What is uintptr_t data type

Question

What is uintptr_t and what can it be used for?

Answer 1

First thing, at the time the question was asked, uintptr_t was not in C++. It's in C99, in <stdint.h>, as an optional type. Many C++03 compilers do provide that file. It's also in C++11, in <cstdint>, where again it is optional, and which refers to C99 for the definition.

In C99, it is defined as "an unsigned integer type with the property that any valid pointer to void can be converted to this type, then converted back to pointer to void, and the result will compare equal to the original pointer".

Take this to mean what it says. It doesn't say anything about size.

uintptr_t might be the same size as a void*. It might be larger. It could conceivably be smaller, although such a C++ implementation approaches perverse. For example on some hypothetical platform where void* is 32 bits, but only 24 bits of virtual address space are used, you could have a 24-bit uintptr_t which satisfies the requirement. I don't know why an implementation would do that, but the standard permits it.

Answer 2

Running the risk of getting another Necromancer badge, I would like to add one very good use for uintptr_t (or even intptr_t) and that is writing testable embedded code. I write mostly embedded code targeted at various arm and currently tensilica processors. These have various native bus width and the tensilica is actually a Harvard architecture with separate code and data buses that can be different widths. I use a test driven development style for much of my code which means I do unit tests for all the code units I write. Unit testing on actual target hardware is a hassle so I typically write everything on an Intel based PC either in Windows or Linux using Ceedling and GCC. That being said, a lot of embedded code involves bit twiddling and address manipulations. Most of my Intel machines are 64 bit. So if you are going to test address manipulation code you need a generalized object to do math on. Thus the uintptr_t give you a machine independent way of debugging your code before you try deploying to target hardware. Another issue is for the some machines or even memory models on some compilers, function pointers and data pointers are different widths. On those machines the compiler may not even allow casting between the two classes, but uintptr_t should be able to hold either. -- Edit -- Was pointed out by @chux, this is not part of the standard and functions are not objects in C. However it usually works and since many people don't even know about these types I usually leave a comment explaining the trickery. Other searches in SO on uintptr_t will provide further explanation. Also we do things in unit testing that we would never do in production because breaking things is good.

Answer 3

There are already many good answers to the part "what is uintptr_t data type". I will try to address the "what it can be used for?" part in this post.

Primarily for bitwise operations on pointers. Remember that in C++ one cannot perform bitwise operations on pointers. For reasons see Why can't you do bitwise operations on pointer in C, and is there a way around this?

Thus in order to do bitwise operations on pointers one would need to cast pointers to type unitpr_t and then perform bitwise operations.

Here is an example of a function that I just wrote to do bitwise exclusive or of 2 pointers to store in a XOR linked list so that we can traverse in both directions like a doubly linked list but without the penalty of storing 2 pointers in each node.

 template <typename T>
 T* xor_ptrs(T* t1, T* t2)
 {
     return reinterpret_cast<T*>(reinterpret_cast<uintptr_t>(t1)^reinterpret_cast<uintptr_t>(t2));
  }

Answer 4

uintptr_t is an unsigned integer type that is capable of storing a data pointer. Which typically means that it's the same size as a pointer.

It is optionally defined in C++11 and later standards.

A common reason to want an integer type that can hold an architecture's pointer type is to perform integer-specific operations on a pointer, or to obscure the type of a pointer by providing it as an integer "handle".

Answer 5

It's an unsigned integer type exactly the size of a pointer. Whenever you need to do something unusual with a pointer - like for example invert all bits (don't ask why) you cast it to uintptr_t and manipulate it as a usual integer number, then cast back.