Program C counting total number of odd digits - c

I'm writing a C program that counts the number of odd digits from user input.
Eg.
Please enter the number: 12345
countOddDigits(): 3
int countOddDigits(int num);
int main()
{
int number;
printf("Please enter the number: \n");
scanf("%d", &number);
printf("countOddDigits(): %d\n", countOddDigits(number));
return 0;
}
int countOddDigits(int num)
{
int result = 0, n;
while(num != 0){
n = num % 10;
if(n % 2 != 0){
result++;
}
n /= 10;
}
return result;
}
The code is not working.
Can someone tell me where does it go wrong?

There were a few mistakes in your code. Here is a working version of your code:
#include <stdio.h>
int countOddDigits(int n);
int main()
{
int number;
printf("Please enter the number: \n");
scanf("%d", &number);
printf("countOddDigits(): %d\n", countOddDigits(number));
return 0;
}
int countOddDigits(int n)
{
int result = 0;
while(n != 0){
if(n % 2 != 0)
result++;
n /= 10;
}
return result;
}
You are mixing n and num together - there is no need for two variables.
n%=10 is just causing mistakes - you need to check the last digit if(n%2!=0) and then move to the next one n/=10, that's all.

Looping variable is not correct. Your outer loop is
while (num !=0)
but the num variable is never decremented; the final statement decrements the n variable. My guess is you want to initialize
int n = num;
while (n != 0 )
{ ...
n/= 10;
}

Related

Why does my code calling a function twice?

#include <stdio.h>
int bolucu(int n){
int temp;
temp=n;
int basamak=0;
while(temp != 0){
temp/=10;
++basamak;
}
int digits = 0;
int m = n;
while (m) {
digits++;
m /= 10;
}
digits /= 2;
int tmp = 0, lower_half = 0;
while (digits--) {
tmp *= 10;
tmp += n % 10;
n /= 10;
}
while (tmp) {
lower_half *= 10;
lower_half += tmp % 10;
tmp /= 10;
}
if (basamak % 2==1){
n/=10;
}
int a;
int b;
a = n;
b=lower_half;
printf("%d %d\n",a,b);
int loopTemp;
for(int i=0;i<10;i++){
a=3*a+2;
b=2*b+3;
if(a>b){
temp=a;
a=b;
b=temp;
}
if(a==b){
printf("Congratulations you caught one!!!\n");
return 1;
break;
}
}
if(a!=b){
printf("10 tries were not enough!\n");
return 2;
}
}
int main()
{
int number;
printf("\nEnter a number with at least two digits: ");
scanf("%d",&number);
bolucu(number);
while(bolucu(number) != 1){
printf("\nEnter a new number: ");
scanf("%d",&number);
printf("%d",bolucu(number));
}
return 0;
}
e.g:
This is terminal screen.
As you can see there is a second one. First one is true but i don't want second one.
How can i get rid of the second calling?
(Also sorry for bad code writing, i'm new)
What im missing here?
And i cant use any library other than stdio.(Like math.h)
The reason for this is because you call the bolucu() function both inside the while loop and in it's condition check. To fix this, call the function and hold it's result in a variable once, and then use that single result in both the check and your print statement. Your main function can be rewritten like so:
int main()
{
int number;
printf("\nEnter a number with at least two digits: ");
scanf("%d", &number);
int result = bolucu(number);
while (result != 1)
{
printf("\nEnter a new number: ");
scanf("%d", &number);
result = bolucu(number);
printf("%d", result);
}
return 0;
}

Find two largest numbers in input

I need to make a program that will perform the following task:
Enter N natural numbers. Complete the input with 0. Output the number
of the maximal number.
I have already done this, and you can see the code below:
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
int main(void) {
int i = 0, num, max_place = -1;
int max = -2147483647;
printf("Start enter numbers, bruh (please end input with 0):\n");
scanf("%d", &num);
while (num != 0) {
if (num >= max) {
max = num;
max_place = i;
}
i++;
scanf("%d", &num);
}
if (max_place == -1) printf("Numbers were not entered");
else printf("\nMax number was on %d place, bruh", max_place + 1);
return 0;
}
The teacher then made the task more difficult – the program needs to print the maximum number and the next maximum after it of the entered numbers.
How can I do it?
If you can use arrays and sort use that way. if not, this is in your code
int main(void) {
int i = 0, num, max_place = -1, second_max_place = -1;
int max = -2147483647;
int second_max = -2147483647;
printf("Start enter numbers, bruh (please end input with 0):\n");
scanf("%d", &num);
while (num != 0) {
if (num == 0) break;
if (num >= max) {
second_max = max;
second_max_place = max_place;
max = num;
max_place = i;
}
if(num < max && num >= second_max){
second_max = num;
second_max_place = i;
}
i++;
scanf("%d", &num);
}
if (max_place == -1) printf("Numbers were not entered");
else{
printf("\nMax number was on %d place, bruh", max_place + 1);
printf("\nSecond Max number was on %d place, bruh", second_max_place + 1);
}
return 0;
}

School exercice in c about prime numbers

Make a program that asks the user for an integer and says if the
number is prime or not. A number greater than 1 is prime if only
is divisible by 1 and by itself. Then, it will tell us what the prime number is.
example:
Enter a number: 8
8 is not first. The first one immediately superior to 8 is 11.
Enter a number: 5
5 is first. The first one immediately above 5 is 7.
I can only solve first part.
Here is my code:
#include <stdio.h>
int main() {
int num, i;
do {
printf("Enter a numer: ");
scanf("%d", & num);
}
while (num < 1);
for (i = 2; i < num; i++) {
if (num % i == 0)
printf("Its prime");
}
if (num % 1 == 0 && num % num == 0)
printf("Not prime");
return 0;
}
Try this logic. Not tested
#include <stdio.h>
int main()
{
int num, i, flag = 0;
printf("Enter a positive integer: ");
scanf("%d", &num);
int isPrime=IsPrime(num)
if(isPrime==0){
numNext=num+1;
int nextPrimeNum=checkNextPrime(numNext);
}
}
int IsPrime(int num){
for(i = 2; i <= num/2; ++i)
{
// condition for nonprime number
if(num%i == 0)
{
flag = 1;
break;
}
}
if (num == 1)
{
flag=1;//neither prime nor composite
}
return flag;
}
int checkNextPrime(int numNext){
int isNextPrime=IsPrime(numNext)
if(isNextPrime==0){
printf("This is the required output :"numNext);
return numNext;
}
else{
numNext=numNext+1;
checkNextPrime(int numNext)
}
}

Using arrays in c to find repeated digits in an number (C)

I need to make a program that checks to see if an entered value has any repeated digits. The user is asked to enter numbers until the entered value is 0. If there are any repeated digits, it displays "repeated digits" and then asks the user to enter another value. If there are no repeated digits, it displays "no repeated digits" and asks the user to enter another number. So far, this is what i have. It terminates the program when 0 is entered, but it always displays "no repeated digits" even if there are some.
#include <stdbool.h>
#include <stdio.h>
int main(void)
{
bool digit_seen[10] = {false};
int digit;
long int n = 0;
printf("Enter a number: ");
scanf("%ld", &n);
while(n >= 0){
if(n==0)
break;
while (n > 0){
digit = n % 10;
if (digit_seen[digit]){
digit_seen[digit] = true;
break;
}
n /= 10;
}
if (n > 0)
printf("Repeated digit: %d\n", digit);
else
printf("No repeated digit\n");
scanf("%ld", &n);
}
return 0;
}
A couple of things:
1: A bool only has two states: true and false. If you trying to build a frequency counter of each digit seen, for the presence of a digit more than once, then you should use a data type that can count to at least two, like a char or short or int, or your own enum.
2: This code:
if (digit_seen[digit]){
digit_seen[digit] = true;
break;
}
Is never going to be evaluated as true since you initialized digit_seen to be false at the start of your main function. What you should be doing is something like this:
#include <stdio.h>
int main(int argc, char *argv[])
{
int digit_seen[10] = {0};
int entry;
int i, flag = 0;
printf("Enter a number: ");
scanf("%ld", &entry);
while(entry > 0)
{
int digit = (entry%10);
digit_seen[digit]++;
if(digit_seen[digit]>=2)
{
printf("Repeated digit: %d\n", digit);
}
entry /= 10;
}
for(i = 0; i < 10; i++)
{
if(digit_seen[i]>1) flag=1;
}
if(!flag)
{
printf("No repeated digits\n");
}
return 0;
}
#include <stdio.h>
int main() {
int seen [10] ={0}; // we set every element for a number is just 0
int N,rem;
printf("Enter the number:");
scanf("%d", &N);
while(N>0){
rem = N%10;
seen[rem]+=1;
N = N/10;
}
int i;
for(i=0;i<10;i++){ // checking the number seen counts
if(seen[i]==0){
continue;
}
printf("%d seen %d times\n",i,seen[i]); // just returned the given numbers informations
}
return 0;
}

C program to find if a number is palindrome or not

I made a C program to check if a number is palindrome or not. I used the following code, but it shows numbers like 12321 as non palindrome. Can you please explain me the mistake in the program below?
#include <stdio.h>
int main()
{
int i, x, n, c, j;
int d=0;
printf ("enter total digits in number: ");
scanf ("%d", &i);
printf ("\nenter number: ");
scanf ("%d", &n);
j=n;
for (x=1; x<=i; x++)
{
c= j%10;
d=c*(10^(i-x))+d;
j=(j-c)/10;
}
if (d==n)
{
printf ("\npalindrome");
}
else
{
printf ("\nnon palindrome");
}
return 0;
}
^ is the xor operator.
In order to raise power, you need to include math.h and call pow
d = (c * pow(10, i - x)) + d;
this algorithm is as simple as human thinking, and it works
#include <stdio.h>
int main() {
int i=0,n,ok=1;
char buff[20];
printf("Enter an integer: ");
scanf("%d", &n); // i am ommiting error checking
n=sprintf(buff,"%d",n); //convert it to string, and getting the len in result
if(n<2) return 0;
i=n/2;
n--;
while(i && ok) {
i--;
//printf("%c == %c %s\n", buff[i],buff[n-i],(buff[i]==buff[n-i])?"true":"false");
ok &= (buff[i]==buff[n-i]);
}
printf("%s is %spalindrome\n",buff, ok?"":"not ");
return 0;
}
// Yet another way to check for palindrome.
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char **argv)
{
int n, rn, tn;
printf("Enter an integer: ");
scanf("%d", &n);
// reverse the number by repeatedly extracting last digit, add to the
// previously computed partial reverse times 10, and keep dropping
// last digit by dividing by 10
for (rn = 0, tn = n; tn; tn /= 10) rn = rn * 10 + tn % 10;
if (rn == n) printf("%d is palindrome\n", n);
else printf("%d is not palindrome\n", n);
}
A loop like this might do:
int src; // user input
int n; // no of digits
int res = 0;
int tmp; // copy of src
// .... read the input: n and src ....
tmp = src;
for(int i = 0; i < n; i ++)
{
int digit = tmp % 10; // extract the rightmost digit
tmp /= 10; // and remove it from source
res = 10*res + digit; // apend it to the result
}
// ...and test if(res == src)...

Resources