CODEWITHSUNDEEP
c++c
Sadık
Sadık

Reputation: 4419

Assign memory address stored in int to int

Assuming the value of uint32_t x is an address. How can I get the value behind this address? My try was to simply assign the address to a pointer.

int*y = x;

But x is not an int pointer, it's just int with an address as value.

Upvotes: 0

Views: 104

Answers (2)

P.W
P.W

Reputation: 26800

In C++ this can be done using reinterpret_cast

8.5.1.10 Reinterpret cast [expr.reinterpret.cast]
...
5. A value of integral type or enumeration type can be explicitly converted to a pointer. A pointer converted to an integer of sufficient size (if any such exists on the implementation) and back to the same pointer type will have its original value; mappings between pointers and integers are otherwise implementation-defined. [Note: Except as described in 6.6.4.4.3, the result of such a conversion will not be a safely-derived pointer value. —end note]

And the rules in 6.6.4.4.3 state:

An integer value is an integer representation of a safely-derived pointer only if its type is at least as large as std::intptr_t and it is one of the following:
—(3.1) the result of a reinterpret_cast of a safely-derived pointer value;
—(3.2) the result of a valid conversion of an integer representation of a safely-derived pointer value;
—(3.3) the value of an object whose value was copied from a traceable pointer object, where at the time of the copy the source object contained an integer representation of a safely-derived pointer value;
—(3.4) the result of an additive or bitwise operation, one of whose operands is an integer representation of a safely-derived pointer value P, if that result converted by reinterpret_cast<void*> would compare equal to a safely-derived pointer computable from reinterpret_cast<void*>(P).

So if x (in the question) has a type at least as large as std::intptr_t and is already an integral representation of a safely derived pointer as per the rules above, you will be able to get the value behind the address stored in x.

Upvotes: 2

eerorika
eerorika

Reputation: 238391

An integer type which is large enough to represent all data pointers can be converted into a pointer using reinterpret_cast or an explicit conversion. The pointer can be indirected to get the pointed value using the indirection operator.

Note that uint32_t is not guaranteed to be large enough to be able to represent all pointer values (and in fact will not be enough on modern 64 bit cpus). uintptr_t is meant precisely for this purpose.

Note that if the pointed address does not contain an object (of compatible type), then behaviour will be undefined.

Upvotes: 3

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