Pointer conversion issues

**bremenpl** · 12-21-2015

Hello there,
I have a question regarding pointers conversions (not sure if i got that right in English). Lets say I am on a 32 bit system, little endian.

There is an 8 bit variable:

Code:

uint8_t var8 = 0x12;

This variable exists somewhere in the memory space.

Now I create a 64 bit type pointer without memory allocation:

Code:

uint64_t* ptr64 = 0;

And assign my 8 bit var address to it:

Code:

ptr64 = (uint64_t*)&var8;

When I will look at the LSB of value under ptr64 it will be 0x12. But what will be under other 56 bits? For my tests, it gave 0's, but I feel i was lucky. By doing this conversion, am i playing with segmentation fault, as I am overlaping some memory next to my pointer?

I would apreciate all help!

**laserlight** · 12-21-2015

The rule is:

Originally Posted by C11 Clause 6.3.2.3 Paragraph 7

A pointer to an object type may be converted to a pointer to a different object type. If the resulting pointer is not correctly aligned for the referenced type, the behavior is undefined. Otherwise, when converted back again, the result shall compare equal to the original pointer. When a pointer to an object is converted to a pointer to a character type, the result points to the lowest addressed byte of the object. Successive increments of the result, up to the size of the object, yield pointers to the remaining bytes of the object.

The way I see it, a pointer to uint64_t is not correctly aligned for a single uint8_t, related to your question about what happens to the rest of the bits. The behaviour is undefined.

**bremenpl** · 12-21-2015

Hello thank you for answer,
But what does it mean that it is not correctly aligned? Do you mean that uint8 is not placed as lsb of uint64 when converted?

**laserlight** · 12-21-2015

Originally Posted by bremenpl

But what does it mean that it is not correctly aligned? Do you mean that uint8 is not placed as lsb of uint64 when converted?

No, my interpretation of the standard is that this means that a pointer to a uint64_t object must be able to access what it points to in multiples of sizeof(uint64_t), i.e., to reinterpret the underlying object as an array of uint64_t. But this cannot be done for a single uint8_t (unless sizeof(uint64_t) == 1, but that would be so unusual that you should have mentioned it), so I reason that this means the behaviour is undefined.

**bremenpl** · 12-21-2015

I think I understand... Also I meants sizeof(uint64_t) = 8. But I dont know why is it undefined? If lets say one creates a union like this:

Code:

typedef union
{
     uint64_t var64;
     uint8_t bar8[8];
} anUnion_t;

anUnion_t myUnion;

I can do:

Code:

uint64_t newVar64 = myUnion.var64;

And this would work too:

Code:

newVar64 = 0;
size_t i = 0;
for (i = 0; i < sizeof(uint64_t); i++)
     newVar64 |= myUnion.var8[i] << (i * sizeof(uint64_t));

Do I understand correctly?

**laserlight** · 12-21-2015

Originally Posted by bremenpl

But I dont know why is it undefined?

Because a uint64_t cannot validly point to a single uint8_t: there are missing bytes. If you have a box that fits 8 identical chocolates, putting in only one such chocolate and claiming that you have a full box of chocolates is a lie.

Originally Posted by bremenpl

Do I understand correctly?

Yes. Strictly speaking the standard does not make any such guarantee for unions, and for pointers it just says that you can safely convert to and from the destination pointer, but in practice it should work as expected.

**bremenpl** · 12-21-2015

Ok, thank you very much for explanation.

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