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C++11 for statement and addresses:

This is a discussion on C++11 for statement and addresses: within the C++ Programming forums, part of the General Programming Boards category; Why is C++11 act different than C++ in reference to the for statements. I was pretty excited when I read ...

  1. #1
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    C++11 for statement and addresses:

    Why is C++11 act different than C++ in reference to the for statements.

    I was pretty excited when I read how they worked in my textbook and now I am left with questions. Why am I getting different addresses?

    Code:
     #include <iostream>
    #include <array>
    
    using namespace std;
    
    void changePointer(int change[])
    {
            cout<<change<<endl;
    }
    
    int main(int argc, const char * argv[])
    {
    
        array <int, 5> drew={34,22,55,56,98};
        
        for(int items : drew)
            cout<<"The arrays is:"<<items<<endl;
            
        for (int change : drew) {
            cout<<"The address is:"<<&change<<endl;   //c++11 is suppose to do the same as below
        }
        
        for (int i=0; i<5; i++) {
            
        changePointer(&drew[i]);     // this did what I thought it would
            
        }
        
        return 0;
    }

  2. #2
    Registered User MutantJohn's Avatar
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    I think it might be because using int[] tells your compiler or w/e to make a static copy of whatever you're passing to it. You're passing the value of an address and where that value is stored is what you're printing, right?

    Is that right? I'm probably wrong but I figured, eh, the internet, right?

    What if you printed changed[0]?

  3. #3
    C++ Witch laserlight's Avatar
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    Quote Originally Posted by jocdrew21
    Why am I getting different addresses?
    In your first and second for loops, a copy of each element is made in order to initialise items and change respectively on each iteration of the loop, hence the address printed is the address of the variable to which the copy was made. In your third for loop, you access each element of the array without copying, hence the address printed is the address of the element.

    If you want change to refer to the current element of the array instead of being a copy, make it a reference variable, e.g.,
    Code:
    for (int& change : drew) {
        cout << "The address is:" << &change << endl;
    }
    Thus, change is an alias for the current element of drew, hence &change will result in the address of the current element of drew.
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  4. #4
    - - - - - - - - oogabooga's Avatar
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    While the third loop is printing the addresses of the individual elements of the int array, the second loop is printing the address of the integer variable chnage, which occupies a fixed position.
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  5. #5
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    Your amazing!!!!

    Hopefully one day I understand programing like you laser light !!!

  6. #6
    CSharpener vart's Avatar
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    looking at the for loop definition I see it would be extracted to something similar to

    Code:
    for(...)
    {
       int change = *&drew[i];
       cout<<"The address is:"<<&change<<endl; 
    }
    so you are printing the address of for-range-declaration variable. Why do you think it should match address of array member?

    For a range-based for statement of the form
    for ( for-range-declaration : expression ) statement
    let range-init be equivalent to the expression surrounded by parentheses89
    ( expression )
    and for a range-based for statement of the form
    for ( for-range-declaration : braced-init-list ) statement
    let range-init be equivalent to the braced-init-list. In each case, a range-based for statement is equivalent
    to
    Code:
    {
      auto && __range = range-init;
      for ( auto __begin = begin-expr,
        __end = end-expr;
        __begin != __end;
        ++__begin ) {
          for-range-declaration = *__begin;
          statement
        }
    }
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  7. #7
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    If I change the value at the location of the variable, I change it everywhere right. Thus it should print it out whenever I output it after I changed it. What am I missing? I thought I did this below.

    Code:
     #include <iostream>
    #include <array>
    
    using namespace std;
    
    int changer(int *a)
    {
        return *a * 4;
    
    }
    
    
    
    int main(int argc, const char * argv[])
    {
    
        array <int, 5> drew={34,22,55,56,98};
        
        for(int items : drew)
            cout<<"The arrays is:"<<items<<endl;
            
        for (int& change : drew) {
            cout<<"The address is:"<<&change<<endl;   
        }
        
        
        for(int &i : *&drew)            //pass a copy and did simple arithmetic
            
            cout<< i*6 <<endl;
        
        cout<<changer(&drew[0])<<endl;              // changed the value with a pointer
        
        for(int items : drew)
            cout<<"The arrays is now:"<<items<<endl;      // Here the value should be the same as above
        
        return 0;
    }

  8. #8
    Registered User whiteflags's Avatar
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    > *a * 4;

    This is not an assignment. You need to store the result, for example:

    *a *= 4;

  9. #9
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    ahhhh... Thanks... Finally I get pointers and pass by reference. It's lovely how programing can make you feel like a complete retard at times.

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