10-19 Exercise 1: International Standard Book Number (ISBN)

Question

The 13-digit International Standard Book Number (ISBN) is a unique code that identifies a book commercially. The last digit is a check digit used for error detection. To calculate its value, each digit of the first twelve digits is alternately multiplied, from left to right, by 1 or 3. The products are summed up and divided by 10. The check digit is the remainder of the division subtracted from 10. If it is 10, it becomes 0. 

ISBN檢查號的計算方法: 前12個數字，奇數項 乘以 1 ， 偶數項乘以 3

For example, assume that the first twelve digits are 978960931961.  

假設號碼前12位是978960931961

(a) (9*1 + 7*3 + 8*1 + 9*3 + 6*1 + 0*3 + 9*1 + 3*3 + 1*1 + 9*3 + 6*1 + 1*3) = 126 

(b) The check digit = 10 – (126 % 10) = 10 – 6 = 4  

檢查號 = 10 - 餘數 = 10 – (126 % 10) = 10 – 6 = 4  

So the valid ISBN is: 9789609319614

Write a program that reads a 13-digit ISBN and checks the last digit to see if it is valid or not. 

寫一個輸入13位ISBN號碼 判斷最後一個數字是否檢查碼即可

#Try 1:

Input:

9789609319614

Output: 

Valid ISBN

 

#Try 2:

Input:

9781482214505

Output: 

Valid ISBN

 

#Try 3:

Input:

9780132936493

Output: 

Invalid ISBN

 

Answer 1

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}

Comments

prog.c: In function 'main':
prog.c:11:4: error: this 'else' clause does not guard... [-Werror=misleading-indentation]
    else
    ^~~~
prog.c:14:8: note: ...this statement, but the latter is misleadingly indented as if it is guarded by the 'else'
        return 0;
        ^~~~~~
cc1: all warnings being treated as errors

Answer 2

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Comments

Case 0: Wrong output
Case 1: Correct output
Case 2: Correct output
Case 3: Correct output

Answer 3

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}

Comments

prog.c: In function 'main':
prog.c:9:4: error: this 'else' clause does not guard... [-Werror=misleading-indentation]
    else
    ^~~~
prog.c:12:8: note: ...this statement, but the latter is misleadingly indented as if it is guarded by the 'else'
        return 0;
        ^~~~~~
cc1: all warnings being treated as errors

Answer 4

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}

Comments

prog.c: In function 'main':
prog.c:9:4: error: this 'else' clause does not guard... [-Werror=misleading-indentation]
    else
    ^~~~
prog.c:12:8: note: ...this statement, but the latter is misleadingly indented as if it is guarded by the 'else'
        return 0;
        ^~~~~~
cc1: all warnings being treated as errors

Answer 5

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int main()
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}

Comments

prog.c:1:9: error: expected '=', ',', ';', 'asm' or '__attribute__' before '<' token
 include <stdio.h>
         ^

Answer 6

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}

Comments

Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
Case 3: Correct output

Answer 7

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}

Comments

Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
Case 3: Correct output

Answer 8

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}

Comments

prog.c: In function 'main':
prog.c:11:16: error: too many arguments for format [-Werror=format-extra-args]
         printf("Valid ISBN",check);}
                ^~~~~~~~~~~~
prog.c:12:22: error: too many arguments for format [-Werror=format-extra-args]
         else {printf("Invalid ISBN",check);}
                      ^~~~~~~~~~~~~~
cc1: all warnings being treated as errors

Answer 9

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#include<stdio.h>
int main()
{
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}

Comments

Case 0: Correct output
Case 1: Correct output
Case 2: Correct output
Case 3: Correct output

Answer 10

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#include  *   ***   * 
int main()
{
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return 0;
}

Comments

prog.c: In function 'main':
prog.c:11:16: error: too many arguments for format [-Werror=format-extra-args]
         printf("Valid ISBN",check);}
                ^~~~~~~~~~~~
prog.c:12:22: error: too many arguments for format [-Werror=format-extra-args]
         else {printf("Invalid ISBN",check);}
                      ^~~~~~~~~~~~~~
cc1: all warnings being treated as errors