The program would ask the user an integer input.
and it counts how many zero the int has.
constraints: use while loop
ex:
input: 2400
count: 2
now I have no problem in that part, only when the user would input a zero.
supposed it counts 1.
ex:
input 0
count: 1
but then the program returns count 0.
here's the code:
int main(){
int n, counter = 0;
printf("Enter the number: ");
scanf("%d", &n);
while(n != 0){
if(n % 10 == 0){
counter ++;
n=n/10;
}else{
break;
}
}
printf("%d", counter);
return 0;
}
Use functions.
int countZeroes(int x)
{
int result = !x; // if x == 0 then result = 1
while(x)
{
result += !(x % 10);
x /= 10;
}
return result;
}
int main(void)
{
printf("%d\n", countZeroes(0));
printf("%d\n", countZeroes(1000));
printf("%d\n", countZeroes(-202020));
}
https://godbolt.org/z/91hKr46eo
You have while(n != 0) this does so when you enter just 0 it doesn't run. So the counter that you have set to 0 at the beginning is still 0
Here is what I would have done :
int main()
{
int num, count = 0;
scanf("%d",&num);
if (num == 0) {
printf("1");
return 0;
}
while(num != 0) //do till num greater than 0
{
int mod = num % 10; //split last digit from number
num = num / 10; //divide num by 10. num /= 10 also a valid one
if(mod == 0) count ++;
}
printf("%d\n",count);
return 0;
}
Just don't forget to consider everything that can happen with a condition that you set
**Fixed it
A different version that prints the integer as a string, and looks for '0' characters in it. Tested.
#include <stdio.h>
#include <string.h>
int main(void)
{
int input = 0;
int zeroes = 0;
char *foundpos, teststring[100];
scanf("%d", &input);
sprintf(teststring, "%d", input);
foundpos = strchr(teststring, '0');
while (foundpos != NULL) {
++zeroes;
foundpos = strchr(foundpos + 1, '0');
}
printf("%d contains %d zeroes", input, zeroes);
}
Just count the zero digits you get between \n chars.
#include <stdio.h>
int main()
{
int ndigs = 0, c;
while ((c = getchar()) != EOF) {
switch (c) {
case '0': ndigs++;
break;
case '\n': printf(" => %d zero digs", ndigs);
ndigs = 0;
break;
}
putchar(c);
}
}
sample output:
$ ./a.out
123001202010
123001202010 => 5 zero digs
^D
$ _
No need to convert digits to a number, to convert it back to decimal digits. You can improve the program counting digits until a nondigit is detected, then output. But there's no need to convert a decimal representation of a number (in base 10) to internal representation to then get the digits you have destroyed (in the conversion) back again to count them.
As earlier mentioned, the problem is with the loop:
while(n != 0){
if(n % 10 == 0){
counter ++;
n=n/10;
}else{
break;
}
}
It doesnt do anything in case n == 0. But replacing it with n > 0 is not a good solution because ints can be negative too.
You should use do{}while() construction instead, it will always do one iteration of loop no matter what condition you put there. Notice that no matter what you get as a number, it is still a number so you can do one iteration of loop either way.
Just do as follows:
do{
if(n % 10 == 0){
counter ++;
n=n/10;
}else{
break;
}
} while( n != 0 );
This should work(if i didnt mess up the braces/semicolumns).
The problem says that the user will input numbers till he inserts 0 (exit) then if there are any prime digits in that given number, the program shall multiply them.
For example:
input : 4657
output : 35 ( 5 * 7 )
Here is what I have tried so far but I cannot pull it off with the multiplication... my code might look a little bit clumsy, I am a beginner :)
int main() {
int number;
int digit, prime;
int i, aux;
int multiplier;
input:
printf("Give Number :");
scanf("%d", &number);
do {
multiplier = 1;
digit = number % 10;
number = number / 10;
printf("%d \n", digit);
prime = 1;
for (i = 2; i <= sqrt(digit); i++) {
if (digit % 1 == 0) {
prime = 0;
}
}
if (prime != 0) {
multiplier = multiplier * digit;
}
} while (number != 0);
printf("The result : %d\n", multiplier);
goto input;
return 0;
}
There are multiple problems in your code:
missing #include <stdio.h>
use of label and goto statement to be avoided at this stage.
testing for prime digits can be done explicitly instead of a broken loop.
digit % 1 == 0 is always true, you should write digit % i == 0 (maybe a typo from copying someone else's code?
i <= sqrt(digit) will not work unless you include <math.h> and still a bad idea. Use i * i < 10 instead.
0 and 1 should not be considered primes.
it is unclear what the output should be if none of the digits are prime, let's assume 1 is OK.
Here is a modified version:
#include <stdio.h>
int main() {
for (;;) {
int number, multiplier;
printf("Give Number :");
if (scanf("%d", &number) != 1 || number <= 0)
break;
multiplier = 1;
while (number != 0) {
int digit = number % 10;
number = number / 10;
if (digit == 2 || digit == 3 || digit == 5 || digit == 7)
multiplier *= digit;
}
printf("The result: %d\n", multiplier);
}
return 0;
}
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 issue I'm having is, I want to take an integer and separate it. For example: The user enters: 23432. The console should print" 2 3 4 3 2. The issue I'm having is storing that digits. For example,
User Input : 2020
assign input to num.
digit = 2020 % 10 = 0 <--- 1st Digit
num = num / 10 = 202
digit2 = num % 10 = 2 <--- 2nd Digit
num = num / 100 = 20.2
temp = round(num) = 20
digit3 = num % 10 = 0 <--- 3rd Digit
digit4 = num / 10 = 2 <---- 4th Digit
The problem with this approach is that its dependent on the user input, I'm working with the range 1-32767, so I wont know how many digit variables to create. Using the structure I've created can someone assist in making it run in a way the no matter what the number is, the digit is saved and printed in the way I've described?
int Rem(int num);
int Div(int num);
int main() {
int num;
printf("Enter an integer between 1 and 32767: ");
scanf("%d", &num);
Rem(num);
Div(num);
printf("%d","The digits in the number are: ");
}
int Rem(int num) {
int rem = num % 10;
return rem;
}
int Div(int num){
int div = num / 10;
return div;
}
The problem with this approach is that its dependent on the user input, I'm working with the range 1-32767, so I wont know how many digit variables to create.
So calculate it. You can do this by increasing a variable by a factor of 10 each time until increasing it one more time would make it larger than your input number:
int num;
printf("Enter an integer between 1 and 32767: ");
scanf("%d", &num);
int div = 1;
while(div * 10 <= num)
div *= 10;
Then you can repeatedly divide your number by this divisor to get each of the digits, dividing the divisor by 10 each time to shift one place at a time:
printf("The digits in the number are: ");
while(div > 0)
{
printf("%d ", (num / div) % 10);
div /= 10;
}
That's a really complicated approach. Why not read the string, and parse the string out like this:
int main(void) {
char buf[256]; // should be big enough, right? 10^256-1
memset(buf, 0, 256];
puts("enter something : ");
if( NULL == fgets(STDIN, 255, buf)) {
perror("Egad! Something went wrong!");
exit(1);
}
// at this point, you already have all the input in the buf variable
for(int i=0; buf[i]; i++) {
putchar( buf[i] ); // put out the number
putchar( ' ' ); // put out a space
}
putchar( '\n' ); // make a nice newline
}
As written above, it allows any character, not just numbers. If you want to restrict to numbers, you could filter the input, or put a check in the for loop ... depending on what you were trying to accomplish.
One way that C allows you do deal with problems this extremely elegantly is via recursion.
Consider a routine that only knows how to print the very last digit of a number, with a preceding space if needed.
void printWithSpaces(int neblod)
{
// Take everything except the last digit.
int mene = neblod / 10;
// Now extract the last digit
int fhtagn = neblod % 10;
// Check if there are leading digits
if (mene != 0)
{
// There are, so do some magic to deal with the leading digits
// And print the intervening space.
putchar(' ');
}
putchar(fhtagn + '0');
}
OK. So that's well and good, except what can we use to "Do some magic to deal with the leading digits"?
Don't we want to just print them as a sequence of digits, with suitable intervening spaces?
Isn't that exactly what void printWithSpaces(int neblod) does?
So we make one change:
void printWithSpaces(int neblod)
{
// Take everything except the last digit.
int mene = neblod / 10;
// Now extract the last digit
int fhtagn = neblod % 10;
// Check if there are leading digits
if (mene != 0)
{
// There are, so print them out
printWithSpaces(mene);
// And print the intervening space.
putchar(' ');
}
putchar(fhtagn + '0');
}
And you're done.
For the curious, the following article on C recursion may provide both an amusing read, and a little insight into my slightly unusual choice of variable names. ;) http://www.bobhobbs.com/files/kr_lovecraft.html
The program I am writing to take a number and display that number as a calculator would display it (shown below) is compiling with no issues, but when I try to run it, I am able to input my number, but nothing happens. It seems like it is "hanging", since no further output is shown as I would have expected. Might anyone know what the problem is?
#include <stdio.h>
#define MAX_DIGITS 20
char segments[10][7] = /* seven segment array */
{{'1','1','1','1','1','1','0'}, /* zero */
{'0','1','1','0','0','0','0'}, /* one */
{'1','1','0','1','1','0','1'}, /* two */
{'1','1','1','1','0','0','1'}, /* three */
{'0','1','1','0','0','1','1'}, /* four */
{'1','0','1','1','0','1','1'}, /* five */
{'1','0','1','1','1','1','1'}, /* six */
{'1','1','1','0','0','0','0'}, /* seven */
{'1','1','1','1','1','1','1'}, /* eight */
{'1','1','1','0','0','1','1'}};/* nine */
char digits[3][MAX_DIGITS * 4]; /* digits array */
int i, j; /* count variables */
int adjust; /* output formatting */
int main(void) {
clear_digits_array();
int digit[20];
for (i = 0; i < 20; i++) {
digit[i] = 0;
}
int count = 20;
int position = 0;
printf("Enter a number: ");
int number = scanf("%d%d%d%d%d%d%d%d%d%d%d%d%d%d%d%d%d%d%d%d",
&digit[0],
&digit[1],
&digit[2],
&digit[3],
&digit[4],
&digit[5],
&digit[6],
&digit[7],
&digit[8],
&digit[9],
&digit[10],
&digit[11],
&digit[12],
&digit[13],
&digit[14],
&digit[15],
&digit[16],
&digit[17],
&digit[18],
&digit[19]); //NOTHING HAPPENS AFTER HERE
printf("Got input, number is %d", number);
while (count > 0) {
printf("Reading digits, count is %d", count);
process_digit(digit[20 - count], position);
position++;
count--;
}
print_digits_array();
printf("\n");
return 0;
}
void clear_digits_array(void) {
/* fill all positions in digits array with blank spaces */
for (i = 0; i < 3; i++) {
for (j = 0; j < (MAX_DIGITS * 4); j++) {
digits[i][j] = ' ';
}
}
}
void process_digit(int digit, int position) {
/* check each segment to see if segment should be filled in for given digit */
for (i = 0; i < 7; i++) {
printf("Processing digit %d at position %d, i is %d", digit, position, i);
if (segments[digit][i] == 1) {
switch (i) {
case 0: digits[0][(position * 4) + 1] = '_';
break;
case 1: digits[1][(position * 4) + 2] = '|';
break;
case 2: digits[2][(position * 4) + 2] = '|';
break;
case 3: digits[2][(position * 4) + 1] = '_';
break;
case 4: digits[2][(position * 4) + 0] = '|';
break;
case 5: digits[1][(position * 4) + 0] = '|';
break;
case 6: digits[1][(position * 4) + 1] = '_';
break;
}
}
}
}
void print_digits_array(void) {
/* print each character in digits array */
for (i = 0; i < 3; i++) {
for (j = 0; j < (MAX_DIGITS * 4); j++) {
printf("%c", digits[i][j]);
}
printf("/n");
}
}
Your code includes:
printf("Enter a number: ");
int number = scanf("%d%d%d%d%d%d%d%d%d%d%d%d%d%d%d%d%d%d%d%d",
&digit[0],
&digit[1],
&digit[2],
&digit[3],
&digit[4],
&digit[5],
&digit[6],
&digit[7],
&digit[8],
&digit[9],
&digit[10],
&digit[11],
&digit[12],
&digit[13],
&digit[14],
&digit[15],
&digit[16],
&digit[17],
&digit[18],
&digit[19]); //NOTHING HAPPENS AFTER HERE
Have you entered twenty separate digits? If not, scanf() is waiting for you to type some more numbers.
Note that the return value from scanf() is the number of successfully converted numbers (0..20 in the example), not the value you entered.
Is that the issue? I tried to make it such that the maximum amount of numbers the user could enter was 20, and if any more were entered they would be ignored, or if fewer were entered, it would only consider those (say, 5 were entered, then only the 5 would be used). Is there an easier way to do this sort of thing then?
Yes, I think that's the issue.
There are probably several easier ways to do it. I'd be tempted to use:
char buffer[22]; // 20 digits, newline, null
if (fgets(buffer, sizeof(buffer), stdin) != EOF)
{
size_t len = strlen(buffer);
if (len >= sizeof(buffer) - 1)
{
fprintf(stderr, "You entered too long a string (maximum 20 digits)\n");
exit(EXIT_FAILURE);
}
if (len > 0)
buffer[--len] = '\0'; // Zap newline — carefully
for (int i = 0; i < len; i++)
{
if (!isdigit(buffer[i]))
{
fprintf(stderr, "Non-digit %c found\n", buffer[i]);
exit(EXIT_FAILURE);
}
}
...and from here, process the digits in buffer, one at a time...
...Use buffer[i] - '0' to get a number 0..9 from the character in buffer...
}
Another option is to use a long long number, but that only gives you up to 18 digits (signed) or 19 digits (unsigned), with some values of 19 or 20 also within range. You'd then strip the digits out of the number using division and modulus operations.
long long value;
if (scanf("%lld", &value) == 1)
{
if (value < 0)
...error processing...
do
{
int digit = value % 10;
value /= 10;
...process digit...
} while (value > 0);
}
This has some merits, but in practice, I'd be tempted to use fgets() to read a line of input, and either sscanf() or strtoll() to convert and validate the number. That isn't as simple as it looks; the error returns from strtoll(), in particular, are many and subtle, and none of the scanf() family handle overflowing numbers gracefully. You could constrain the scanf() though with %18lld so that no more than 18 digits are read, but that would mean that if the user typed 19 or more digits, you'd get the leftovers on the next attempt to read with scanf(). So, handling scanf() is not simple either, especially if you need to convert multiple numbers in a single run of the program.
With those caveats out of the way, you can usually do a 'good enough' job with a sensible person providing the input. It is the process of making a program bomb-proof (foolproof — as in, proof against fools) that is hard. I find meaningful error reporting easier when I can report the whole string that was read in (as with fgets()); with scanf(), you can't see the characters that were entered and consumed before something went wrong.
use GDB, here is introduction for it http://www.gnu.org/software/gdb/
look into this tutorial too http://www.cs.cmu.edu/~gilpin/tutorial/
int number = scanf("%d%d%d%d%d%d%d%d%d%d%d%d%d%d%d%d%d%d%d%d",
&digit[0],
&digit[1],
&digit[2],
&digit[3],
&digit[4],
&digit[5],
&digit[6],
&digit[7],
&digit[8],
&digit[9],
&digit[10],
&digit[11],
&digit[12],
&digit[13],
&digit[14],
&digit[15],
&digit[16],
&digit[17],
&digit[18],
&digit[19]);
I don't think this is a good idea use loop here
for (i = 0; i < 20; i++)
scanf("%d",&digit[i]);
And if you need the number
then do like this
int number = i; when loop finishes.
You can also try this
char buf[12];
while((c=getchar())!=EOF && i < 20)
{
buf[j++] =c;
if((c == '\n' || c == ' ' || c == '\t') && (sscanf(buf,"%d",&digit[i])) == 1)
{
i++;
j = 0;
}
}
For EOF you will have to press CTRL+D
In this way you can take 20 or less integers