I thought pointers were pointers...

Quote:

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Does this differ in C++? In C++ are arrays not glorified pointers?

To demonstrate:

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I was hoping you would demonstrate the difference yourself using sizeof as was mentioned earlier, but you did not :p

To demonstrate:

Code:

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#include <iostream>

int main() {
    char pointer[] = "keira";
    char* otherPointer = pointer;
    std::cout << sizeof(pointer) << " " << sizeof(otherPointer) << std::endl;
    return 0;
}

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On my system, the output is "6 4". Basically, C++ has the same distinction between arrays and pointers as C.

Alright thank you for your help everyone and sorry I'm so thick headed... This is a concept I thought I had completely locked down after spending many months in assembly and C++.

Quote:

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Originally Posted by keira

Why is the size of c being read as the size of a 32bit pointer instead of the size of the buffer it points to? What is different about the pointers?

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Nothing is different. c is a pointer. Therefore sizeof(c) is the size of the pointer. Why would you expect anything different?

Quote:

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I don't understand where you got the idea that I was comparing dynamically allocated memory with local memory...

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When you use pointer you usually use dynamically allocated memory. What is the local memory then? You mean arrays?

Local would be stack memory, dynamic would be heap memory.

The expression: char* string = "keira"; Creates both the pointer string and the consecutive ordering of bytes that translate to "keira" on the stack.

Basically to answer my original question, this was what i was looking for: (taken from http://c-faq.com/aryptr/aryptrequiv.html)

That is, whenever an array appears in an expression, the compiler implicitly generates a pointer to the array's first element, just as if the programmer had written &a[0]. (The exceptions are when the array is the operand of a sizeof or & operator, or is a string literal initializer for a character array. [footnote] See questions 6.23, 6.12, and 1.32, respectively.)