I have a task to sort an integer array into an even number array and odd number array. Then i have to show what digits has been placed in which. However, in my code not all places of the array are occupied, so in the end i receive random numbers when I want to show what the arrays odds and evens contain. Instead of random numbers I would like to have literally nothing added in their place.
I did the following:
int main()
{
int evens[10], whole[10], odds[10], i;
printf("Enter 10 integer(/whole) numbers\n");
for (i = 0; i < 10; i++)
{
scanf("%d", &whole[i]);
if (whole[i] % 2 == 0)
{
evens[i] = whole[i];
else odds[i] = whole[i];
}
}
printf("Your even numbers are the following:\n");
for (i = 0; i < 10; i++)
{
printf("%d\n", evens[i]);
}
printf("Your odd numbers are the following:\n");
for (i = 0; i < 10; i++)
{
printf("%d\n", odds[i]);
}
return 0;
}
and then I get this output for having entered digits from 1 to 10:
Your even numbers are the following:
-1832565259
2
1985901418
4
4200864
6
4200958
8
74
10
Your odd numbers are the following:
1
4200864
3
6356652
5
1986281400
7
1985964450
9
1985901427
So how do i get an odds/even array without these random digits like 1985964450 in between? Is there a command to add literally nothing instead?
You should have a counter of odds and a counter of evens.
int oddcount = 0;
int evencount = 0;
When you decide that a number is even, you use this counter to know where in the array it should go. For example:
if (whole[i] % 2 == 0) {
evens[evencount] = whole[i];
evencount++;
}
Notice that evencount not only gives you the number of even numbers, but since arrays indices begin at zero, it also tells you what is the position of the next even number.
Then you modify your for loops at the end to use the actual number of even and odd numbers that were typed. You can even check for zero and print a specific message such as No even numbers supplied.
Also, unless you have been specifically asked to keep the input numbers in an array, you don't need whole. You can do just like so:
int input;
for (i=0; i<10; i++)
{
scanf("%d", &input);
if (input %2 == 0)
/* ... */
else
/* ... */
}
As a final remark, you should indent your code. Indentation is simply augmenting the number of spaces before the code (like I did inside the ifs above). It is very important to indent your code because it makes the structure of your code clear. For a more comprehensive discussion on this, read here: Importance of code indentation.
It is better to have two variables that represents indexes, one you use to add odd numbers, other to add even numbers. Then you will have two arrays without redundant data :)
int evensIndex = 0;
int oddsIndex = 0;
for (i=0; i<10; i++)
{
scanf("%d", &whole[i]);
if (whole[i] %2 == 0)
{
evens[evensIndex] = whole[i];
evensIndex++;
}
else
{
odds[oddsIndex] = whole[i];
oddsIndex++;
}
}
int main()
{
int evens[10], temp, odds[10], i;
int oddIndex = 0, evenIndex = 0;
printf("Enter 10 integer(/whole) numbers\n");
for (i=0; i<10; i++)
{
scanf("%d", &temp);
if(temp%2)
odd[oddIndex++]=temp;
else
even[evenIdex++]=temp;
}
printf("Your even numbers are the following:\n");
for (i=0; i<10; i++)
printf("%d\n", evens[i]);
printf("Your odd numbers are the following:\n");
for (i=0; i<10; i++)
printf("%d\n", odds[i]);
return 0;
}
Related
i am new to programing, i want to know that how we can find the odd digits in a number.
the condition in this program is we should only use concept of arrays.I tried a code for this as follows:
#include <stdio.h>
int main()
{
int A[50],i,x,y,n,sum=0;
scanf("%d",&n);
printf("the value is %d\n",n);
for(i=0;i<n;i++)
scanf("%d",&A[i]);
for(i=0;i<n;i++){
x=A[i]%10;
if(x%2!=0)
sum=sum+x;
A[i]=A[i]/10;
printf("the sum of odd numbers is %d\n",sum);
}
return 0;
}
but in this the code is checking for only one digit of the first number in the loop and then next time it is going to check the digit of second number.
so, i need to write a code to check all digits in the number and then it goes to next number and check all digits and should continue the same process in the loop.So, for this how should i modify my code?
You were missing a loop that would iterate through every digit of A[i] - the inner while loop below,
#include <stdio.h>
int main()
{
int A[50], i, x, y, n, sum=0;
printf("How many numbers will you input?\n");
scanf("%d",&n);
printf("the value is %d\n",n);
for(i=0; i<n; i++) {
scanf("%d",&A[i]);
}
for(i=0; i<n; i++) {
sum = 0;
while (A[i] > 0) {
x = A[i]%10;
if(x%2 != 0) {
sum = sum + x;
}
A[i] = A[i]/10;
}
printf("the sum of odd numbers is %d\n",sum);
}
return 0;
}
The exact algorithm for iterating through each digit in a nice form can be found in this post - although for a different language. Here, apart from the while loop, you also need to reset the sum each time unless you want a cumulative sum over all provided numbers.
Note that I changed the formatting a bit - more space, extra braces, and a message about what you're prompting the user to input.
int temp;
int sum = 0;
temp = number;
do {
lastDigit = temp % 10;
temp = temp / 10;
sum += (lastDigit %2 != 0) ? lastDigit : 0;
} while(temp > 0);
I'm new to C and need some help to understand how this piece of code works. I know that it reads the values that the user writes, puts them into an array, and then prints them out.
But I don't understand why I need two "counters" (i and j) to do this. Can someone help me to figure it out?
#include<stdio.h>
int main ()
{
int A[5];
int i=0;
int j=0;
while (i < 5)
i++;
printf("Enter your %d number\n", i);
scanf("%d", &A[i]);
}
while (j < 5)
{
j++;
printf ("\n%d\n", A[j]);
}
}
Technically, what you have aren't two counters, but two loops. If you wanted, to, you could just reuse i for the second loop as well, by doing something like this:
while (i < 5)
i++;
printf("Enter your %d number\n", i);
scanf("%d", &A[i]);
}
i = 0;
while (i < 5)
{
i++;
printf ("\n%d\n", A[i]);
}
As for why you have two loops, the reason is simple. The first loop (using i in your code), reads the 5 integers into the array A. After the first loop concludes, your array A holds the 5 int values, which you could've used however you wanted. In your case, you want to print those values. So what you do is use a loop for looping over the array elements and printing the values to the screen, one by one.
You don't need it, you can simply reset the first and reuse it. However you must increment your index only after having using it otherwise you will overflow the limit of the array :
#include<stdio.h>
int main ()
{
int A[5];
int i=0;
while (i < 5) {
printf("Enter your %d number\n", i);
scanf("%d", &A[i]); // the last must be 4 not 5
i++; //<== increment here
}
i=0;
while (i < 5)
{
printf ("\n%d\n", A[i]); //idem
i++;
}
}
I'm a beginner to C, and am trying to sort user inputted numbers into odd and even arrays. I don't understand why my code isn't working.
Cheers.
This is my code, I don't understand my mistake.
int x[]= {};
int i=0;
int d=0;
int j=0;
int even[12]={};
int odd[12]={};
printf("Enter amount of numbers: "); // asking user for amount of numbers
scanf("%d", &d);
for (j=0; j<d; j++){
printf("Enter number %d: ", i+1); // scanning input into 'x' array
scanf("%d", x[i]);
}
printf("Even numbers: ");
for (i=0; i<d; i++) {
if (x[i] % 2 == 0) { // sorting into even array
even[i]=x[i];
printf("%d \n", even[i]);
}
}
printf("\n Odd numbers: ");
for (i=0; i<d;i++){
if (x[i] % 2 != 0) { // sorting into odd array
odd[i]=x[i];
printf("%d \n", odd[i]);
}
}
This error message keeps coming up:
$ ./main
Enter amount of numbers: 4
Enter number 1: 6
Segmentation fault (core dumped)
int x[]= {}; doesn't work because it would hold no elements. But initializing it with {} doesn't work in C anyway, do this instead:
int x[24] = {0}; // first element explicitely set to 0, the rest default-initialized to 0
You also need to put {0} for even and odd. If it's compiling for you with {} then it's possible that you're compiling it as a C++ program, or perhaps your compiler just tolerates it anyway (but it won't work on every C compiler).
scanf needs the address of the int, so instead of scanf("%d", x[i]); you need scanf("%d", &x[i]);. But i is the wrong iterator for this for (j = 0; j < d; j++) loop. Instead do this:
for (j = 0; j < d; j++) {
printf("Enter number %d: ", j + 1); // scanning input into 'x' array
scanf("%d", &x[j]);
}
Also note that the way you're doing this, half the array will be left at 0. So for instance if I imputted the values 1 through 6, then odd contains the values 1 0 3 0 5 0.
I am making simple example of little game about guessing numbers.
And I want to build a function which check the numbers and make two values as follows:
1) hits-the number of digits that contain in both number and in same place for both numbers.
2) misses-the number of the digits which contain in both number but not in the same place.
For example:
int systemNumber=1653;
int userGuess=5243;
in this example, in both numbers there are the digits 5 and 3. In both numbers the digit 3 in the same place. But, the digit 5 in systemNumber is not in the same place as userNumber. So, we have here 1 hit and 1 miss.
I've written the code for it with arrays, and I'd like to know if there is a way that I will be able to do this without array and strings.
Here is my code. Please, if you have any improvement for my code, I'd like to know it :)
#include <stdio.h>
#include <stdlib.h>
void checkUserCode(int num1[4], int num2[4]); // declare the function which check the guess
int hits=0, misses=0; // hits and misses in the guess
int main(void)
{
int userCode=0;
int userCodeArray[4];
int systemCodeArray[4]={1, 4, 6, 3};
int i=0;
// printing description
printf("welcome to the guessing game!\n");
printf("your goal is to guess what is the number of the system!\n");
printf("the number have 4 digits. Each digit can be between 1 to 6\nGood Luck!\n");
// input of user guess
printf("enter number: ");
scanf("%d", &userCode);
for (i=3; i>=0; i--)
{
userCodeArray[i]=userCode%10;
userCode=userCode/10;
}
checkUserCode(systemCodeArray, userCodeArray);
printf("there are %d hits and %d misess", hits, misses); // output
return 0;
}
/*
this function gets two arrays and check its elements
input (parameters): the two arrays (codes) to check
output (returning): number of hits and misses
if the element in one array also contains in the other array but not the same index: add a miss
if the element in one array also contains in the other array and they have the same index: add a hits
*/
void checkUserCode(int num1[4], int num2[4])
{
int i=0, j=0;
for (i=0; i<4; i++)
{
for (j=0; j<4; j++)
{
if(num1[i]==num2[j])
{
if (j==i)
hits++;
else
misses++;
}
}
}
}
Here is an example I wrote a while ago, which I tweaked for your problem:
I basically uses two for loops, the outer loop going over the first number, 1653, and the inner loop going over the second number, 5243. It basically compares each individual number in the first number against all the numbers in the second number.
Depending on the counters, it evaluates if equal numbers have been matched in the same positions, using modulo %10 to compare each number.
This is the code:
#include <stdio.h>
#include <stdlib.h>
int
main(void) {
int num1 = 1653;
int num2 = 5243;
int pos1, pos2, hit = 0, miss = 0, i, j;
pos1 = 0;
for (i = num1; i > 0; i /= 10) {
pos2 = 0;
for (j = num2; j > 0; j /= 10) {
if (i % 10 == j % 10) {
if (pos1 == pos2) {
hit++;
} else {
miss++;
}
}
pos2++;
}
pos1++;
}
printf("hits = %d\n", hit);
printf("misses = %d\n", miss);
return 0;
}
Even though this question has been asked a million times I just haven't found an answer that actually helps my case, or I simply can't see the solution.
I've been given the task to make a program that takes in a whole number and counts how many times each digit appears in it and also not showing the same information twice. Since we're working with arrays currently I had to do it with arrays of course so since my code is messy due to my lack of knowledge in C I'll try to explain my thought process along with giving you the code.
After entering a number, I took each digit by dividing the number by 10 and putting those digits into an array, then (since the array is reversed) I reversed the reverse array to get it to look nicer (even though it isn't required). After that, I have a bunch of disgusting for loops in which I try to loop through the whole array while comparing the first element to all the elements again, so for each element of the array, I compare it to each element of the array again. I also add the checked element to a new array after each check so I can primarily check if the element has been compared before so I don't have to do the whole thing again but that's where my problem is. I've tried a ton of manipulations with continue or goto but I just can't find the solution. So I just used **EDIT: return 0 ** to see if my idea was good in the first place and to me it seems that it is , I just lack the knowledge to go back to the top of the for loop. Help me please?
// With return 0 the program stops completely after trying to check the digit 1 since it's been checked already. I want it to continue checking the other ones but with many versions of putting continue, it just didn't do the job. //
/// Tried to make the code look better. ///
#include <stdio.h>
#define MAX 100
int main()
{
int a[MAX];
int b[MAX];
int c[MAX];
int n;
int i;
int j;
int k;
int counter1;
int counter2;
printf("Enter a whole number: ");
scanf("%i",&n);
while (1)
{
for (i=0,counter1=0;n>10;i++)
{
a[i] = n%10;
n=n/10;
counter1+=1;
if (n<10)
a[counter1] = n;
}
break;
}
printf("\nNumber o elements in the array: %i", counter1);
printf("\nElements of the array a:");
for (i=0;i<=counter1;i++)
{
printf("%i ",a[i]);
}
printf("\nElements of the array b:");
for (i=counter1,j=0;i>=0;i--,j++)
{
b[j] = a[i];
}
for (i=0;i<=counter1;i++)
{
printf("%i ",b[i]);
}
for (i=0;i<=counter1;i++)
{
for(k=0;k<=counter1;k++)
{
if(b[i]==c[k])
{
return 0;
}
}
for(j=0,counter2=0; j<=counter1;j++)
{
if (b[j] == b[i])
{
counter2+=1;
}
}
printf("\nThe number %i appears %i time(s)", b[i], counter2);
c[i]=b[i];
}
}
The task at hand is very straightforward and certainly doesn't need convoluted constructions, let alone goto.
Your idea to place the digits in an array is good, but you increment counter too early. (Remember that arrays in C start with index 0.) So let's fix that:
int n = 1144526; // example number, assumed to be positive
int digits[12]; // array of digits
int ndigit = 0;
while (n) {
digits[ndigit++] = n % 10;
n /= 10;
}
(The ++ after ndigit will increment ndigit after using its value. Using it as array index inside square brackets is very common in C.)
We just want to count the digits, so reversing the array really isn't necessary. Now we want to count all digits. We could do that by counting all digits when we see then for the first time, e.g. in 337223, count all 3s first, then all 7s and then all 2s, but that will get complicated quickly. It's much easier to count all 10 digits:
int i, d;
for (d = 0; d < 10; d++) {
int count = 0;
for (i = 0; i < ndigit; i++) {
if (digit[i] == d) count++;
}
if (count) printf("%d occurs %d times.\n", d, count);
}
The outer loop goes over all ten digits. The inner loop counts all occurrences of d in the digit array. If the count is positive, write it out.
If you think about it, you can do better. The digits can only have values from 0 to 9. We can keep an array of counts for each digit and pass the digit array once, counting the digits as you go:
int count[10] = {0};
for (i = 0; i < ndigit; i++) {
count[digit[i]]++;
}
for (i = 0; i < 10; i++) {
if (count[i]) printf("%d occurs %d times.\n", i, count[i]);
}
(Remember that = {0} sets the first element of count explicitly to zero and the rest of the elements implicitly, so that you start off with an array of ten zeroes.)
If you think about it, you don't even need the array digit; you can count the digits right away:
int count[10] = {0};
while (n) {
count[n % 10]++;
n /= 10;
}
for (i = 0; i < 10; i++) {
if (count[i]) printf("%d occurs %d times.\n", i, count[i]);
}
Lastly, a word of advice: If you find yourself reaching for exceptional tools to rescue complicated code for a simple task, take a step back and try to simplify the problem. I have the impression that you have added more complicated you even you don't really understand instead.
For example, your method to count the digits is very confused. For example, what is the array c for? You read from it before writing sensible values to it. Try to implement a very simple solution, don't try to be clever at first and go for a simple solution. Even if that's not what you as a human would do, remeber that computers are good at carrying out stupid tasks fast.
I think what you need is a "continue" instead of a return 0.
for (i=0;i<=counter1;i++) {
for(k=0;k<=counter1;k++) {
if(b[i]==c[k]) {
continue; /* formerly return 0; */
}
for(j=0,counter2=0; j<=counter1;j++)
if (b[j] == b[i]){
counter2+=1;
}
}
Please try and see if this program can help you.
#include <stdio.h>
int main() {
unsigned n;
int arr[30];
printf("Enter a whole number: ");
scanf("%i", &n);
int f = 0;
while(n)
{
int b = n % 10;
arr[f] = b;
n /= 10;
++f;
}
for(int i=0;i<f;i++){
int count=1;
for(int j=i+1;j<=f-1;j++){
if(arr[i]==arr[j] && arr[i]!='\0'){
count++;
arr[j]='\0';
}
}
if(arr[i]!='\0'){
printf("%d is %d times.\n",arr[i],count);
}
}
}
Test
Enter a whole number: 12234445
5 is 1 times.
4 is 3 times.
3 is 1 times.
2 is 2 times.
1 is 1 times.
Here is another offering that uses only one loop to analyse the input. I made other changes which are commented.
#include <stdio.h>
int main(void)
{
int count[10] = { 0 };
int n;
int digit;
int elems = 0;
int diff = 0;
printf("Enter a whole number: ");
if(scanf("%d", &n) != 1 || n < 0) { // used %d, %i can accept octal input
puts("Please enter a positive number"); // always check result of scanf
return 1;
}
do {
elems++; // number of digits entered
digit = n % 10;
if(count[digit] == 0) { // number of different digits
diff++;
}
count[digit]++; // count occurrence of each
n /= 10;
} while(n); // do-while ensures a lone 0 works
printf("Number of digits entered: %d\n", elems);
printf("Number of different digits: %d\n", diff);
printf("Occurrence:\n");
for(n = 0; n < 10; n++) {
if(count[n]) {
printf(" %d of %d\n", count[n], n);
}
}
return 0;
}
Program session:
Enter a whole number: 82773712
Number of digits entered: 8
Number of different digits: 5
Occurrence:
1 of 1
2 of 2
1 of 3
3 of 7
1 of 8