How to print the value of char pointers using cout?

This is a discussion on How to print the value of char pointers using cout? within the C++ Programming forums, part of the General Programming Boards category; Code: #include <iostream> int main() { char *ch = "123"; std::cout << static_cast<void *>(ch); return 0; } I know I ...

  1. #1
    C/C++Newbie Antigloss's Avatar
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    How to print the value of char pointers using cout?

    Code:
         #include <iostream>
    
         int main()
         {
             char *ch = "123";
             
             std::cout << static_cast<void *>(ch);     
             
             return 0;
         }
    I know I can do it this way. But are there better ways to do that?

  2. #2
    Registered User
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    No. and If you consider how often you'll need to be printing pointer values it shoudln't really bother you.

  3. #3
    Registered User
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    Instead of using the C++ hammer, you can use the hammer that comes from C ....

    Code:
    #include <iostream>
    
    int main()
    {
             char *ch = "123";
             
             std::cout << ((void *)ch);     
             
             return 0;
    }
    This is functionally equivalent in this case, but a little terser. If that meets your definition of "better", go for it.

  4. #4
    C/C++Newbie Antigloss's Avatar
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    No. Sure I don't think less typing is better.

    Thanks guys~~ :-)

  5. #5
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    maybe not "better" but "different"
    Code:
     #include <iostream>
    
         int main()
         {
             char *ch = "123";
             
             printf("%p\n",ch);     
             
             return 0;
         }

  6. #6
    C/C++Newbie Antigloss's Avatar
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    Sure I know it. That's why I wrote "using cout" at the title

  7. #7
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    [rant]printf() is about as simple as it can be done. cout (and all of iostreams/fstream c++ library) is just plain-old ugly, overly-complicated and horribly implemented. I refuse to use it in all but the simplest cases. [/rant] But that doesn't answer your question

  8. #8
    semi-colon generator ChaosEngine's Avatar
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    Wink

    Quote Originally Posted by Ancient Dragon
    [rant]printf() is about as simple as it can be done. cout (and all of iostreams/fstream c++ library) is just plain-old ugly, overly-complicated and horribly implemented. I refuse to use it in all but the simplest cases. [/rant]
    I agree. Why would you use a type-safe and extensible library when you can use some lovely vararg function that is the cause of more buffer overflows and security loopholes then you can count?

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