This code is supposed to recieve to arrays and then call function to return them in 1 array but I don't know how to print the last array returned from the function thanks in advance ???
and now I write anything because it says that the post is mostly code :D :D
#include <stdio.h>
#include <stdlib.h>
int join_arrays(int *array1, int *array2, int arr1_size, int arr2_size);
int main() {
int size_arr1, size_arr2, i, num1 = 1, s;
printf("Please enter the size of the first array: ");
scanf("%d", &size_arr1);
int arr1[size_arr1];
printf("start fill your first array: \n");
for (i = 0; i < size_arr1; i++) {
printf("enter element number %d: ",num1);
scanf("%d", &arr1[i]);
num1++;
}
num1 = 1;
printf("Please enter the size of the second array: ");
scanf("%d", &size_arr2);
int arr2[size_arr2];
int *ptr1_last;
printf("start fill your second array: \n");
for (i = 0; i < size_arr2; i++) {
printf("enter element number %d: ", num1);
scanf("%d", &arr2[i]);
num1++;
}
ptr1_last = join_arrays(arr1, arr2, size_arr1, size_arr2);
printf("sorted array= \n");
for (s = 0; s < (size_arr1 + size_arr2); s++) {
printf("%d\n", ptr1_last);
}
return 0;
}
int join_arrays(int *array1, int *array2, int arr1_size, int arr2_size) {
int counter_arr1, counter_arr2, m = 0;
int last_arr[arr1_size + arr2_size];
for (counter_arr1 = 0; counter_arr1 < arr1_size; counter_arr1++) {
last_arr[counter_arr1]=array1[counter_arr1];
}
for (counter_arr2 = counter_arr1; counter_arr2 < (arr1_size + arr2_size); counter_arr2++) {
last_arr[counter_arr2] = array2[m];
m++;
}
return last_arr[0];
}
Modified the code to create the receiving array in main and pass a pointer to it to the merge function because the local array last_arr would no longer exist when the function returned in your code.
#include <stdio.h>
#include <stdlib.h>
//Prototype changed to include a pointer to the receiving array, also no longer returns a value.
void join_arrays(int *last_arr, int *array1,int *array2,int arr1_size,int arr2_size);
int main()
{
int size_arr1,size_arr2,i,num1=1,s;
printf("Please enter the size of the first array: ");
scanf("%d",&size_arr1);
int arr1[size_arr1];
printf("start fill your first array: \n");
for(i=0; i<size_arr1; i++)
{
printf("enter element number %d: ",num1);
scanf("%d",&arr1[i]);
num1++;
}
num1=1;
printf("Please enter the size of the second array: ");
scanf("%d",&size_arr2);
int arr2[size_arr2];
int *ptr1_last;
printf("start fill your second array: \n");
for(i=0; i<size_arr2; i++)
{
printf("enter element number %d: ",num1);
scanf("%d",&arr2[i]);
num1++;
}
int last_arr[size_arr1 + size_arr2]; //Create receiving array here
join_arrays(last_arr, arr1,arr2,size_arr1,size_arr2); //And pass it to the function.
printf("merged array= \n");
for(s=0;s<(size_arr1+size_arr2);s++)
{
printf("%d\n", last_arr[s]);
}
return 0;
}
void join_arrays(int *last_arr, int *array1,int *array2,int arr1_size,int arr2_size)
{
int counter_arr1, m=0;
for(counter_arr1=0; counter_arr1<arr1_size; counter_arr1++)
{
last_arr[counter_arr1]=array1[counter_arr1];
}
for(; counter_arr1<(arr1_size+arr2_size); counter_arr1++)
{
last_arr[counter_arr1]=array2[m];
m++;
}
}
With that function you return only the first element of the last_array, you should create a global array so it's visible in all functions, or return a pointer of the last_array[0] position in memory
Related
function for get array from the user
#include <stdio.h>
void getArray()
{
printf("Enter size of array: ");
scanf("%d",&n);
printf("Enter %d elements in the array : ", n);
for(i=0;i<n;i++)
{
scanf("%d", &a[i]);
}
}
function for display array
void displayArray(){
printf("\nElements in array are: ");
for(i=0;i<n;i++)
{
printf("%d ", a[i]);
}
}
Both functions are called in the main
int main(){
int a[1000],i,n;
getArray();
displayArray();
return 0;
}
The problem is how to pass the array that we get from the user to the display array function and both functions can be called in the main and also the array want to declare in the main function
An example that does not handle input errors.
In order for your functions to have knowledge of the array, you must send them its address as well as its size.
#include <stdio.h>
int getArray(int a[], int size_max)
{
int n;
printf("Enter size of array: ");
while(1)
{
scanf("%d",&n);
if(n>size_max) printf("The size must be less than %d: ", size_max);
else break;
}
printf("Enter %d elements in the array : ", n);
for(int i=0; i<n; i++) scanf("%d", &a[i]);
return n;
}
void displayArray(int a[], int n)
{
printf("\nElements in array are: ");
for(int i=0; i<n; i++) printf("%d ", a[i]);
}
int main()
{
int a[1000];
int n = getArray(a, 1000);
displayArray(a, n);
return 0;
}
You can pass that shared variable as argument. Also return and use the returned value if reference not having valid data. Or else you need to pass this argument as reference instead of value like this.
#include <stdio.h>
int[] getArray(int a[])
{
printf("Enter size of array: ");
scanf("%d",&n);
printf("Enter %d elements in the array : ", n);
for(i=0;i<n;i++)
{
scanf("%d", &a[i]);
}
return a;
}
function for display array
void displayArray(int a[]){
printf("\nElements in array are: ");
for(i=0;i<n;i++)
{
printf("%d ", a[i]);
}
}
Both functions are called in the main
int main(){
int a[1000],i,n;
a = getArray(a);
displayArray();
return 0;
}
I can't figure out how to print array elements from my function into the main program so if some can examine this code and help me fix it I would appreciate it. The program is supposed to take the length of the array from user input and then ask for its elements and print them out afterward.
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int arrayN(int N) {
printf("Input array lenght: ");
scanf("%d",&N);
if(N>2) {
return N;
} else {
return 0;
}
}
int arrayelements(int array[], int array_length) {
int loop, i, N;
array_length = arrayN(N);
printf("Enter elements of the array: \n");
for(int i = 0; i < array_length; ++i) {
scanf("%d", &array[i]);
}
for(loop = 0; loop < array_length; loop++) {
printf("%d ", array[loop]);
}
}
int main() {
int N, array[], array_length;
int b = arrayelements(array[], array_length);
int a = arrayN(N);
printf("Array length is: %d \n", a);
printf("Elements of array are: %d \n", b);
return 0;
}
I reworked your example code. Hope it is what you want.
Focus lied on fixing the array declaration issues, memory allocation and
user input.
#include <stdio.h>
#include <stdlib.h>
#include <stdarg.h>
#include <ctype.h>
int userinput_integer(const char *fmt, ...){
int N, rv = 0;
va_list va;
va_start(va, fmt);
vprintf(fmt, va);
while(1){
rv = scanf("%d", &N);
if (1 == rv) break;
printf("Input error! The input >>");
do{
rv = fgetc(stdin);
if (isprint(rv)) putchar(rv);
}while(rv != EOF && rv != '\n');
printf("<< is not a valid integer.\nPlease try again: ");
}
va_end(va);
return N;
}
int userinput_arraylength(void) {
int N;
N = userinput_integer("Input array lenght: ");
if(N>2) {
return N;
} else {
printf("Invalid length\n");
return 0;
}
}
int userinput_arrayelements(int *array, int N) {
printf("Enter elements of the array: \n");
for(int i = 0; i < N; ++i) {
array[i] = userinput_integer("%d: ", i);
}
return N;
}
void print_arrayelements(int *array, int N){
for(int i = 0; i < N; ++i) {
printf("%d ", array[i]);
}
}
int main() {
int N, *array;
N = userinput_arraylength();
array = malloc(N * sizeof(*array));
if (NULL == array){
printf("Allocation error!\n");
exit(-1);
}
N = userinput_arrayelements(array, N);
printf("Array length is: %d \n", N);
printf("Elements of array are:\n");
print_arrayelements(array, N);
free(array);
return 0;
}
Fistly, declaration of array is not correct.
It should be array[] = {0}
Secondly, you cannot call your array elements function before arrayN function, the size of array should be entered first
And in the array elements() there is no need to call the size function you can directly pass the size of array when calling the array elements ()
Here's the code:
#include <stdio.h>
#include<malloc.h>
int *getarray()
{
int size;
printf("Enter the size of the array : ");
scanf("%d",&size);
int *p= malloc(sizeof(size));
printf("\nEnter the elements in an array");
for(int i=0;i<size;i++)
{
scanf("%d",&p[i]);
}
return p;
}
int main()
{
int *ptr;
ptr=getarray();
int length=sizeof(*ptr);
printf("Elements that you have entered are : ");
for(int i=0;ptr[i]!='\0';i++)
{
printf("%d ", ptr[i]);
}
return 0;
}
This is a piece of code to add the same number multiple times to an empty array but when I am printing the now non empty array, I am getting some other values:
#include<stdio.h>
#include<stdlib.h>
void sort_0(int arr[100], int i, int n){
int final_array[100], c=0;
// Count the number of '0' in the array
for(i=0;i<n;i++){
if(arr[i] == 0){
c++;
}
}
// Add the c number of '0' to the final_array
for(i=0;i<c;i++){
scanf("%d",final_array[i]);
}
for(i=0;i<c;i++){
printf("%d ", final_array[i]);
}
}
int main(){
int arr[100], i, n;
// Entering the size of the array
scanf("%d", &n);
// Entering n elements into the array
for(i=0;i<n;i++){
scanf("%d", &arr[i]);
}
sort_0(arr,i,n);
return 0;
}
In the above code, the number of times 0 appears in the array is counted. Then the count is taken as the range and 0 is adding to the empty array final_array count times.
If c = 5, the final_array = {0,0,0,0,0}
Expected Output:
arr = {0,1,4,3,0}
Output = 2
I am not getting any output
Since you don't know how much 0 you'll need to add to your array_final I figured out that a better solution could be to create that array after you have the number of 0 of the first array. Also, I see no reason why you were passsing i to the function since you can simply define it in the function itself.
void sort_0(int arr[10], int n, int* c){
int i;
for(i=0;i<n;i++){
if(arr[i] == 0){
(*c)+= 1;
}
}
}
int main (void) {
int size;
printf("Enter array size: ");
scanf("%d", &size);
int arr[size];
for (int i=0;i<size;i++) {
scanf("%d",&arr[i]);
}
int c = 0;
sort_0(arr, size, &c);
printf("C is: %d\n",c);
int* final_array;
if ((final_array=malloc(c * sizeof(int)))==NULL) // should always check malloc errors
{
perror("malloc");
return -1;
}
for (int i=0;i<c;i++) {
final_array[i]= 0;
}
printf("{");
for (int i=0;i<c-1;i++) {
printf("%d,", final_array[i]);
}
printf("%d}\n",final_array[c-1]);
return 0;
}
I am trying to make a c program where user will input 2 array and using pointer and user defined function. But it really not showing right answer. The main function will take the input from the user and it will send it to calculator function. where it will sum those array. And it will return the sum value to main function.
int calculator(int *my_array1, int *my_array2)
{
int sum=0,sum1=0,sum2=0;
for (int i = 0; i < strlen(my_array1); i++)
{
sum1+=(*(my_array1+i));
}
for (int i = 0; i < strlen(my_array2); i++)
{
sum2+=(*(my_array1+i));
}
sum=sum1+sum2;
return sum;
}
int main()
{
int size1,size2,i;
printf("Enter First Array Size: ");
scanf("%d",&size1);
int array1[size1];
printf("Enter Array's Value: ");
for(i=0; i<size1; i++)
{
scanf("%d",&array1[i]);
}
printf("Enter Second Array Size: ");
scanf("%d",&size2);
int array2[size2];
printf("Enter Array's Value: ");
for(i=0; i<size2; i++)
{
scanf("%d",&array2[i]);
}
printf("\nSum: %d",calculator(array1,array2));
return 0;
}
You can not use strlen for int* because it is not NULL terminated instead do this
#include<stdio.h>
int calculator(int *my_array1, int *my_array2,int len1,int len2)
{
int sum=0,sum1=0,sum2=0;
for (int i = 0; i < len1; i++)
{
sum1+=(*(my_array1+i));
}
for (int i = 0; i < len2; i++)
{
sum2+=(*(my_array1+i));
}
sum=sum1+sum2;
return sum;
}
int main()
{
int size1,size2,i;
printf("Enter First Array Size: ");
scanf("%d",&size1);
int array1[size1];
printf("Enter Array's Value: ");
for(i=0; i<size1; i++)
{
scanf("%d",&array1[i]);
}
printf("Enter Second Array Size: ");
scanf("%d",&size2);
int array2[size2];
printf("Enter Array's Value: ");
for(i=0; i<size2; i++)
{
scanf("%d",&array2[i]);
}
printf("\nSum: %d",calculator(array1,array2,size1,size2));
return 0;
}
Note: you are using variable length arrays it is in c99 only
So I have a problem:
My program has to find the most common element in an array. We have to make this array with calloc. My code is working but I am asked to move part of my code to a function in which I have to create an array and enter numbers.
int main() {
int n, i, dazn;
int *A;
dazn = 0;
printf("Iveskite naturalu skaiciu: \n");// enter how many elements in an array we will have
scanf("%d", &n);
A = (int*)calloc(n, sizeof(int)); // Starting from here...
for (i = 0; i < n; i++) {
printf("Iveskite skaiciu \n");
scanf("%d", &A[i]); //
}
... // and ending here have to be in a function
... // sorting and finding the most common element
As instructed, move the code to a separate function:
#include <stdio.h>
#include <stdlib.h>
int *allocate_and_read_array(int size) {
int *A = (int*)calloc(size, sizeof(int));
if (A != NULL) {
for (int i = 0; i < size; i++) {
printf("Iveskite skaiciu \n");
if (scanf("%d", &A[i]) != 1) {
printf("Missing values\n");
break;
}
}
}
return A;
}
int main(void) {
int n, i, dazn;
int *A;
dazn = 0;
printf("Iveskite naturalu skaiciu: \n");// enter how many elements in an array we will have
scanf("%d", &n);
A = allocate_and_read_array(n);
... // sort the array
... // find the most common element
free(A);
return 0;
}