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Write a function num_digits(n) that returns the number of digits in n ( a positive integer). Hint: to determine the number of digits in a number n, divide it by 10 repeatedly. when n reaches 0, the number of divisions indicates how many digits n originally had.

寫一個num_digits(n)函數 計算n有幾位數

#include <stdio.h>
int num_digits(int n)
{
   /*INSERT YOUR CODE HERE*/

   /*在這裡寫你的程式*/
   /*END OF YOUR CODE*/
}

int main(void)
{
   int n;
   scanf("%d", &n);
   printf("%d has %d digit(s)",n,num_digits(n));
   return 0;
}

Please complete this program by only insert your code between those tags:

   /*INSERT YOUR CODE HERE*/ 
   /*END OF YOUR CODE*/

Example input

12347

Example output

12347 has 5 digit(s)

Remember: You may correct the cases, but your code always be revised!

[Exercise] Coding (C) - asked in Chapter 9: Functions by (5.2k points)
ID: 35784 - Available when: 2017-12-07 18:00 - Due to: Unlimited

edited by | 9.6k views

45 Answers

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answered by (-281 points)
0 0
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Case 1: Correct output
Case 2: Correct output
Case 3: Correct output
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answered by (-249 points)
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Case 1: Correct output
Case 2: Correct output
Case 3: Correct output
0 like 0 dislike
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answered by (-32 points)
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Case 1: Correct output
Case 2: Correct output
Case 3: Correct output
0 like 0 dislike
Hidden content!
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answered by (-127 points)
0 0
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Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
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Hidden content!
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answered by (-127 points)
0 0
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Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
0 like 0 dislike
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answered by (-255 points)
0 0
Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
Case 3: Correct output
0 like 0 dislike
Hidden content!
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answered by (54 points)
0 0
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Case 1: Correct output
Case 2: Correct output
Case 3: Correct output
0 like 0 dislike
Hidden content!
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answered by (54 points)
0 0
Case 0: Wrong output
Case 1: Wrong output
Case 2: Wrong output
Case 3: Wrong output
0 0
#include <stdio.h>
int num_digits(int n)
{
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{
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   printf("%d has %d digit(s)",n,num_digits(n));
   return 0;
}
0 like 0 dislike
Hidden content!
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