Implementing space allocated for a character array

I hope the title did a decent job of summarizing my issue. I am trying to extend a two dimensional algorithm to a 3 dimensional solution, and am having some trouble storing my slice files. I have an image object I, which has a member pt that is a pointer to an array of unsigned chars. My code looks like this:

Code:

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istream& operator>>(istream& is, Image& I)
{
        if(I.pt) {
                cout << "test\n";
                delete [] I.pt;
                I.width = 0;
                I.height = 0;
        }

        .
        .
        .

        // Now we are ready to read the data
        I.pt = new unsigned char [I.getDataSize()];
        I.pt[1000] = (unsigned char)1;
        cout << I.pt[1000] << endl;
        cout << sizeof(I.pt) / sizeof(unsigned char) << endl;
        cout << I.getDataSize() << endl;
        cout << sizeof(I.pt) << endl;
        cout << sizeof(unsigned char) << endl;

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and, my output is:

Code:

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8
94633984
8
1

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This is some pointless code that I am displaying hoping to resolve a larger issue. I am expecting to see:

Code:

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1
94633984
94633984
94633984
1

---

What am I not understanding about the size of space pointed to by I.pt after the new statement?

Quote:

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cout << sizeof(I.pt) / sizeof(unsigned char) << endl;

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Prints the size of a pointer, divided by the size of an unsigned char. Since sizeof(unsigned char) = 1, the answer depends on the pointer size on your system. Commonly 4 or 8. In your case, it seems to be 8, so I'd be guessing you're on a 64-bit system.

Quote:

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cout << I.getDataSize() << endl;

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Probably gonna get the size of the array pointed to by pt. Your output seems to corroborate this.

Quote:

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cout << sizeof(I.pt) << endl;

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Prints the size of a pointer, thus 8.

Quote:

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cout << sizeof(unsigned char) << endl;

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Prints the size of an unsigned char, thus 1.

So, the output you are getting is correct.

The first cout statement is what is confusing me the most (i.e. why I'm seeing a blank line rather than '1'). Could you address that output, or lack thereof? Thanks

You are sending an unsigned char to cout, hence it is interpreted as a printable character and will be printed as such.

Thanks a bunch! My pointer had the correct values stored all along.