Question about memory address

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  1. #1
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    Question about memory address

    Hi I have another question. I was reading my tutorial and there's a sample program to count the number of characters an user entered with a pointer array.
    Code:
    // Ex3_10.cpp
    // Counting string characters using a pointer
    #include <iostream>
    using namespace std;
    
    int main()
    {
       const int MAX = 80;                 // Maximum array dimension
       char buffer[MAX];                   // Input buffer
       char* pbuffer=buffer;               // Pointer to array buffer
       cout << endl                        // Prompt for input
            << "Enter a string of less than "
            << MAX << " characters:"
            << endl;
       cin.getline(buffer, MAX, '\n');     // Read a string until \n
    
       while(*pbuffer)                     // Continue until \0
          pbuffer++;
    
       cout << endl
            << "The string \"" << buffer
            << "\" has " << pbuffer-buffer  << " characters.";
    
       cout << endl;
       return 0;
    }
    What I don't get is the bolded part, pbuffer-buffer. How does that output the number of characters? Wouldn't that show the bytes of memory the string takes up since they point to memory address?
    Thanks in advance for whoever answers this.

  2. #2
    Registered User manofsteel972's Avatar
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    A char is equal to 1 byte always. So it works out to be the same.
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  3. #3
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    Exclamation Pointer Arithmetic

    A char is equal to 1 byte always.
    WRONG! (Sorry, manofsteel972)

    The language standard requires that a char hold at least one byte (8 bits).

    Actually, this has to do with Pointer Arithmetic. This is a feature of the language that lets you, for example, incriment pointers without worrying about the size-of the object being pointed-to.

  4. #4
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    The language standard requires that a char hold at least one byte (8 bits).
    That may be what the standard says, but how exactly the compiler implements the standard is... well.... compiler specific. Most compilers that I know of use 1 byte and 1 byte only.

  5. #5
    and the hat of wrongness Salem's Avatar
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    It doesn't matter what type you have, you get the same answer

    Code:
    int array[10];
    int *p1 = array;
    int *p2 = &array[8];
    ptrdiff_t n = p2 - p1;
    n will have the value 8, no matter what type you choose.
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