C++ reinterpret_cast

Question

In running this program:

#include <iostream>
int main()
{
char *name = "abc";
int i = reinterpret_cast<int>(name);
std::cout<<i<<std::endl;
return 0;
}

I got the following output:

4202656

What does this number represent? Is it a memory address? But, memory address of what? Isn't "abc" stored as an array of characters in memory?

Thanks.

The reinterpret_cast operator produces a value of a new type that has the same bit pattern as its argument. Which compiler are you using ? g++ gives an error for the casting. — DumbCoder, Jan 28 '11 at 11:21
@DumbCoder. When I type `char *name = "abc";`, I get the **same** output — Simplicity, Jan 28 '11 at 11:23

score 4 · Answer 1 · answered Jan 28 '11 at 11:22

It is undefined. sizeof(int) might not be equal to sizeof(char*). I'm not sure if strict aliasing rules apply here as well.

In practice however, assuming their sizes are indeed equal (most 32-bit platforms), 4202656 would represent the address of the first character in the array. I would do this more cleanly this way:

#include <iostream>
int main()
{
   const char *name = "abc"; // Notice the const. Constant string literals cannot be modified.
   std::cout << static_cast<const void*>(name) << std::endl;
}

score 3 · Answer 2 · answered Jan 28 '11 at 11:22

3

It is probably the address of the character 'a'.
Though I don;t think this is guaranteed (i.e. an int may not be long enough to hold the address).

answered Jan 28 '11 at 11:22

Martin York

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York. When I type char *name = "abc";, I get the same output – Simplicity Jan 28 '11 at 11:24
@user588855: Don't understand your comment. That's the same as your code. – Martin York Jan 28 '11 at 11:38

score 0 · Answer 3 · edited May 23 '17 at 09:58

0

You probably want to look at the question: casting via void* instead of using reinterpret_cast

The short answer is that it could be anything at all.

edited May 23 '17 at 09:58

Community

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answered Jan 28 '11 at 11:26

Clinton

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The standard does say. The result should be unsurprising to somebody that understands the underlying memory architecture. [ Note: it is intended to be unsurprising to those who know the addressing structure of the underlying machine. — end note ] – Martin York Jan 28 '11 at 11:41

score 0 · Answer 4 · answered Jan 28 '11 at 11:26

This the memory address of the first character of "abc", so "a". Because an array is a pointer who point to the first value of the array.
If you do cout << *(name++) normaly "b" is printed.

So when cast name, you try to cast the adress who point to "a"

C++ reinterpret_cast

4 Answers4