Skip to content

Stage 7 : ABI and XEXE Format (2 Hours)

Learning Objectives

  • Familiarise with the Application Binary Interface(ABI) of eXpOS.
  • Modify the INIT program to comply with the eXpOS ABI.

Pre-requisite Reading

Read and Understand the eXpOS Virtual Address Space Model and XEXE Executable File Format from eXpOS ABI Documentation before proceeding further.

In this stage we will rewrite the user program and OS startup code of Stage 6 in compliance with expos ABI.

Modifying INIT

The INIT program must be modified to comply with the XEXE executable format. The executable format stipulates that the first 8 words of the file must contain a header. The rest of the file contains the program instructions. The OS is expected to load the file into logical pages starting from page 4. Thus the first disk block of the program is loaded into logical address starting from 2048, second (if the file size exceeds 512 words) to logical addresses starting from 2560 and so forth.

Since the first instruction starts after the 8 word header, the first instruction in the program will be loaded into memory address 2056. Since each instruction requires two words, the second instruction will start at memory address 2058 and so on. Thus the jump addresses in the INIT program must be designed with this in mind.

The INIT program complying to ABI is given below. The code is given in bold and the corresponding addresses are added for reference.


2048  0
2049  2056
2050  0
2051  0
2052  0
2053  0
2054  0
2055  0
2056  MOV R0, 1
2058  MOV R2, 5
2060  GE R2, R0
2062  JZ R2, 2074
2064  MOV R1, R0
2066  MUL R1, R0
2068  BRKP
2070  ADD R0, 1
2072  JMP 2058
2074  INT 10

Modifications to OS Startup Code

1) Load Library Code from disk to memory

loadi(63,13);
loadi(64,14);

The eXpOS ABI stipulates that the code for a shared library must be loaded to disk blocks 13 and 14 of the disk. During OS startup, the OS is supposed to load this code into memory pages 63 and 64. This library code must be attached to logical page 0 and logical page 1 of each process. Thus, this code will be shared by every application program running on the operating system and is called the common shared library or simply the library.

The library provides a common code interface for all system calls. This means, to invoke a system call, the application can call the corresponding library function and the library will in turn invoke the system call and return values back to the application. The library also implements some functions like dynamic memory allocation and de-allocation from the heap area.

The dynamic memory allocation functions of the library manage the heap memory of the application program. The ABI stipulates that each application must be provided 2 pages of memory for the heap. These two pages must be attached to logical pages 2 and 3 of the application.

Note here that the library code is not part of the application's XEXE executable file. The library code is "attached" to the address space of the application when the application is loaded into memory for execution. Since the ABI stipulates the the library will be loaded to logical pages 0 and 1, the application "knows" the logical address of the library routines and will contain call to these routines, though the routines are not present in the application's code.

Thus, the OS must do the following to ensure correct run time linkage of library code to each application.

i. The library code must be pre-loaded to disk blocks 13 and 14 before OS startup.The library code can be found in the expl folder in eXpOs package.Load it into the XSM disk using the xfs-interface

load --library ../expl/library.lib

ii. During OS start-up, this code must be loaded to memory pages 63 and 64.

iii. When each application is loaded for execution, the logical pages 0 and 1 must be mapped to physical pages 63 and 64.

iv. Two physical pages must be allocated for the application's heap and attached to logical pages 2 and 3.

2) Modify the Page table entries according to ABI.

The SPL constantPAGE_TABLE_BASE holds the value 29696. A total of 16 page tables can be stored starting from this address. Each page table will be 20 entries. For each user process, one page table will be allocated.

Here since we are running the first user program, we will use the first few entries of this memory region for setting up the page table for the INIT process.

As noted, the first disk block of the INIT program (block 7) must be loaded to logical page 4. Similarly, block 8 must be loaded to logical page 5. The ABI stipulates that two pages must be allocated for the stack at logical pages 8 and 9.

The following code sets page table entries for logical page 4 and 5(for code area), logical page 8 and 9(for user stack), logical pages 2 and 3(for heap) and logical pages 0 and 1(for library). Since pages 0 to 75 are reserved for the use of the OS kernel, the first four free pages (76,77,78 and 79) will be allocated for stack and heap area. See Memory Organisation. Note that the code and library pages must be kept read only where as stack and heap must be read-write.(see page tablesettings for details).

//Library
[PTBR+0] = 63;
[PTBR+1] = "0100";
[PTBR+2] = 64;
[PTBR+3] = "0100";

//Heap
[PTBR+4] = 78;
[PTBR+5] = "0110";
[PTBR+6] = 79;
[PTBR+7] = "0110";

//Code
[PTBR+8] = 65;
[PTBR+9] = "0100";
[PTBR+10] = 66;
[PTBR+11] = "0100";
[PTBR+12] = -1;
[PTBR+13] = "0000";
[PTBR+14] = -1;
[PTBR+15] = "0000";

//Stack
[PTBR+16] = 76;
[PTBR+17] = "0110";
[PTBR+18] = 77;
[PTBR+19] = "0110";

3) Since the total address space of a process is 10 pages, PTLR register must be set to value 10.

PTLR = 10;

4) The second entry of the header of an executable file will contain an entry point value. This is the address of the first instruction to be executed when the program is run.

Hence, you must initialise IP to the second word in the header. Since the first code page is loaded into memory page 65, the address of the second word in header is calculated as (65 * 512) + 1. This value is stored to the top of the user stack. The machine on executing IRET instructions pops this value from the stack and sets IP to that value.

SP = 8*512;
[76*512] = [65 * 512 + 1];

Making Things Work

  1. Compile and load the modified OS startup Code.
  2. Load the modified user program.
  3. Run the machine in debug mode.

Assignment 1

Change the user program to compute cubes of the first five numbers.

Back to top