I'm trying to sort elements in an array from smallest to largest that a user inputs along with the size. This is my code so far:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define MAXVALUES 20
void sort(int *A[], int n) {
int i, tmp, j;
for (i = 0; i <= (n - 2); i++) {
for (j = (i + 1); j <= (n - 1); j++) {
if (*A[i] > *A[j]) {
tmp = *A[i];
*A[i] = *A[j];
*A[j] = tmp;
}
}
}
return;
}
int main(int argc, char *argv[]) {
int n, A[MAXVALUES], i;
printf("Enter an array and no. of elements in array: ");
scanf("%d%d", &A[MAXVALUES], &n);
sort(A[MAXVALUES], n);
printf("Ordered array is: \n");
for (i = 0; i <= (n - 1); i++) {
printf(" %d", A[i]);
}
return 0;
}
The compiler compiles it without any errors but it stops working after I put in the inputs. I've yet to quite grasp the theory behind arrays and pointers so could someone tell me where in my code I'm going wrong?
scanf("%d%d", &A[MAXVALUES], &n); is the problem
That's not how you read an array.
First you read the n, after that inside a loop you read every element like
scanf("%d", &A[i]); where i is the index from 0 to n
EDIT:
scanf("%d", &n);
int i = 0;
for(i = 0; i < n; i++)
{
scanf("%d", &A[i]);
}
This is what you want.
You can't use scanf() to read in a whole array at once.
This:
scanf("%d%d", &A[MAXVALUES], &n);
Makes no sense; it passes scanf() the address of the element after the last one in A, causing undefined behavior. The sort() call is equally broken.
To read in multiple numbers, use a loop. Also, of course you must read the desired length first, before reading in the numbers themselves.
Firstly I'll tell you a couple of things about your sorting function. What you want, is take an array (which in C is representable by a pointer to the type of elements that are in that array, in your case int) and the array's size(int) (optionally you can take a sorting method as an argument as well, but for simplicity's sake let's consider you want to sort things from lowest to highest). What your function takes is a pointer to an array of integers and an integer (the second argument is very much correct, or in other words it's what we wanted), but the first is a little more tricky. While the sorting function can be wrote to function properly with this argument list, it is awkward and unnecessary. You should only pass either int A[], either int *A. So your function header would be:
void sort1(int *A, int n);
void sort2(int A[], int n);
If you do this however, you have to give up some dereferencing in the function body. In particular I am referring to
if (*A[i] > *A[j]) {
tmp = *A[i];
*A[i] = *A[j];
*A[j] = tmp;
} which should become
if (A[i] > A[j]) {
tmp = A[i];
A[i] = A[j];
A[j] = tmp;
}
You should check out the operators [] and *(dereference) precedence http://en.cppreference.com/w/c/language/operator_precedence
Now, to adapt to these changes (and further correct some code) your main would look like so:
int main(int argc, char *argv[])
{
int A[MAXVALUES], n;
do{
printf("n="); scanf("%d", &n);
}while(n < 0 || n > 20);
int i;
for(i = 0; i < n; ++i)
scanf("%d", &A[i]); //this will read your array from standard input
sort(A, n); //proper sort call
for(i = 0; i < n; ++i)
printf(" %d", A[i]);
printf("\n"); //add a newline to the output
return 0;
}
Related
I want to copy my array over to a pointer, so I can sort the array without changing the original. I'm not getting any errors, and am at a dead end. I've tried printing the pointer's contents, but it's always junk data. If anyone could help me it'd be much appreciated.
#include <stdio.h>
void bubbleSort(char *array[], int arrIndex);
void displaySort (char *array[], int arrIndex);
int main (void)
{
int index;
int count = 0;
printf("Enter number of people (0 - 50): ");
scanf("%d", &index);
index -= 1;
char userLastFirst[25][index];
int userAge[index];
//defining pointer
char *namePtr[25][index];
//do while loop, loops while count is less than or equal to the index
do{
printf("Enter name %d (last, first): ", count);
scanf(" %[^\n]s", userLastFirst[count]);
printf("Enter age %d: ", count);
scanf("%d", &userAge[count]);
// printf("\n\n%s %d\n\n", userLastFirst[count], userAge[count]);
count++;
} while (count <= index);
//assigning values to pointer
for(int i = 0; i < index; i++){
namePtr[25][i] = &userLastFirst[25][i];
}
//does not print values of pointer
for(int i = 0; i < index; i++){
printf("value of ptr[%d] = %s\n", i, namePtr[i]);
}
bubbleSort(*namePtr, index);
displaySort(*namePtr, index);
return 0;
}
void bubbleSort(char *array[], int arrIndex)
{
for (unsigned int pass = 0; pass < arrIndex - 1; ++pass) {
for (int i = 0; i < arrIndex - 1; ++i) {
if (array[i] < array[i + 1]) {
int temp = *array[i];
*array[i] = *array[i + 1];
*array[i + 1] = temp;
}
}
}
}
void displaySort(char *array[], int arrIndex)
{
for(int i = 0; i < arrIndex; i++){
printf("%s",*array[i]);
}
}
Sorry it's a lot, but I've looked everywhere, and am not exactly sure what the problem is. I suspect it's how I pass pointers thru the functions, or how I'm using * and & but am not totally sure what I'm doing wrong. The for loop below the one that assigns the array values to the pointer is for testing. It's supposed to print out the values I had given it in the previous for loop, but it doesn't. It just prints junk data.
There are couple of issues here, and since I don't know what exactly you want your final code to look like, I am going to give a few examples of what is wrong and how you could possibly fix that.
char userLastFirst[25][index];
This should be
char userLastFirst[index][25];
You don't want 25 people with names up to index characters in them, but rather index people with names up to 25 characters int them.
char *namePtr[25][index];
should be
char namePtr[index][25];
Same as before + you don't need the * here. Since you decided to go with vla let's stick with it. You would use * and more specifically char ** if you went with malloc/calloc.
namePtr[25][i] = &userLastFirst[25][i];
rather than doing this awkward copying, try:
strcpy(namePtr[i], userLastFirst[i]);
it copies the entire string for you, rather than just a single character. You will need to #include the <string.h> library for that.
void bubbleSort(char *array[], int arrIndex)
The first bubble sort argument should be:
char array[][25] // the same in displaySort
if you want to do it without the hassle of malloc/calloc.
And also don't go with int as your temp type. Rather do char temp[25] and copy them around with strcpy.
And don't compare strings with <, it doesn't work in C. Use strcmp for that.
And that's it for doing it without malloc/calloc, here's an example code:
#include <stdio.h>
#include <string.h>
void bubbleSort(char array[][25], int size);
void displaySort(char array[][25], int size);
int main (void)
{
int num_of_people;
printf("Enter number of people (0 - 50): ");
scanf("%d", &num_of_people);
char original_array[num_of_people][25];
char copied_array[num_of_people][25];
for (int i = 0; i < num_of_people; i++) {
printf("Enter name %d (last, first): ", i);
scanf(" %[^\n]s", original_array[i]);
}
for(int i = 0; i < num_of_people; i++){
strcpy(copied_array[i], original_array[i]);
}
bubbleSort(copied_array, num_of_people);
displaySort(copied_array, num_of_people);
return 0;
}
void bubbleSort(char array[][25], int size)
{
for (int i = 0; i < size - 1; ++i) {
for (int j = 0; j < size - 1 - i; ++j) {
if (strcmp(array[j], array[j + 1]) > 0) {
char temp[25];
strcpy(temp, array[j]);
strcpy(array[j], array[j + 1]);
strcpy(array[j + 1], temp);
}
}
}
}
void displaySort(char array[][25], int size)
{
for(int i = 0; i < size; i++){
printf("%s\n",array[i]);
}
}
and the program work like this:
Enter number of people (0 - 50): 4
Enter name 0 (last, first): Kowalski, Jan
Enter name 1 (last, first): Kowalska, Anna
Enter name 2 (last, first): Nowak, Miłosz
Enter name 3 (last, first): Amper, Ohm
Amper, Ohm
Kowalska, Anna
Kowalski, Jan
Nowak, Miłosz
I hope this is something you wanted to achieve.
#include<stdio.h>
#include<math.h>
int perfectSquare(int arr[], int n);
int main()
{
int n , arr[n];
printf("number of elements to store in array");
scanf("%d", &n);
for (int i = 0; i < n; i++)
{
printf("enter %d number", i+1);
scanf("%d", &arr[i]);
}
perfectSquare(arr, n);
return 0;
}
int perfectSquare(int arr[], int n)
{
int i;
int a;
for (i = 0; i <= n; i++) //i=4 arr[4]==9 //arr[1]=2 i=1
{
a=sqrt((double)arr[i]); //a=3 //a=1.454=1
if ( a*a==arr[i] ) //a==3*3==9==arr[4] //a*a=1!=arr[2]
printf("%d", arr[i]);
}
}
I am new to coding and I am currently learning c. I came up with this code but it doesn't work can someone tell me what is the problem with this code?
There are a couple of issues with this exercise, but generally you're on the right track. Here, a version of your example with some possible corrections:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
void perfect_square(int arr[], int n);
int main(void)
{
int i, n, *arr;
printf("number of elements to store in array: ");
scanf("%d", &n);
if (n <= 0)
return -1;
arr = malloc(n*sizeof(int));
if (arr == NULL)
return -2;
for (i = 0; i < n; i++) {
printf("enter number %d: ", i+1);
scanf("%d", &arr[i]);
}
perfect_square(arr, n);
free(arr);
arr = NULL;
return 0;
}
void perfect_square(int arr[], int n)
{
int i, a;
for (i = 0; i < n; i++) {
a = (int)sqrt((double)arr[i]);
if (a*a == arr[i])
printf("%d ", arr[i]);
}
}
Some hints:
Arrays, that have an unknown size at compile time are usually allocated with malloc(), and must be deallocated again with free() (see also: alloca(), calloc(), realloc()). (In "more recent versions of C" there is also the possibility to use variable length arrays, but those can limit the portability of the code).
Always make sure to check the start value and end condition of for-loops, to prevent out of bound errors.
And try to consistently format the code, use good names and nice indentation to improve read-, maintain-, reusablilty.
So basically what my c program needs to do is find out if the first digit of a number is odd, and do so for every element of a 12- element array. Also I need to make the program in the manner that finding out if the first element of a single number is odd needs to be written in a special function outside of main(). This program is really easy to make only in main() but as far as I know you need to use pointers for an array if you want to do something with it outside of main() and I'm not really good at that tbh. So any help would do me a big favour I guess.
This is what I've done so far:
#include <stdio.h>
int function(int p[], int n)
{
int i;
int x;
for (i = 0; i < n; i++)
{
while (p[i] >= 10)
;
{
p[i] /= 10;
}
if (p[i] % 2 != 0)
{
x++;
}
}
return x;
}
int main()
{
int n = 12;
int array[n];
int i;
for (i = 0; i < n; i++)
{
scanf("%d", &array[i]);
}
int r;
r = function(array[n], n);
printf("%d", r);
return 0;
}
And this is what my apparent errors are:
main.c:31:22: warning:
passing argument 1 of ‘function’ makes pointer from integer without a cast [-Wint-conversion]
main.c:3:9: note: expected ‘int *’ but argument is of type ‘int’
So yeah as I said, any help would do good. Also keep in mind that I'm in the first year of the first semester of college and that we can't really use anything outside of <stdio.h> or <stdlib.h> to write our code.
No pointers needed, unless you want to modify the array passed.
Some issues:
while(p[i] >= 10);{
p[i] /= 10;
}
The above code runs in an infinite loop, after which it runs p[i] /= 10; once.
Most C programmers have issues with missing ;. You have the opposite problem: A ; where it shouldn't be.
To put it simply, that intruding ; tells the compiler that the while loop doesn't run any code, so the compiler actually thinks your code means this:
while(p[i] >= 10) {
// Do nothing
}
p[i] /= 10;
int r;
r = function(array[n], n);
printf("%d", r);
That r variable is pointless. Unless you don't want to directly pass the return value from function
If the value of n is 12 then array[n] is the 12th element of array. Which doesn't exist because array only has elements 0-11
This is how I would write the code if I wanted to pass the whole array to my function:
printf("%d", function(array, n));
Here is a working version of the code you want
#include <stdio.h>
int count_odd_fdigits();
int main() {
const int ARRAY_SIZE = 12;
int array[ARRAY_SIZE];
int i;
for (i=0; i < ARRAY_SIZE; i++) {
scanf("%d", &array[i]);
}
printf("Numbers with an odd first digit: %d", count_odd_fdigits(array, ARRAY_SIZE));
return(0);
}
int count_odd_fdigits(int numbers[], int limit) {
int i;
int count = 0;
for (i=0; i < limit; ++i) {
while (numbers[i]/10 > 0)
numbers[i] /= 10;
if (numbers[i] % 2 != 0)
++count;
}
return(count);
}
(Run it online: https://onlinegdb.com/yEPr_mYgna)
I was given an assignment to write a code which takes in numbers as input from the user and provides the sum of it, specifically by the use of pointer arithmetic i.e. no array subscripting a[i] is allowed.
Below is the code that I wrote, which got compiled and even ran. But almost always it gives the sum of the input numbers as 0. I tried to fix it, but to no avail. Thus, I am asking for help, any help is greatly appreciated.
#include<stdio.h>
#define N 5
int sum_array( const int *p, int n)
{
int sum, a[N];
sum = 0;
for(p=&a[0]; p<&a[N]; p++)
sum += *p;
return sum;
}
int main()
{
int a[N], *i,x;
printf("Enter %d Numbers: ", N);
for(i=a; i<a+N; i++)
scanf("%d", i);
// all the input values get scanned as i or the array a
x= sum_array(i,N);
printf("the sum is %d\n", x);
return 0;
}
Beware, you are declaring array int a[N] in both main and sum_array. They are in different scopes, so they are different arrays (and the one from sum_array is never initialized so reading it invokes Undefined Behaviour).
The correct way is to pass the array along with its used length:
Here is a fixed version:
#include<stdio.h>
#define N 5
int sum_array( const int *a, int n) // a points to a array of at least n elements
{
int sum = 0; // initialize at definition time
for(const int *p=a; p<&a[n]; p++) // have the pointer p take all values from a
sum += *p;
return sum;
}
int main()
{
int a[N], *i,x;
printf("Enter %d Numbers: ", N);
for(i=a; i<a+N; i++)
scanf("%d", i);
// all the input values get scanned as i or the array a
x= sum_array(a,N); // pass the array address, not a pointer past last element
printf("the sum is %d\n", x);
return 0;
}
Finally it is mainly a matter of taste, but I was too often burnt for trying to add an instruction in a for loop without braces, so I strongly recommend using always braces for loops:
for(i=a; i<a+N; i++) {
scanf("%d", i);
}
int sum_array( const int *p, int n)
{
int sum = 0, i = 0;
for(i = 0; i < n ; i++)
sum += *(p+i);
return sum;
}
int main(void)
{
int a[N], i = 0, x = 0;
printf("Enter %d Numbers: ", N);
for(i=0; i<N; i++)
scanf("%d", a+i);
// all the input values get scanned as i or the array a
x= sum_array(a,N);
printf("the sum is %d\n", x);
return 0;
}
In x= sum_array(i,N); i is the iterator of your loop so after the loop has finished it points to the first position after the array.
You should pass the original array instead x= sum_array(a,N);
In the sum function you just throw away the passed pointer and replace it with your local a[].
int sum_array( const int *p, int n)
{
int sum = 0;
int *end = &p[n]; // first element after the array.
for(; p<end; p++) // just use p because you don't need the reference to the start of the array
{
sum += *p;
}
return sum;
}
but as you said that array notation is not allowed you can change it as follows
#include "stdio.h"
#define N 5
int sum_array( const int *p, int n)
{
int sum = 0;
const int *end = p+n; // first element after the array.
for(; p<end; p++)
{
sum += *p;
}
return sum;
}
int main()
{
int *a, *i, x;
a = malloc(N * sizeof(*a));
if (a == NULL)
exit(-1);
printf("Enter %d Numbers: ", N);
for(i=a; i<a+N; i++)
{
scanf("%d", i);
}
// all the input values get scanned as i or the array a
x= sum_array(a,N); // pass the array address, not a pointer past last element
printf("the sum is %d\n", x);
return 0;
}
in general, when programming, the code should be kept as simple as possible while still being complete.
The program criteria shows no need to keep a number after it is applied to the sum of the numbers, So in the proposed code, the input number is only kept long enough to be applied to the sum, then it is discarded.
The following proposed code:
cleanly compiles
performs the desired functionality
is kept very simple
properly checks for; and handles any errors
And now the proposed code:
#include <stdio.h> // scanf(), printf(), fprintf(), stderr
#include <stdlib.h> // exit(), EXIT_FAILURE
#define N 5
int main( void )
{
int num = 0;
int sum = 0;
printf("Enter %d Numbers: ", N);
for(size_t i=0; i<N; i++)
{
if( scanf("%d", &num) != 1 )
{
fprintf( stderr, "failed to read number\n" );
exit( EXIT_FAILURE );
}
// implied else, scanf successful
sum += num;
}
printf( "the sum is %d\n", sum );
return 0;
}
How to read space-separated integers representing the array's elements and sum them up in C?
I used the below code but it reads all the elements in a new line:
#include <math.h>
#include <stdio.h>
int main() {
int i = 0, N, sum = 0, ar[i];
scanf("%d" , &N);
for (i = 0; i < N; i++) {
scanf("%d", &ar[i]);
}
for (i = 0; i < N; i++) {
sum = sum + ar[i];
}
printf("%d\n", sum);
return 0;
}
Your array ar is defined with a size of 0: the code invokes undefined behavior if the user enters a non zero number for the number of items.
Furthermore, you should check the return value of scanf(): if the user enters something not recognized as a number, your program will invoke undefined behavior instead of failing gracefully.
Here is a corrected version:
#include <stdio.h>
int main(void) {
int i, N, sum;
if (scanf("%d", &N) != 1 || N <= 0) {
fprintf(stderr, "invalid number\n");
return 1;
}
int ar[N];
for (i = 0; i < N; i++) {
if (scanf("%d", &ar[i]) != 1) {
fprintf(stderr, "invalid or missing number for entry %d\n", i);
return 1;
}
}
sum = 0;
for (i = 0; i < N; i++) {
sum += ar[i];
}
printf("%d\n", sum);
return 0;
}
Note that the program will still fail for a sufficiently large value of N as there is no standard way to check if you are allocating too much data with automatic storage. It will invoke undefined behavior (aka stack overflow).
You should allocate the array with malloc() to avoid this:
#include <stdio.h>
#include <stdlib.h>
int main(void) {
int i, N, sum;
int *ar;
if (scanf("%d", &N) != 1 || N <= 0) {
fprintf(stderr, "invalid number\n");
return 1;
}
ar = malloc(sizeof(*ar) * N);
if (ar == NULL) {
fprintf(stderr, "cannot allocate array for %d items\n", N);
return 1;
}
for (i = 0; i < N; i++) {
if (scanf("%d", &ar[i]) != 1) {
fprintf(stderr, "invalid or missing number for entry %d\n", i);
return 1;
}
}
sum = 0;
for (i = 0; i < N; i++) {
sum += ar[i];
}
printf("%d\n", sum);
free(ar);
return 0;
}
Finally, there is still a possibility for undefined behavior if the sum of the numbers exceeds the range of type int. Very few programmers care to detect such errors, but it can be done this way:
#include <limits.h>
...
sum = 0;
for (i = 0; i < N; i++) {
if ((sum >= 0 && arr[i] > INT_MAX - sum)
|| (sum < 0 && arr[i] < INT_MIN - sum)) {
fprintf(stderr, "integer overflow for entry %d\n", i);
return 1;
}
sum += ar[i];
}
#include <math.h>
#include <stdio.h>
int main()
{
int i=0,N,sum=0;
scanf("%d" ,&N);
int ar[N];
for(i=0; i<N; i++)
scanf("%d",&ar[i]);
for(i=0; i<N; i++)
sum=sum+ar[i];
printf("%d\n", sum);
return 0;
}
This should be the code.
You have initially declared the array of size 0 (because i=0).
Even though you declared the array of size 0, when I ran it on my machine, it actually executed successfully with the correct output.
This is generally due to undefined behavior which means that we can only guess the output when the code is correct. If the code has undefined behavior, then it can do whatever it wants (and in the worst case the code will execute successfully giving the impression that it's actually correct).
Declaring a Variable Size Array (VLA) is optional in C11 standard. Thus, it depends on the implementation of the compiler whether it will support VLA or not. As pointed out by #DavidBowling in comments, if the compiler does support, then declaring a VLA of size 0 can invoke undefined behavior (which you should avoid in all cases). If it doesn't support, then this will simply give a compilation error and you'll have to declare the array size as some integer constant (example, int arr[100];).
#include <math.h>
#include <stdio.h>
int main()
{
int i=0,N,sum=0;
scanf("%d" ,&N);
int ar[N];
for(i=0; i<N; i++)
{
scanf("%d",&ar[i]);
}
for(i=0; i<N; i++)
{
sum=sum+ar[i];
}
printf("%d\n", sum);
return 0;
}
You should declare the array after accepting the value of N.
#include <math.h>
#include <stdio.h>
int main()
{
int i=0,N,sum=0;
scanf("%d" ,&N);
int ar[N];
for(i=0; i<N; i++)
{
scanf("%d",&ar[i]);
sum=sum+ar[i];
}
printf("%d\n", sum);
return 0;
}
As this is a very simple question I'll expand it a bit to include some good programming practices.
1. Analyze the problem
We have to complete two tasks here:
Read and store the numbers to array.
Sum the array elements.
Of course you can both read and calculate the same time, but we ❤ the SoC design principle. This will help you later with bigger programs.
2. Create the program structure
In this state we have to consider what function to use, as we already solved the data structure problem (we use array).
Of course, we always can put the whole procedure in main function but this would break the SoC principle.
The main principle here is:
I create a function for a separate procedure.
So we'll have to build two functions. Let's consider the following example:
ReadArrayData will be used to read the data from the standard input (your keyboard in other words) to array. But what will declare as parameters? We surely have to pass the array and the array size. The return type of this function will be void (we don't have to return something).
Keep in mind that if you pass array to function you can manipulate it as you please and keep these changes in your main program. This is because the arrays are passed always by reference to a function.
In the end this will be your function prototype:
void ReadArrayData(int arraySize, int array[]);
CalculateArraySum will be used to calculate the sum of the array elements. The function prototype will be the same as for ReadArrayData with the difference that the returning type will be int (we return the sum).
int CalculateArraySum(int arraySize, int array[]);
3. Write your program
#include <stdio.h>
void ReadArrayData(int arraySize, int array[]);
int CalculateArraySum(int arraySize, int array[]);
int main(void) {
int N;
printf("Give the array size: ");
scanf("%d", &N);
int array[N];
ReadArrayData(N, array);
int sumOfArrayElements = CalculateArraySum(N, array);
printf("The sum of array elements is %d.\n", sumOfArrayElements);
return 0;
}
void ReadArrayData(int arraySize, int array[]) {
printf("Give %d elements: ", arraySize);
for (int i = 0; i < arraySize; ++i) {
scanf("%d", &array[i]);
}
}
int CalculateArraySum(int arraySize, int array[]) {
int sum = 0;
for (int i = 0; i < arraySize; ++i) {
sum += array[i];
}
return sum;
}
I know this was a large scaled answer, but I saw you are new to computer programing. I just wanted to present you the main functionality to solve all kinds of problems. This was just a small introduction. In the end, you have to remember what steps we take to solve a problem. With time and as you solve many problems you will learn many many other things.