making a bug apparent

This is a discussion on making a bug apparent within the C++ Programming forums, part of the General Programming Boards category; My instructor said that if the writer of the code below made full use a language feature in C++ from ...

  1. #1
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    making a bug apparent

    My instructor said that if the writer of the code below made full use a language feature in C++ from one of the following:

    References
    The const qualifier
    NULL vs. 0
    bool
    The friend specifier
    Copy constructors

    the bug would be apparent to see.... do you guys have any idea what it is?
    I know the problem is clearly at the get method and the itsChars variable.. but what feature would make the bug apparent?

    Code:
    #include <cstdio>
    #include <cstring>
    #include <cstdlib>
    class CharList {
    public:
    CharList(char* s) {
    itsChars = new char[strlen(s)+1];
    strcpy(itsChars, s);
    }
    //
    // Iterate over the characters, returning a NUL when done
    //
    char get() {
    if (*itsChars)
    return *itsChars++;
    else
    return 0;
    }
    //
    // Return the nth character
    //
    char at(int n) {
    if (n < length())
    return itsChars[n];
    else
    return 0;
    }
    int length() { return strlen(itsChars); }
    private:
    
    char *itsChars;
    };
    //
    // debugging routine -- print the characters in the list
    //
    void print(const char *label, CharList& L)
    {
    printf("%s", label);
    const char *fmt = "%c";
    while (char c = L.get()) {
    printf(fmt, c);
    fmt = ", %c";
    }
    puts("");
    }
    int main(int argc, char **argv)
    {
    if (argc != 2) {
    fprintf(stderr, "Usage: %s STRING\n", argv[0]);
    exit(1);
    }
    CharList L(argv[1]);
    print("argv[1]: ", L);
    printf("Ready to enter loop...\n");
    for (int i = 0; i < L.length(); i++) {
    printf("in loop\n");
    printf("L[%d] = '%c'\n", i, L.at(i));
    }
    printf("Done with loop...\n");
    }
    Last edited by -EquinoX-; 02-27-2010 at 11:23 PM.

  2. #2
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    Logically, the act of printing out an object should not change that object. Otherwise, if you print it out twice, there will be different results each time.

    Make the second argument of print() a const reference. Then you will see what your instructor is talking about. The compiler will complain bitterly, but that is actually the point. Your code is broken but, if you don't follow your instructor's guidance, the compiler can't provide helpful information to recognise that.
    Last edited by grumpy; 02-27-2010 at 05:28 PM.
    Right 98% of the time, and don't care about the other 3%.

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    Could you give me more details why making the second argument a const reference would make it work?

  4. #4
    Dae
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    Quote Originally Posted by -EquinoX- View Post
    Could you give me more details why making the second argument a const reference would make it work?
    It doesn't make it work. It helps you find the bug in your print function. const (constant) means the function can't change it. Given that, what do you think you're doing? You're changing it. Printing shouldn't change the subject. That's your bug.
    Warning: Have doubt in anything I post.

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  5. #5
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    Properly indenting the code would certainly make the code more readable and make bugs more apparent.
    I don't waste my time looking through unindented code, but this line definitely looks like a bug. I don't know any compiler that wouldn't give you an error there:
    Code:
    C SC 397a Assignment 5; page 2
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    Quote Originally Posted by Dae View Post
    It doesn't make it work. It helps you find the bug in your print function. const (constant) means the function can't change it. Given that, what do you think you're doing? You're changing it. Printing shouldn't change the subject. That's your bug.
    so are you saying that by putting a const in the function parameter gives an indication that the print shouldn't change the subject?

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    Declaring a variable (or function parameter) as "const" means that your promising to the compiler not to modify it. You're saying you want it to be "read-only". Then later you break your promise by attempting to modify it, which the compiler catches and calls you a liar! So it doesnt really matter the context (i.e. function X passes const variable Y, function Z modifies it, etc). All that matters is if your trying to do anything but read a const variable.

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    so putting a const in that function parameter will make the bug apparent because the compiler will fail?

  9. #9
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    Quote Originally Posted by -EquinoX-
    so putting a const in that function parameter will make the bug apparent because the compiler will fail?
    Yes, there will be a compile error, unless you do some nasty things that make the compiler ignore the bug like using const_cast to cast away the const-ness inappropriately.
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  10. #10
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    @-EquinoX-
    500+ posts, and that's the best code formatting you can foist upon us?
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    Dae mention the term "subject", what is he actually referring to?

  12. #12
    C++まいる!Cをこわせ!
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    An analogy, I believe. Basically, saying that your function shouldn't be changing something, yet it does.
    Now, why is your indentation POC?
    Quote Originally Posted by Adak View Post
    io.h certainly IS included in some modern compilers. It is no longer part of the standard for C, but it is nevertheless, included in the very latest Pelles C versions.
    Quote Originally Posted by Salem View Post
    You mean it's included as a crutch to help ancient programmers limp along without them having to relearn too much.

    Outside of your DOS world, your header file is meaningless.

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