I cannot seem to find my error .
If I were to key in 28212 or just any number that contain multiple digits , it would give me the output "No repeated digits".
// Check numbers for repeated digits
#include <stdio.h>
#include <stdbool.h>
int main (void)
{
bool digit_seen[10] = {false} ;
int digit ;
long n;
printf("Enter a number: ");
scanf("%1d", &n);
while(n > 0){
digit = n % 10 ;
if (digit_seen[digit])
break;
digit_seen[digit] = true;
n /= 10;
}
if (n > 0)
printf("Repeated digits\n");
else
printf("No repeated digits\n");
return 0;
}
You have used %1d which is only reading one character from the integer input. Since the input has only one character it is always printing No repeated digits
Since n is long, you should use %ld.
Related
Write a program to input 10 numbers and find the largest positive even number .In case no such number is present
a relevant message should be printed.
I'm stuck while writing the code for printing the relevant message (if there is no such number present). What to do ?
Here's how I tried...
#include <stdio.h>
int main(){
int a,lar;
printf("Enter the first number");
scanf("%d",&a);
lar = a;
for (int i = 2; i <=10; i++)
{
printf("Number %d : ",i);
scanf("%d",&a);
if (a > 0 && (a % 2) == 0)
{
if (a > lar)
{
lar = a;
}
}
}
printf("The largest positive even number is : %d",lar);
}
Also the code is returning wrong value if I enter all odd numbers.
You can use "lar" as a flag showing whether there was a positive even number in the input or not. To do that, you set it first to a negative number (e.g. -1). If there is even a single positive even number in the input, that number will overwrite lar and it won't be negative at the end of the loop.
#include <stdio.h>
int main(){
int a,lar;
lar = -1;
for (int i = 1; i <=10; i++)
{
printf("Number %d : ",i);
scanf("%d",&a);
if (a > 0 && (a % 2) == 0)
{
if (a > lar)
{
lar = a;
}
}
}
if(lar > 0)
{
printf("The largest positive even number is : %d",lar);
}
else
{
print("There are no positive even numbers.")
}
}
I'm writing a program that takes in your student number(8 digits long), prints each digit on its own new line, and then gets the sum of all the digits in the number
(E.g. Student Number - 20305324, Sum - 19)
#include <stdio.h>
#include <string.h>
int main(void) {
char student_number[8];
int i = 0;
int sum = 0;
printf("Enter your student number: ");
scanf("%s", student_number);
// ensures input is only 8 digits - WORKS
while (strlen(student_number) < 8 || strlen(student_number) > 8){
printf("Enter your student number: ");
scanf("%s", student_number);
}
// prints each digit of the student number on a new line - WORKS
while (student_number[i] != '\0'){
printf("%c\n", student_number[i]);
i++;
}
// sum all the digits in the student number and print - DOESN'T WORK
for (i=0;i<8;i++){
sum = sum + student_number[i];
printf("%d\n", sum);
}
printf("Sum of the numbers is %d", sum);
}
OUTPUT
The problem I'm encountering is when my for loop attempts to add each digit in the student number. The output I expect here is 19, but for some reason the sum evaluates to some bizarre number like 403
}
Would someone mind pointing out where exactly the fault in my for loop is or if it is elsewhere? Thanks :)
Firstly, your array char student_number[8]; cannot hold 8-character string because there are no room for terminating null character. You must allocate one more element.
Then, you should convert the characters to corresponding numbers. Character codes for digits are defined to be continuous, so this can be done by subtracting '0' from the character code.
Also you should set a limit of length of string to read via scanf() to avoid buffer overrun. One more good practice is checking the return values of scanf() to see if something is successfully read.
Fixed code:
#include <stdio.h>
#include <string.h>
int main(void) {
char student_number[10]; // *** allocate enough elements (one more than needed to catch too long input)
int i = 0;
int sum = 0;
printf("Enter your student number: ");
if(scanf("%9s", student_number) != 1){ // *** limit the length to read and check the result
fputs("read error\n", stderr);
return 1;
}
// ensures input is only 8 digits - WORKS
while (strlen(student_number) < 8 || strlen(student_number) > 8){
printf("Enter your student number: ");
if(scanf("%9s", student_number) != 1){ // *** limit the length to read and check the result
fputs("read error\n", stderr);
return 1;
}
}
// prints each digit of the student number on a new line - WORKS
while (student_number[i] != '\0'){
printf("%c\n", student_number[i]);
i++;
}
// sum all the digits in the student number and print -DOESN'T WORK
for (i=0;i<8;i++){
sum = sum + (student_number[i] - '0'); // *** convert characters to numbers before adding
printf("%d\n", sum);
}
printf("Sum of the numbers is %d", sum);
}
When you read characters as a string, the values of the char objects are codes for the characters. Your C implementation is likely using ASCII codes, in which 48 is the code for “0”, 49 is the code for “1”, 65 is the code for “A”, and so on.
To convert a code x for a digit to the value of the digit, use x - '0'.
I think that the task was to read the number not the string.
void printDigitsAndSum(unsigned number)
{
unsigned mask = 1;
unsigned sum = 0;
while(number / (mask * 10)) mask *= 10;
while(mask)
{
printf("%u\n", number / mask);
sum += number / mask;
number %= mask;
mask /= 10;
}
printf("Sum: %u\n", sum);
}
int main(void)
{
unsigned number;
if(scanf("%u", &number) == 1)
printDigitsAndSum(number);
else printf("Wrong number\n");
}
https://godbolt.org/z/1edceh
This is my code:`
#include <stdio.h>
void main() {
int n;
int count = 0;
printf("Enter an integer: ");
scanf("%d", &n);
// iterate until n becomes 0
// remove last digit from n in each iteration
// increase count by 1 in each iteration
while (n != 0) {
n /= 10; // n = n/10
++count;
}
printf("Number of digits: %lld", count);
}
I am able to run the code finely but when I enter 15 or 16 digits of number as input then it always shows me that the number of digits is 10. And another problem with this code is that suppose if I input 000 then I want the output to be 3 digits but this code is not able to do that as the condition in the while loop becomes instantly false. So how write a code that enables me to take upto 100 or 1000 digits as input and also enables me to input 0s as well.
Note: This program should be solved using a loop and in C language
I found a answer to the question here in stackoverflow written in c++ that I couldn't even understand as I am a beginner and I am learning C.
Link to the answer:
How can I count the number of digits in a number up to 1000 digits in C/C++
Instead of reading a number, read a string and count the digits:
#include <stdio.h>
int main() {
char buffer[10000];
int n;
printf("Enter an integer: ");
if (scanf("%9999s", buffer) == 1) {
for (n = 0; buffer[n] >= '0' && buffer[n] <= '9'; n++)
continue;
printf("Number of digits: %d\n", n);
}
return 0;
}
You can also use the scanf() scanset feature to perform the test in one step:
#include <stdio.h>
int main() {
char buffer[10000];
int n;
printf("Enter an integer: ");
if (scanf("%9999[0-9]%n", buffer, &n) == 1) {
printf("Number of digits: %d\n", n);
}
return 0;
}
32 bits signed integer has the max value equals 2,147,483,647 so, if you input a bigger one, it will not be stored. I'd make it receiving a string and get its length, like so:
#include <stdio.h>
#include <string.h>
int len = strlen("123123123123132");
You are taking int variable and you are trying to count a number like whose digit is 100 or 1000. it will not fit with int. so take input as a string and count the length of string.
I'm self-studying C and I'm trying to make 2 programs for exercise:
the first one takes a number and check if it is even or odd;
This is what I came up with for the first one:
#include <stdio.h>
int main(){
int n;
printf("Enter a number that you want to check: ");
scanf("%d",&n);
if((n%2)==0)
printf("%d is even.",n);
else
printf("%d is odd.",n);
return 0;
}
the second one should take n numbers as input and count the number of even numbers, odd numbers, and zeros among the numbers that were entered. The output should be the number of even numbers, odd numbers, and zeros.
I would like to ask how to implement the loop in this case: how can I set an EOF value if every integer is acceptable (and so I cannot, say, put 0 to end)? Can you show me how to efficiently build this short code?
#include <stdio.h>
int main(void) {
int n, nEven=0, nOdd=0, nZero=0;
for (;;) {
printf("\nEnter a number that you want to check: ");
//Pressing any non-numeric character will break;
if (scanf("%d", &n) != 1) break;
if (n == 0) {
nZero++;
}
else {
if (n % 2) {
nEven++;
}
else {
nOdd++;
}
}
}
printf("There were %d even, %d odd, and %d zero values.", nEven, nOdd, nZero);
return 0;
}
Check the return value of scanf()
1, 1 field was filled (n).
0, 0 fields filled, likely somehtlig like "abc" was entered for a number.
EOF, End-of-file encountered (or rarely IO error).
#include <stdio.h>
int main(void) {
int n;
for (;;) {
printf("Enter a number that you want to check: ");
if (scanf("%d",&n) != 1) break;
if((n%2)==0)
printf("%d is even.",n);
else
printf("%d is odd.",n);
}
return 0;
}
Or read the count of numbers to subsequently read:
int main(void) {
int n;
printf("Enter the count of numbers that you want to check: ");
if (scanf("%d",&n) != 1) Handle_Error();
while (n > 0) {
n--;
printf("Enter a number that you want to check: ");
int i;
if (scanf("%d",&i) != 1) break;
if((i%2)==0) {
if (i == 0) printf("%d is zero.\n",i);
else printf("%d is even and not 0.\n",i);
}
else
printf("%d is odd.\n",i);
}
return 0;
}
hey look at this
#include<stdio.h>
#include<conio.h>
void main()
{
int nodd,neven,num,digit ;
clrscr();
printf("Count number of odd and even digits in a given integer number ");
scanf("%d",&num);
nodd = neven =0; /* count of odd and even digits */
while (num> 0)
{
digit = num % 10; /* separate LS digit from number */
if (digit % 2 == 1)
nodd++;
else neven++;
num /= 10; /* remove LS digit from num */
}
printf("Odd digits : %d Even digits: %d\n", nodd, neven);
getch();
}
You can do something like this:
#include <stdio.h>
int main(){
int n,evenN=0,oddN=0,zeros=0;
char key;
do{
clrscr();
printf("Enter a number that you want to check: ");
scanf("%d",&n);
if(n==0){
printf("%d is zero.",n);
zeros++;
}
else if((n%2)==0){
printf("%d is even.",n);
evenN++;
}
else{
printf("%d is odd.",n);
oddN++;
}
puts("Press ENTER to enter another number. ESC to exit");
do{
key = getch();
}while(key!=13 || key!=27) //13 is the ascii code fore enter key, and 27 is for escape key
}while(key!=27)
clrscr();
printf("Total even numbers: %d",evenN);
printf("Total odd numbers: %d",oddN);
printf("Total odd numbers: %d",zeros);
return 0;
}
This program ask for a number, evaluate the number and then ask to continue for another number or exit.
I've created a program to determine largest number, but my lecturer says it isn't perfect, can anybody make it perfect?
#include <stdio.h>
int main () {
double a,b=0,n, i;
printf("limit of n input: ");
scanf ("%lf",&n);
for (i=1;i<=n;i++) {
scanf("%lf",&a);
if (a>b) b=a;
}
printf("%.2lf", b);
return 0;
}
If by "not perfect" she meant "doesn't properly handle negative numbers or an empty set", then you'd want to
Treat n<1 as a special case (why should 0 be the largest of an empty set?)
Read the first number outside of the loop, so you're not making as assumption as to the smallest possible number
I would do it that way, sorry for the mass of text. I think it is coming from the typical Objective-C style programming with long words:
#include <stdio.h>
int clean_stdin() {
while (getchar()!='\n');
return 1;
}
int main(int argc, char *argv[]) {
char c;
signed int count = 0; // number of numbers to scan
unsigned int fireErrorMessage = 0;
do {
if (fireErrorMessage == 1) {
printf("You entered not a positive natural number. Please enter a number >0 Examples: 1 22 4012\n"); // output for the user
}
if (fireErrorMessage == 0) {
fireErrorMessage = 1;
}
printf("How many integers do you want to insert (Inser a number >0)? ");
} while (((scanf("%d%c", &count, &c) != 2 || c != '\n') && clean_stdin()) || count < 1);
signed int indexOfNumber; // for index, declared outside because of output at the end
signed int highestNumberIndex;
double highestNumber; // saving the highest value in a helper variable
fireErrorMessage = 0;
for (indexOfNumber = 1; indexOfNumber <= count; indexOfNumber++) {
double scannedNumber;
do {
if (fireErrorMessage == 1) {
printf("You entered not a number. Please enter a number. Examples: 3.0 -1 14\n"); // output for the user
}
if (fireErrorMessage == 0) {
fireErrorMessage = 1;
}
printf("Input number %d: ", indexOfNumber); // output for the user
} while (((scanf("%lf%c", &scannedNumber, &c) != 2 || c != '\n') && clean_stdin()));
fireErrorMessage = 0;
if (indexOfNumber == 1 || scannedNumber > highestNumber) {
highestNumberIndex = indexOfNumber;
highestNumber = scannedNumber;
}
}
printf("Highest input number on index %d, the value is about %.2lf\n", highestNumberIndex, highestNumber);
return 0;
}
Output
How many integers do you want to insert (Inser a number >0)? aa5
You entered not a positive natural number. Please enter a number >0 Examples: 1 22 4012
How many integers do you want to insert (Inser a number >0)? -3
You entered not a positive natural number. Please enter a number >0 Examples: 1 22 4012
How many integers do you want to insert (Inser a number >0)? 3
Input number 1: aa
You entered not a number. Please enter a number. Examples: 3.0 -1 14
Input number 1: -50.0001
Input number 2: 51a
You entered not a number. Please enter a number. Examples: 3.0 -1 14
Input number 2: -1.00
Input number 3: -0.1
Highest input number on index 3, the value is about -0.10
This code caters for negative, not a number input for the loop index as well as negative and not a number inputs inside the loop. Thanks
#include <stdio.h>
#include <math.h>
int main () {
int n, i;
double a,b=0;
printf("limit of n input: ");
scanf ("%lf",&n);
if(n < 0){
printf("value of n cannot be negative");
return 0;
}
else if (n == 0)
return 0;
else if (isnan(n))
return 0;
else{
for (i=1;i<=n;i++)
{
scanf("%lf",&a);
if(!isnan(a) && a > 0)
{
if (a>b) b=a;
}
}
printf("%.2lf", b);
return 0;
}
}