CODEWITHSUNDEEP
c++arrayspointerscasting
NGambit
NGambit

Reputation: 1181

C++ pointer vs array and pointer casting

I am trying the following code:

unsigned long * foo = (unsigned long *) 0x200000;

So, as I understand it, foo points to the unsigned long 0x200000. Now, if I try,

std::cout<<foo[0];

I thought this should print the value 0x200000 (may be in decimal). Because, foo[0] = *(foo + 0) = 0x200000. But, it actually prints 0.

What I am missing here?

Upvotes: 1

Views: 233

Answers (2)

Daniel Fischer
Daniel Fischer

Reputation: 183878

You misunderstand,

unsigned long * foo = (unsigned long *) 0x200000;

interprets 0x200000 as the address of an unsigned long. Since it is extremely unlikely that that is the address of an unsigned long that you can legitimately access in your programme,

std::cout<<foo[0];

is almost certainly undefined behaviour, and likely causes a segfault. In your case, though, accessing that memory location didn't cause a segfault and the bits there were interpreted and printed as an unsigned long, they happened to be all 0.

Upvotes: 5

SwiftMango
SwiftMango

Reputation: 15284

type * pointerName = addressItPointsTo;

pointerName[0] is same as *pointerName, means dereference the pointer and get the value at the address it points too. It just happens to be 0, however it usually cause a segmentation fault.

Upvotes: 0

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