So my task is to make an array that accepts 10 characters. If the characters entered by the user are greater than 10, then an error is dispayed. If the 10 characters entered contain a letter, it displays another error.
Therefore, the array can only have 10 numbers and nothing else, if the numbers entered are less or more than 10, error is displayed as well as if there are letters in the array.
My code accepts both numbers and letters, as i cannot figure out how to display error when letters are entered.
void getTenDigitPhone(char telNum[])
{
int i;
int z = 1;
do
{
scanf("%s", telNum);
if (strlen(telNum) != 10)
{
printf("Enter a 10-digit phone number: ");
z = 1;
}
else if (strlen(telNum) == 10)
{
return telNum;
}
} while (z == 1);
}
You just need to check that telNum contains only digits:
for (int i = 0; i < 10; i++)
if (!isdigit(telNum[i])) {
// handle error because a non-digit was found.
}
I'm not going to do your homework for you but this should give you the idea.
You can use the function isdigit(x).
This returns true (non-zero) if x is a digit and returns false (zero) if not.
You have to check digit by digit.
I'm going to give an answer because you have posted your current code as your effort. As other answers you should use isdigit(x) function.
...
else if (strlen(telNum) == 10)
{
int i;
char err = 0;
for (i = 0; i < 10; i++) {
if (!isdigit(telNum[i])) {
// Your error here
printf("Non-digit character found");
err = 1;
break;
}
}
if (err == 0) {
return telNum;
}
}
...
Related
Write a program that takes an integer and
prints the number of trailing zeroes.
Example:
Enter the number: 24100
Trailing zeroes: 2
I have no Idea what condition to create to determine the number of zeroes in a number.
You could use a modulus trick here:
int input = 24100;
int num_zeroes = 0;
while (input > 0) {
if (input % 10 == 0) {
num_zeroes = num_zeroes + 1;
input = input / 10;
}
else {
break;
}
}
printf("%d", num_zeroes); // 2
int cnt = 0; while( input && input % 10 == 0) cnt++, input /= 10;
But, if you think you'll have access to SO when you are sitting the proficiency tests when applying for jobs, you are sadly mistaken.
Expected Input:
123456
Expected output:
12
34
56
I have tried in this way
#include <stdio.h>
int main()
{
// Put variables for further proceed
int number, remainder, quotient = 1, numberUpdate, count = 0;
int value = 10000;
/*This for input validation. User can give exactly 6 digit as an input If
user breaks this condition then It
will redirect back to take input again */
while (1)
{
int countUpdate = 0;
int quotientUpdate = 1;
printf("Enter a number: ");
scanf("%d", &number);
numberUpdate = number;
while (quotientUpdate != 0)
{
quotientUpdate = numberUpdate / 10;
numberUpdate = quotientUpdate;
countUpdate++;
}
if (countUpdate > 6 || countUpdate < 6)
{
printf("It allows exactly 6 digits\n");
}
else
{
break;
}
}
//This for finding the pair of two consecutive digits.
while (quotient != 0)
{
count++;
if (count == 4)
{
break;
}
quotient = number / value;
remainder = number % value;
if (count != 2)
printf("%d\n", quotient);
if (count == 1)
{
number = remainder;
}
else
{
number = quotient;
}
if (count == 1)
{
value = value / 1000;
}
if (count == 3)
{
remainder = remainder * 10 + 6;
printf("%d\n", remainder);
}
}
return 0;
}
My problem is: I have made this for the exact input 6 digits. From my code, I did not get the expected output. Output comes from my code like:
If a user gives an input 987654
Output shows:
98
76
56
But my expectation is:
98
76
54
Here is another problem: this code does not work for less than 6 or greater than 6 digits. But I want to solve this problem for any number of digit.
Can you help me identifying and solving my problem?
Your solution is a bit overcomplicated.
If you want to use integers, you could do it like this (untested).
Depending on range for your number, you might change to long long.
#include <stdio.h>
int main(void)
{
int number;
int digits = 1;
while (digits & 1)
{ // loop until we get an even number
printf("Enter a number: ");
int ret = scanf("%d", &number);
if (ret != 1)
continue;
// count number of digits
digits = 0;
while (number != 0)
{
number /= 10;
digits++;
}
if (digits & 1)
printf("Please enter even number of digits.\n");
}
// If we are here, we have 2, 4, 6, ... digits
// Calculate divider to chop first 2 digits
int divider = 1;
while (digits > 2)
{
divider *= 100;
digits -= 2;
}
// chop remaining digits and print 2 of them
while (divider)
{
pair = (number / divider) % 100;
printf("%d\n", pair);
divider /= 100;
}
return 0;
}
Another option would be to use strings instead of numbers and then simply print 2 characters per line.
I've updated your code a bit, it should be working and handle the "0" digit within the code. For the "0" digit at the beginning of the code, you should input a string and not a number.
#include <stdio.h>
int main()
{
// Put variables for further proceed
int number, remainder, quotient = 1, numberUpdate, count = 0;
int countUpdate = 0;
int value = 10000;
/*This for input validation. User can give exactly 6 digit as an input If
user breaks this condition then It
will redirect back to take input again */
while (1)
{
int quotientUpdate = 1;
printf("Enter a number: ");
scanf("%d", &number);
numberUpdate = number;
while (quotientUpdate != 0)
{
quotientUpdate = numberUpdate / 10;
numberUpdate = quotientUpdate;
countUpdate++;
}
if (countUpdate < 2 || countUpdate % 2 != 0)
{
printf("Even number of digits only\n");
}
else
{
break;
}
}
count = countUpdate / 2;
numberUpdate = number;
int d[count];
for (int i = 0; i < count; i++)
{
d[i] = numberUpdate % 100;
numberUpdate /= 100;
}
for (int i = count - 1; i >= 0; i--)
{
if (d[i] < 10) printf("0");
printf("%d\n", d[i]);
}
return 0;
}
Before proposing my solution, I'll try to explain what's wrong in your code.
Analysis of the original code
First of all, since you have currently the fixed length limitation, your loop that checks if the number has exactly 6 digits can be omitted; the same check can be performed just checking the range:
if (number < 1000000 || number > 999999)
{
printf("It allows exactly 6 digits\n");
}
else
{
break;
}
The core of your logic is in the loop while (quotient != 0). It contains a lot of strange attempts you perform in order to compensate the previous mistake. It leads to the final reminder with a single digit instead of two, so you try to compensate it with this line
remainder = remainder * 10 + 6;
this obviously works only if the last digit is 6.
The root of the problem is in this row:
if (count == 1)
{
value = value / 1000;
}
But why 1000? value represents the divider in the next loop, so you want it to obtain a reminder with two digit less (instead of 3), so the correct division is value = value / 100;.
All the subsequent correction come after this one. The other answers widely cover the correct solution storing the input within an integer.
A solution involving strings
Since you need a solution with any number of digits, you must be aware that using an int you'll be able to manage at most 10 digits (because the maximum value of an integer is INT_MAX (2147483647).
Using an integer you'll only be limited by the size of the string buffer you choose.
That's the code. Our only limitation is forcing the user to insert only an even number of digits:
#include <stdio.h>
#include <string.h>
#include <ctype.h>
int main(void)
{
// Put variables for further proceed
char number[101] = { 0 };
int isValid = 0, count = 0;
/*Input validation */
while(!isValid)
{
count = 0;
isValid = 1;
char *p = number;
printf("Enter a number: ");
scanf("%100s", number);
/* Check the validity of the string */
while(*p != '\0')
{
count++;
if(!isdigit(*p))
{
isValid = 0;
break;
}
p++;
}
if( !(isValid = isValid && ( count % 2 == 0 ) ) )
printf("Please insert an even number of digits (numbers only)\n");
}
/* Print the digits*/
for(int i=0; i<count; i+=2)
printf("%c%c\n", number[i], number[i+1] );
return 0;
}
I defined an array of 101 characters (100 + one for the string terminator) and I say scanf to store up to 100 characters (%100s)
I complicated a bit the input validator just to avoid to loop twice through the string (the first using strlen(), needed to check the even digits requirement,and the second to check the digits-only requirement)
In the end I just print two characters at a time using %c format specifier reading them from the string number
The output:
Enter a number: 1234567 // Test the check on an input with an odd number of digits
Please insert an even number of digits (numbers only)
Enter a number: 1234567A // Test the check on an input containing a non-digit char
Please insert an even number of digits (numbers only)
Enter a number: 1234567890123456 // Well formed input
12
34
56
78
90
12
34
56
Here is my solution to this problem. Hope it satisfy your requirement.
#include <stdio.h>
int main()
{
// Put variables for further proceed
int number, remainder, quotient = 1, numberUpdate, temp,count = 0;
int value = 1;
printf("Enter a number: ");
scanf("%d", &number);
numberUpdate = number;
temp = number;
if(number < 100) {
printf("%d",number);
} else {
while(numberUpdate > 100) {
value = value*100;
numberUpdate = numberUpdate/100;
}
while (temp > 0)
{
temp = number/value;
number = number%value;
value = value/100;
printf("%d\n",temp);
}
}
return 0;
}
The question is that show the digits which were repeated in C.
So I wrote this:
#include<stdio.h>
#include<stdbool.h>
int main(void){
bool number[10] = { false };
int digit;
long n;
printf("Enter a number: ");
scanf("%ld", &n);
printf("Repeated digit(s): ");
while (n > 0)
{
digit = n % 10;
if (number[digit] == true)
{
printf("%d ", digit);
}
number[digit] = true;
n /= 10;
}
return 0;
}
But it will show the repeated digits again and again
(ex. input: 55544 output: 455)
I revised it:
#include<stdio.h>
int main(void){
int number[10] = { 0 };
int digit;
long n;
printf("Enter a number: ");
scanf("%ld", &n);
printf("Repeated digit(s): ");
while (n > 0)
{
digit = n % 10;
if (number[digit] == 1)
{
printf("%d ", digit);
number[digit] = 2;
}
else if (number[digit] == 2)
break;
else number[digit] = 1;
n /= 10;
}
return 0;
}
It works!
However, I want to know how to do if I need to use boolean (true false), or some more efficient way?
To make your first version work, you'll need to keep track of two things:
Have you already seen this digit? (To detect duplicates)
Have you already printed it out? (To only output duplicates once)
So something like:
bool seen[10] = { false };
bool output[10] = { false };
// [...]
digit = ...;
if (seen[digit]) {
if (output[digit])) {
// duplicate, but we already printed it
} else {
// need to print it and set output to true
}
} else {
// set seen to true
}
(Once you've got that working, you can simplify the ifs. Only one is needed if you combine the two tests.)
Your second version is nearly there, but too complex. All you need to do is:
Add one to the counter for that digit every time you see it
Print the number only if the counter is exactly two.
digit = ...;
counter[digit]++;
if (counter[digit] == 2) {
// this is the second time we see this digit
// so print it out
}
n = ...;
Side benefit is that you get the count for each digit at the end.
Your second version code is not correct. You should yourself figured it out where are you wrong. You can try the below code to print the repeated elements.
#include<stdio.h>
int main(void){
int number[10] = { 0 };
int digit;
long n;
printf("Enter a number: ");
scanf("%ld", &n);
printf("Repeated digit(s): ");
while (n > 0)
{
digit = n % 10;
if (number[digit] > 0)
{
number[digit]++;;
}
else if (number[digit] ==0 )
number[digit] = 1;
n /= 10;
}
int i=0;
for(;i<10; i++){
if(number[i]>0)
printf("%d ", i);
}
return 0;
}
In case you want to print the repeated element using bool array (first version) then it will print the elements number of times elements occur-1 times and in reverse order because you are detaching the digits from the end of number , as you are seeing in your first version code output. In case you want to print only once then you have to use int array as in above code.
It is probably much easier to handle all the input as strings:
#include <stdio.h>
#include <string.h>
int main (void) {
char str[256] = { 0 }; /* string to read */
char rep[256] = { 0 }; /* string to hold repeated digits */
int ri = 0; /* repeated digit index */
char *p = str; /* pointer to use with str */
printf ("\nEnter a number: ");
scanf ("%[^\n]s", str);
while (*p) /* for every character in string */
{
if (*(p + 1) && strchr (p + 1, *p)) /* test if remaining chars match */
if (!strchr(rep, *p)) /* test if already marked as dup */
rep[ri++] = *p; /* if not add it to string */
p++; /* increment pointer to next char */
}
printf ("\n Repeated digit(s): %s\n\n", rep);
return 0;
}
Note: you can also add a further test to limit to digits only with if (*p >= '0' && *p <= '9')
output:
$./bin/dupdigits
Enter a number: 1112223334566
Repeated digit(s): 1236
Error is here
if (number[digit] == true)
should be
if (number[digit] == false)
Eclipse + CDT plugin + stepping debug - help you next time
As everyone has given the solution: You can achieve this using the counting sort see here. Time complexity of solution will be O(n) and space complexity will be O(n+k) where k is the range in number.
However you can achieve the same by taking the XOR operation of each element with other and in case you got a XOR b as zero then its means the repeated number. But, the time complexity will be: O(n^2).
#include <stdio.h>
#define SIZE 10
main()
{
int num[SIZE] = {2,1,5,4,7,1,4,2,8,0};
int i=0, j=0;
for (i=0; i< SIZE; i++ ){
for (j=i+1; j< SIZE; j++){
if((num[i]^num[j]) == 0){
printf("Repeated element: %d\n", num[i]);
break;
}
}
}
}
My aim is to accept 4-digit numbers, and 4-character strings (string should not contain digits or special characters)
If an invalid input is given the program should not terminate and it must allow the user to enter the details and continue until he wish to terminate.
I am able to find whether the input is a digit.
if(scanf("%d",&input)!=1)
{
printf("enter the number please");
... // I have option to re enter using while and flags
}
else
{
// I continue my work
...
}
To check it is four digits I have tried using the commands
i=0;
num = input;
while(num>0)
{
i = i+1;
num = num/10;
}
if(i==4){
...//I continue
}
else
printf("please enter four digit");
I have no idea of checking the same for characters. (I know how to check its length using strlen())
Please help me with the code in C. (Also help me to reduce/optimize the above logic to check whether the input is a 4-digit number)
I believe you want 2 inputs a number and a string. You can do that as
int number= 0;
char string[10] = { 0 };
do {
printf("please enter four digit");
scanf("%d", &number);
if(number >=1000 && number<= 9999)
break;
} while(1);
do {
printf("please enter four character string");
fgets(string, sizeof(string), stdin);
if(strlen(string) == 4)
break;
} while(1);
To check it is four digit number you can simply put a check whether the number lies between 1000 and 9999. (I am assuming you don't want the number to start with 0.)
strtol can help:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
char s[32], *p;
int x;
fgets(s, sizeof(s), stdin);
if ((p = strchr(s, '\n')) != NULL)
*p = '\0';
x = (int)strtol(s, &p, 10);
if ((p - s) == 4) {
printf("%d\n", x);
} else {
printf("Please enter four digit\n");
}
return 0;
}
char input[16];
int ok = 1, k = 0;
if (scanf("%s", input) > 0 && strlen(input) == 4) {
// check if it's a word
for (; k < 4; k++)
if (!isalpha(input[k])) {
// check if it's a number
for (int k = 0; k < 4; k++)
if (!isdigit(input[k]))
ok = 0;
break;
}
}
else ok = 0;
if (!ok)
printf("invalid input, please enter a 4-digit number or 4-letter word");
else {
printf("valid input");
...
}
You can use gets()1 fgets() to get the whole line and check line length. If the first character is between '0' and '9' then check the remaining if they are 3 numbers too. If the first character is a valid character in string then check the 3 remaining chars if it's also valid in string.
1See Why is the gets function so dangerous that it should not be used?
Force user to input a positive integer and put user in a loop till they do.
So I want everything including characters not allowed just over > 0
I tried the following:
while (i < 0) do {
printf("please input a number that's positive");
scanf("%d", i);
}
For positive integer use the following code
int i;
do
{
printf("please input a number that's positive");
scanf("%d", &i);
}while (i < 0);
The c language provides no error checking for user input. The user is expected to enter the correct data type. For instance, if a user entered a character when an integer value was expected, the program may enter an infinite loop or abort abnormally.
Both of these while functions manage the numbers, the int k is the set integer which can only be set below 20, the first while loop makes a statement that calls for another scan if the number is greater than 20
and the second loop prints a k*k box.
Hope this helps.
int main ( )
{
int i, j,k;
printf("Please enter Box size:\n\n");
scanf("%d",&k);
while(k>20){
printf("Please enter a value below 20\n\n");
scanf("%d"),&k;
}
while(k<=20)
{
for (i = 0; i < k; i++)
{
printf("\n");
for (j = 0; j < k; j++)
{
printf("#");
}
}
return 0;
}
}
I would do this: declare char term and int wrong = 0.
do {
printf("Enter a number: ");
fflush(stdin);
if (scanf("%d%c", &n, &term) != 2 || term != '\n' || n <= 0) {
printf("Only positive numbers.\n");
wrong = 1;
}
else {
wrong = 0;
//do something here if correct;
}
} while (wrong);
The code above detects invalid input if the user entered a mixture of characters and numbers, or negative numbers (and zero).
However, it doesn't detect if the user entered trailing zeros in front followed by valid digits eg. 001or 00000738. If anyone else could figure this out, please share below. Thanks! :)
Here is another alternative of a function which takes in a char str[20] (of say, maybe 20 elements), analyses the string to check for positive integers, and returns a 0 or 1 accordingly. Lastly, convert that string to an integer using atoi().
int checkPositiveIntegers(char str[]) {
char *ptr = str;
if (*ptr == '-' || *ptr == '0') //checks for negative numbers or zero
return 1;
else {
do {
if (isdigit(*ptr) == 0) { //checks for non-digit at ptr location; isdigit() returns 0 if non-digit
return 1;
break;
}
ptr++;
} while (*ptr != '\0' && *ptr != '\n');
return 0; //returns 0 if positive integer
}
}
So the function only accepts positive numbers from 1 to 9,999,999,999,999,999,999 (up to 19 digits if char str[] holds 20 elements).
However, if you converted the string back to int n = atoi(str);, the maximum value it could reach would be 2,147,483,647 since n is declared as a signed integer. Play around with different datatypes for exploration.
Hope this helps! :)