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Constant and pointer

Discussion in 'C' started by shabbir, Oct 9, 2006.

  1. shabbir

    shabbir Administrator Staff Member

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    Introduction



    In C you can effectively change the value of the constant variable. Just compile the following program and you will see the output.

    Code:
    void main()
    {
    	int const i=123;
    	int *ip;
    	
    	/* Changing constant integer i.e. const int/int const */
    	
    	printf(" %d %x \n",i,&i);
    	ip=&i;
    	*ip=456;
    	printf(" %d %x %x %d \n",i,&i,ip,*ip);
    	
    }
    Remember to name the file as <filename>.c and not <filename>.cpp You can also download the attachment.

    But when you put the same code in the cpp file the compiler would give you an error message. I am talking with respect to MS compiler.

    Getting into the constants and pointer



    This was about the introduction and how .c and .cpp compiler treated the const. Now I would like to go deep into what does const means in cpp and what are the combination you can use with the object pointer and how the complicated the situation can become.

    const int *p
    int const *p
    int * const p
    const * int p // Warning
    * const int p // Error
    * int const p // Error

    Lets take each one of them and see what it means how it behaves

    const int *p & int const *p



    Read it as "Pointer to constant int". For how to read it refer to the last section of the article, How to read. Now the expression means its a pointer pointing to the constant value. As an example
    Code:
    cout<<"Pointer to constant int"<<endl;
    {
    	int i = 123;
    	const int * p = &i;
    	cout <<"p = i = "<<*p<<endl;
    	p = &i;
    //	*p = 321; [URL=http://www.go4expert.com/articles/difference-constant-pointers-pointer-t26816/]Pointer to constant[/URL] objects so cant change values
    	cout <<"p = i = "<<*p<<endl;
    }
    cout<<"Pointer to constant int"<<endl;
    {
    	int i = 123;
    	int const * p = &i;
    	cout <<"p = i = "<<*p<<endl;
    	p = &i;
    //	*p = 321; Pointer to constant objects so cant change values
    	cout <<"p = i = "<<*p<<endl;
    }
    

    int * const p



    Read it as "Constant pointer". For how to read it refer to the last section of the article, How to read. Now the expression means that the pointer is constant
    Code:
    cout<<"Constant pointer"<<endl;
    {
    	int i = 123;
    	int * const p = &i;
    	cout <<"p = i = "<<*p<<endl;
    //	p = &i; Constant Pointer so unable to change
    	*p = 321;
    	cout <<"p = i = "<<*p<<endl;
    }
    

    const * int p



    This is also not a valid expression and would generate a warning and should be avoided.

    Code:
    cout<<"Pointer to constant int"<<endl;
    {
    	int i = 123;
    	const * int p = &i; 
    	// warning: 'int ' storage-class or type specifier(s) unexpected here; ignored
    	cout <<"p = i = "<<*p<<endl;
    	p = &i;
    //	*p = 321; Pointer to constant objects so cant change values
    	cout <<"p = i = "<<*p<<endl;
    }
    

    * const int p & * int const p



    Declaring a variable in this manner will generate and error.

    How to Read



    You have to read pointer declarations right-to-left.

    const int * p means "p points to an int that is const"
    int * const p means "p is a const pointer to an int"
     

    Attached Files:

  2. aisha.ansari84

    aisha.ansari84 New Member

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    i m really confused with such sorts ,but your article gave some concept
     
  3. peterdrew

    peterdrew New Member

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    It's very good; thanks!
     
  4. fashionbop

    fashionbop New Member

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  5. Scripting

    Scripting John Hoder

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    Really cool, shabbir.

    THX for this
     
  6. shabbir

    shabbir Administrator Staff Member

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    The pleasure is mine.
     
  7. qingqing

    qingqing Banned

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    Thanks for the blog loaded with so many information. Stopping by your blog helped me to get what I was looking for.
     
  8. gurumanoh

    gurumanoh New Member

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    good work...
     
  9. wdliming

    wdliming New Member

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    good!! I like it!!
     
  10. kien_vn

    kien_vn New Member

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    It's very clear to understand - "What is pointer"
     
  11. kumardashish

    kumardashish New Member

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    hi..
    your first program gives a warning because you r trying to store the address of const int variable into int type pointer.
    here is the warning::" different 'const' qualifiers"
    this is the reason you can change the value of const variable.if you create const type pointer variable then it gives an error : l-value specifier...
    try this code..
    Code:
    void main()
    {
        int const i=123;
        int *const ip;
        
        /* Changing constant integer i.e. const int/int const */
        
        printf(" %d %x \n",i,&i);
        ip=&i;
        *ip=456;
        printf(" %d %x %x %d \n",i,&i,ip,*ip);
        
    }
     
    Last edited: Oct 21, 2010
  12. itharsh_04

    itharsh_04 New Member

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