Is pointer just an integer?

Question

If I know the address of an data object, could I store the address as an integer and operate the integer as a pointer?

For example,

void main(){
    long a = 101010;
    long *p = &a;
    long b = p;
    printf("%lld\n", *(long*)b);
}

Is it always safe?

Comments: long b = p; produces a warning:

Initialization makes integer from pointer without a cast

However, the program prints 101010.

Answer 1

It's not guaranteed by the standard that such cast would always work.

To store a pointer in an integral type, use intptr_t (or its unsigned cousin uintptr_t). It's guaranteed to convert void * pointers to such types and convert back, resulting the same value.

Note that these types are optional.

Answer 2

sizeof(long) is typically defined by the compiler (subjected to the underlying HW architecture).

sizeof(long*) is subjected to the size of the virtual memory address space on your platform.

For example, with the Visual Studio compiler for 64-bit operating system and x64-based processor:

• sizeof(long) == 4
• sizeof(long*) == 8

Therefore:

• With long b = p, only the 4 least significant bytes of p are copied into b
• With *(long*)b, you are potentially attempting to access an invalid memory address

In this example, if the 4 most significant bytes of p are zero, then "no harm is done". But since it is not guaranteed to be the case, when sizeof(long) != sizeof(long*) this code is generally unsafe.

In addition to that, even if sizeof(long) == sizeof(long*), you should still refrain from using this type of conversions (pointer-to-integer), in order to keep your code portable to other platforms.

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UPDATE

Please note that printf("%lld\n", *(long*)b) is also unsafe.

You should basically use "%ld" for long values and "%lld" for long long values.

If sizeof(long) < sizeof(long long), then it may lead to a memory access violation during runtime.