This was discussed in the first portion of this tutorial: please review before proceeding. Getting the Virtual Address The next thing we need to do is to compile and run the program, which we’ll debug, on Windows. When running the program on Windows, the following will be displayed, because we’ve coded the “int 3” software interrupt into the C++ code. When the “int 3” instruction is reached, the interrupt will be invoked, which will cause WinDbg to pause the execution of whole Windows operating system. WinDbg will then present us with a message about “break instruction exception,” as can be seen on the picture below:

Let’s now list the whole assembly function that corresponds to the above C++ code. The assembly code can be seen in the output below. The first instruction is the “int 3” interrupt instruction that was used to stop the Windows operating system and invoke a debugger. Now we can use the Windbg t command to step into, or the p command to step over, the instructions manually. [plain] kd> u 0x004113ae L40 virtualphysical+0x113ae: 004113ae cc int 3 004113af 6a04 push 4 004113b1 e8c1fdffff call virtualphysical+0x11177 (00411177) 004113b6 83c404 add esp,4 004113b9 898520ffffff mov dword ptr [ebp-0E0h],eax 004113bf 83bd20ffffff00 cmp dword ptr [ebp-0E0h],0 004113c6 741a je virtualphysical+0x113e2 (004113e2) 004113c8 8b8520ffffff mov eax,dword ptr [ebp-0E0h] 004113ce c70000000000 mov dword ptr [eax],0 004113d4 8b8d20ffffff mov ecx,dword ptr [ebp-0E0h] 004113da 898d18ffffff mov dword ptr [ebp-0E8h],ecx 004113e0 eb0a jmp virtualphysical+0x113ec (004113ec) 004113e2 c78518ffffff00000000 mov dword ptr [ebp-0E8h],0 004113ec 8b9518ffffff mov edx,dword ptr [ebp-0E8h] 004113f2 8955ec mov dword ptr [ebp-14h],edx 004113f5 c745f80a000000 mov dword ptr [ebp-8],0Ah 004113fc 8b45ec mov eax,dword ptr [ebp-14h] 004113ff c70014000000 mov dword ptr [eax],14h 00411405 8bf4 mov esi,esp 00411407 8b45f8 mov eax,dword ptr [ebp-8] 0041140a 50 push eax 0041140b 68b0574100 push offset virtualphysical+0x157b0 (004157b0) 00411410 ff15b8824100 call dword ptr [virtualphysical+0x182b8 (004182b8)] 00411416 83c408 add esp,8 00411419 3bf4 cmp esi,esp 0041141b e816fdffff call virtualphysical+0x11136 (00411136) 00411420 8bf4 mov esi,esp 00411422 8b45ec mov eax,dword ptr [ebp-14h] 00411425 8b08 mov ecx,dword ptr [eax] 00411427 51 push ecx 00411428 683c574100 push offset virtualphysical+0x1573c (0041573c) 0041142d ff15b8824100 call dword ptr [virtualphysical+0x182b8 (004182b8)] 00411433 83c408 add esp,8 00411436 3bf4 cmp esi,esp 00411438 e8f9fcffff call virtualphysical+0x11136 (00411136) 0041143d 8bf4 mov esi,esp 0041143f ff15bc824100 call dword ptr [virtualphysical+0x182bc (004182bc)] 00411445 3bf4 cmp esi,esp 00411447 e8eafcffff call virtualphysical+0x11136 (00411136) 0041144c 33c0 xor eax,eax 0041144e 5f pop edi 0041144f 5e pop esi 00411450 5b pop ebx 00411451 81c4e8000000 add esp,0E8h 00411457 3bec cmp ebp,esp 00411459 e8d8fcffff call virtualphysical+0x11136 (00411136) 0041145e 8be5 mov esp,ebp 00411460 5d pop ebp 00411461 c3 ret [/plain] We’re particularly interested in the virtual address that’s been used for the variables x and y in the C++ code. The values of those variables are stored at address 0x004113f5, where we can see a constant 0xAh being saved to it, and at address 0x004113ff, where a constant 0x14h is being saved to it. We could step through the program instruction by instruction with p or t command, but we’d rather set two breakpoints on the interesting addresses with the bp command like this: [plain] kd> bp 004113f5 kd> bp 004113ff kd> bl 0 e 004113f5 0001 (0001) virtualphysical+0x113f5 1 e 004113ff 0001 (0001) virtualphysical+0x113ff [/plain] The first two commands set breakpoints on the addresses 0x004113f5 and 0x004113ff, while the third cl command displays all the set breakpoints, which are the breakpoints we’ve just set. Then we can use the g command to run the program, upon which the breakpoint 0 will be hit as can be seen on the picture below:

Along with the actual breakpoint ID and its address, the current line that’s about to be executed is also displayed. On that line we can see that we’re interested in the address [ebp-8], so we must calculate it by printing the value of register EBP, subtracting 8 from it, and dumping the memory at that address. Let’s first display the value of EBP and dump the memory:

The variable x is located at the address 0x0012ff60 and its content is 0xcccccccc. After running the command that stores 0xAh at that address, the content should be 0x0000000A, as can be seen on the picture below:

We’ve gotten our first virtual address, which is 0x0012ff60. The same steps can be repeated for the y variable as well. All of the commands can be seen on the picture below:

The value 0x14 was written to the address 0x00345988, which is the second address we’re interested in. Thus, we’ve successfully gotten the two virtual addresses that we wanted. The virtual addresses are summarized below:

variable x: 0x0012ff60 variable y: 0x00345988

Getting the Linear Address The first thing that we must do is figure out in which segment the address actually is. It’s clear that the variable x, which address is 0x0012ff60, is on the stack segment, since we’re initializing the variable on the stack. Let’s print all of the values of stack segments with the use of the r command. The results can be seen on the picture below:

The registers are actually 16-bit, so the values correspond to the following:

SS (Stack Segment) : 0x0023 CS (Code Segment) : 0x001b DS (Data Segment) : 0x0023 ES (Extra Segment) : 0x0023 GS (Data Segment) : 0x0000 FS (Data Segment) : 0x003b

Right now you may be confused and thinking: okay, if segmentation is in use, why do the segment registers hold the same value no matter which process is currently being debugged? Shouldn’t every process have its own segment register values? After all, the code and data between the processes is not shared. This isn’t true for shared libraries, but right now we’re not talking about DLL files, just executables. You can be scratching your head and trying to figure out what’s happening, but all in all the reason is very simple. Let’s take a look at the stack segment register for example: 0x0023 can be transformed into binary form, which in this case is: 0000 0000 0010 0011. The first two least significant bits are used for protection, but we won’t go into that right now. The third bit is used to declare whether we should be looking for the descriptors in the global or local descriptor table. In this case the third bit is set to 0 (the 0 is bold in the binary representation), which means that the appropriate descriptor is located in the global descriptor table. The rest of the bits make up an offset into the GDT to specify the right descriptor to be used: those bits are 0000000000100, which can easily be represented with 0x4 in hexadecimal form. Let’s use the same formula to get the offset of all segment registers:

SS: 0x4 CS: 0x3 DS: 0x4 ES: 0x4 GS: 0x0 FS: 0x7

Let’s also use the “dg 0 40” command to print the first part of the GDT table, which can be seen on the picture below:

Let’s take the virtual address that we’ve gotten in the previous step and convert it to a linear address. The virtual address can be converted to a linear address by taking the base address from the GDT descriptor table (of an appropriate index that’s specified by one of the segment registers) and adding the virtual address to it. The base address of the first five segment descriptors that span the entire linear address space is 0x00000000 and the virtual addresses are 0x0012ff60 and 0x00345988, which can be directly translated like this:

variable x : 0x00000000 + 0x0012ff60 = 0x0012ff60 variable y : 0x00000000 + 0x00345988 = 0x00345988

This proves that segmentation is being used because it must be used and cannot be turned off, but it doesn’t actually do anything. We can conclude that virtual addresses are the same as linear addresses and no translation is necessary to translate from one to the other. Conclusion In this tutorial, we’ve looked at how to figure out whether PAE is enabled, but we’ve also started to look at an example and resolved virtual to linear addresses. We saw that the Windows operating system doesn’t actually use segmentation, since the virtual addresses are the same as linear addresses. References: [1] x86 memory management and Linux kernel, accessible at [2] [3] W4118: segmentation and paging, accessible at [4] Common WinDbg Commands, accessible at [5] Understanding !PTE , Part 1: Let’s get physical, accessible at [6] Understanding !PTE, Part2: Flags and Large Pages, accessible at [7] Part 3: Understanding !PTE – Non-PAE and X64, accessible at