如何找到 c 中 int digits 的位数最多为 100 或 1000 位？

Question

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++

Answer 1

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.

Answer 2

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; 
}

Answer 3

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");

Answer 4

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++