(int x, int y, int a, int b)
{
int j;
j = a;
if( a > 0)
{
if(*x < *(y - 1);)
{
b = *(y - 1);
*x = b;
}
ft_recurs(*x,*(y - 1),a - 1, b);
}
else
{
return *x;
}
}
void ft_sort_int_tab(int *tab, int size)
{
int memo;
int i;
while(--size >= 0)
{
i = size;
tab[size] = ft_recurs(tab[i], tab[size], i, memo);
}
}
In the first approach, you did improper when calling API again:
void ft_recure(int *a, int *b, int j, int k)
{
if (--j >= 0)
{
if (a[k] < b[j])
{
a[k] = b[j];
}
else
{
ft_recure(a[k], b[j], j, k);
}
}
else
return a[k];
}
a and b input to API ft_recure is a pointer but in ft_recure(a[k], b[j], j, k); it is value. You should correct this as: ft_recure(&a[k], &b[j], j, k); if you expect to input the address of k and j elements.
In your alternative usage:
int ft_recurs(int x, int y, int a, int b)
{
int j;
j = a;
if( a > 0)
{
if(*x < *(y - 1);)
{
b = *(y - 1);
*x = b;
}
ft_recurs(*x,*(y - 1),a - 1, b);
}
else
{
return *x;
}
}
The input is value but in the function you are using *x andn *(y-1) is not really a proper way. May be you could try int ft_recurs(int x[], int y[], int a, int b). But if so, you also need to provide pointer address at ft_recurs(*x,*(y - 1),a - 1, b); and then the issue come back to similar to first approach.

Expression must have arithmetic type

While trying to write a code to find the intersection of two arrays,I came across a problem. I can't seem to modify a pointer inside a function.
Inside my find_intersection I get the error while doing the realloc function,compiler states that "counter" has no arithmetic value.
Any explanation on what went wrong here?
#include <stdio.h>
#include <stdlib.h>
int quick_sort(int*, int, int);
void swap(int*, int*);
int partition(int *, int, int);
int input_array_dyn(int*n);
int *find_intersection(int*, int*, int*, int, int,int *);
main()
{
int size1, size2, *counter, i=0;
int *arr1 = input_array_dyn(&size1);
int *arr2 = input_array_dyn(&size2);
quick_sort(arr1, 0, size1 - 1);
quick_sort(arr2, 0, size2 - 1);
int *arr3 = (int*)calloc(size2, sizeof(int));
arr3= find_intersection(arr1, arr2, arr3, size1, size2, &counter);
printf("The size of the new array is:%d\n", counter);
while (i < counter)
{
printf("%d\n", arr3[i]);
i++;
}
free(arr1);
free(arr2);
free(arr3);
}
int *find_intersection(int *arr1, int *arr2, int *arr3, int SA, int SB, int *counter)
{
int i = 0, j = 0, n = 0;
*counter = 0;
while (i < SA &&j < SB)
{
if (arr1[i] < arr2[j])
i++;
else if (arr2[j] < arr1[i])
j++;
else
{
arr3[n] = arr1[i];
i++;
n++;
j++;
}
}
counter = n;
arr3 = (int*)realloc(arr3, counter*sizeof(int));/*error here*/
return arr3;
}
int input_array_dyn(int*n)
{
int i;
int *a;
printf("Enter the size of the array:\n");
scanf("%d", n);
a = (int*)calloc(*n, sizeof(double));
assert(a);
printf("Enter the array elements:%d.\n", *n);
for (i = 0; i < *n; i++)
scanf("%d", a + i);
return a;
}
void swap(int* a, int* b)
{
int t = *a;
*a = *b;
*b = t;
}
int partition(int *arr, int low, int high)
{
int pivot = arr[high];
int i = (low - 1);
for (int j = low; j <= high - 1; j++)
{
if (arr[j] <= pivot)
{
i++;
swap(&arr[i], &arr[j]);
}
}
swap(&arr[i + 1], &arr[high]);
return (i + 1);
}
int quick_sort(int *arr, int low, int high)
{
if (low < high)
{
int pi = partition(arr, low, high);
quick_sort(arr, low, pi - 1);
quick_sort(arr, pi + 1, high);
}
}
In find_intersection() counter is a pointer to an int. To change it's value you need to use *counter instead of counter.
return arr3; attempts to return a pointer to int while function is declared to return just int. counter is a pointer to an int while you are using it as an regular int setting to 0 and so on.

min heap program stopped working

I am doing a simple minheap program but when size of heap is 4 i get an error program stopped working.
i checked heap size 2,3,5,6,7 program is working fine.
Why am i getting this error only when heapsize is 4?
I am using codeblocks 16.01, windows 10, gcc compiler.
minHeap.c
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
void MinHeapfy(int A[], int i, int N)
{
int l,r,temp,smallest;
l = 2*i;
r = 2*i + 1;
if((l<=N) && (A[l]<A[i]))
{
smallest = l;
}
else
{
smallest = i;
}
if((r<=N) && (A[r]<A[smallest]))
{
smallest = r;
}
if(smallest != i)
{
temp = A[smallest];
A[smallest] = A[i];
A[i] = temp;
MinHeapfy(A,smallest,N);
}
}
void BuildHeap(int A[],int N)
{
int i;
int f = floor(N/2);
printf("f %d", f);
for(i=f;i>=1;i--)
{
printf("i %d\n",i);
MinHeapfy(A,i,N);
}
}
int main()
{
int *A,N,T,data,q,i=1;
scanf("%d",&N);
A = (int *)calloc(N,sizeof(int));
for(i=1;i<=N;i++)
{
scanf("%d",&data);
A[i]=data;
}
BuildHeap(A,N);
for(i=1;i<=N;i++)
{
printf("%d ",A[i]);
}
return 0;
}
I can share some things that I learned till now:
void BuildHeap(int *A, int N);
void MinHeapfy(int *A, int i, int N);
In this 2 functions you just send a pointer not an array(difference between int *A and int A[5])
int main()
{
/*here it is better for you to declare like this
to see more easier which is which pointer and variable
(PS: at the moment you don't use q)*/
int N,T,data,i;//,q
int *A;
scanf("%d",&N);
A = (int *)calloc(N,sizeof(int));
for(i=0;i<N;i++)
{
scanf("%d",&data);
A[i]=data;
}
BuildHeap(A,N);
for(i=0;i<N;i++)
{
printf("%d ",A[i]);
}
return 0;
}
The 2 fors that you use must start from 0 to N
When you write:
A[0] <=> address pointed by A + 0 * sizeof(data type of A)
So 0 is the first address of the array
and the last address is N-1
A[N-1] <=> address pointed by A + (N-1) * sizeof(data type of A)

Permutations recursive of repeated elements?

so i get two parameters, one is N [ how big the array would be] and nr_vals [ this is the range, if nr_vals == 2, then range would be ( 0~1)]. I'm getting an error in swap function EXC_BAD_ACCESS. and the permutations are not printing right. any ideas?
My Swap Function ;
void swap(int *a, int *b)
{
int temp;
temp = *a;
*a = *b;
*b = temp;
}
Main fun;
void perm_rec_1(int N, int nr_vals){
int array[N];
int tempAr = 0;
for (int arrayFiller = 0; arrayFiller <= N; arrayFiller++)
{
if (arrayFiller == nr_vals){
tempAr = 0;
}
array[arrayFiller] = tempAr;
tempAr++;
}
prem_rec_help(N, nr_vals, array);
}
Secondary;
void prem_rec_help(int N, int nr_vals, int array[])
{
int tempArray[N];
copy_array(array, tempArray, N);
show(tempArray, N);
int M = 0;
int solid = N;
last_helper(tempArray, array, M, N, solid, nr_vals);
}
Recursive Function;
void last_helper(int array[],int temp[], int M, int N, int soild, int nr_vals ){
if (M != N)
{
last_helper(array,temp, M+1, N, soild, nr_vals);
}
for ( int i = 0; i < nr_vals; i++)
{
temp[M] = i;
show(temp, soild);
}
M--;
if(M == 0)
{
for( int swaper = 0; swaper <= soild; swaper++)
{
swap(array[swaper], array[swaper+1]);
copy_array(array, temp, soild);
if(swaper == soild)
{
return;
}else{
last_helper(array,temp, M, N, soild, nr_vals);
}
}
}
}

Print all the permutations using recursion

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);
}
}
}

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