RFC: A Simple Course in C

This is a discussion on RFC: A Simple Course in C within the C Programming forums, part of the General Programming Boards category; Hello All, I am writing a book on C programming and would welcome your review/comments and suggestions on the same. ...

  1. #1
    Registered User
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    RFC: A Simple Course in C

    Hello All,

    I am writing a book on C programming and would welcome your review/comments and suggestions on the same. You can download the latest draft on the following link.

    A Simple Course in C

    Warm Regards,
    Constantine

  2. #2
    Deathray Engineer MacGyver's Avatar
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    Quote Originally Posted by johnxconstantin View Post
    Hello All,

    I am writing a book on C programming and would welcome your review/comments and suggestions on the same. You can download the latest draft on the following link.

    A Simple Course in C

    Warm Regards,
    Constantine
    Do I get a cookie for spotting this?

    Code:
    void main() { printf(“hello world\n”); return 0; }

  3. #3
    Making mistakes
    Join Date
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    OK, some things I've spotted:
    Code:
    main()
    {
      char a;
      int i=10;
      printf(“\nThe value of i is %d”,i);
      a=getchar();
      if(a==’a’)
      {
        int i=1;
        printf(“\nThe value of i is %d”,i);
      }
      printf(“\nThe value of i is %d”,i);
    }
    It should be "int main(void)", your way is the older K&R style. And "a" should be of type int to check for EOF (explain this further)

    Code:
    #include <stdio.h>
    void multi_table(int i);
    main()
    {
      int i;
      multi_table(i);
      printf("\nValue of i is %d",i);
    }
    void multi_table(int i)
    {
      printf("\n 1 x %d = %d",i,i*1);
      printf("\n 2 x %d = %d",i,i*2);
      printf("\n 3 x %d = %d",i,i*3);
      printf("\n 4 x %d = %d",i,i*4);
      printf("\n 5 x %d = %d",i,i*5);
      printf("\n 6 x %d = %d",i,i*6);
      printf("\n 7 x %d = %d",i,i*7);
      printf("\n 8 x %d = %d",i,i*8);
      printf("\n 9 x %d = %d",i,i*9);
      printf("\n10 x %d = %d",i,i*10);
      i=0;
    }
    "i" is garbage in main. You didn't initialize it. The "i = 0" statement changes only a _copy_ of i.

    "The compiler would place variables of this type into the read‐only memory of the program." That's not guaranteed, const values can be changed by pointers:

    Code:
    const int i = 0;
    int *pointer = &i;
    *pointer = 10;
    "Generally when a program is implied, the compiler optimizes certain expressions by assuming that a variable’s value is unchanging if it does not occur on the left side of an assignment".

    Maybe that's correct and I'm a nerd, but i++ is not an assignment. And it may be changed by pointers.

    "auto is used to create a variable, usually a temporary variable, within a block of code. The variable is only available to the block of code it has been defined within and cannot be referenced outside of the block of code in which it is being used."

    Note that auto is the default.

    In general you are using some bad coding practices. And compile your programs first with all warnings enabled.

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