I have an assembly row which the following information:

EBP = 006FFB50
SS  = 002B  

When I looked on this row:

mov eax,dword ptr ss:[ebp+8]  

I assumed that EBP + 8 = 006FFB50 + 8 = 006FFB58.
But according to x64DBG, this is the result:

dword ptr [ebp+8]=[006FFB58]=006FFC98  

I don't understand how it was calculated.
Why ss:[006FFB58] is equal to 006FFC98?

Picture for reference with more information:
enter image description here

1 Answer 1


Registers ss, cs, ds, es, gs, fs are special. They are called segment registers and contain not addresses but selectors.

A selector is used by the CPU as a reference to a segment - area of memory with a specific base (start address), limit (end address) and permissions.

Selectors and segments are set up by the OS and in theory there may be many different segments, however in practice all modern OSes use flat memory segments (0 to 0xFFFFFFFF for 32-bit processes) for the standard code and data segments (ss, cs, ds)1. This means that in the expression ss:[ebp+8], only the value of EBP is used for calculating the address. In your case it is indeed correct that

EBP + 8 = 006FFB50 + 8 = 006FFB58

which matches the value shown in brackets.

However, the value after the = sign in the debugger is not the result of the calculation but the value which is present in memory at that address. If you open a memory dump and go to address 006FFB58, you should see 006FFC98 there.

The brackets in the debugger hint signify memory dereference, similarly to the assembly syntax.

1 gs and fs are treated differently and are usually used for Thread Local Storage (TLS) block which is different for every thread and does not start at 0 so e.g. fs:0 does not map to the RAM address 0.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.