Presentation is loading. Please wait.

Presentation is loading. Please wait.

Chapter 10 The Assembly Process. What Assemblers Do Translates assembly language into machine code. Assigns addresses to all symbolic labels (variables.

Similar presentations


Presentation on theme: "Chapter 10 The Assembly Process. What Assemblers Do Translates assembly language into machine code. Assigns addresses to all symbolic labels (variables."— Presentation transcript:

1 Chapter 10 The Assembly Process

2 What Assemblers Do Translates assembly language into machine code. Assigns addresses to all symbolic labels (variables and labels on statements). Determines initial state of the memory (program and data) in order to execute a program.

3 Macros Some assemblers allow the programmer to define a sequence of instructions to be associated with a keyword. The statements in the macro replace the statement that uses the macro name. Functions much like an include directive. Some assemblers are advanced and allow for symbolic replacement of arguments.

4 Smart Assemblers Typical assemblers translate the assembly language to machine language in a 1 to 1 method. A smart assembler is a little bit like what a compiler does; translates a single statement into multiple machine language statements. MIPS Assembly Language uses a smart assembler, True Assembly Language is 1- to-1 with machine language.

5 True Assembly Language Defines the machine architecture. I.e., what must be built into the machine. TAL arithmetic and logic instructions –Must have 3 operands –MAL allowed 2 for an abbreviated instruction. –MAL allowed for 3 registers or 2 registers and an immediate –TAL requires a different instruction for an immediate operand

6 Immediate Operands MAL allows the following –Add $t1,$t2,$t3 –Add $t1, $t2, 8 In TAL, the first is fine, but for the immediate operand, you would use –Addi $t1, $t2, 8 Other immediate instructions are: andi, lui, ori, xori. –Must work with only 16 bits.

7 Shift Recall the 3 shift instructions in MAL: –Srl –Sra –Sll Can have 3 register operands or 2 registers and a count (immediate). In TAL, if you want to use 3 register operands, you must use srlv, srav, sllv –Shift a variable amount

8 Multiplication In MAL, you could use: –Mul $t1, $t2, $t3 But we know that multiplying two 32 bit values could produce a 64 bit value. The machine (and TAL) multiply the two operands and place the result in a special set of registers called HI and LO So the mul instruction above is translated as –Mult $t2, $t3 –Mflo $t1 #move from lo to reg. t1 –Should check HI for non zero value

9 Division Similar to multiplication, MAL has: –Div $t1, $t2, $t3 –Rem $t1, $t2, $t3 The machine (and TAL) places the quotient in LO and the remainder in HI So we have the above translating into: –Div $t2, $t3 –Mflo $t1 #place the quotient into t1 –Div $t2, $t3 –Mfhi $t1 #place the remainder into t1

10 Unsigned Arithmetic May want to look at the bits as bigger unsigned values rather than twos complement Or, you may want to ignore overflow. Unsigned arithmetic ignores overflow. Addu, subu, multu, divu, addiu

11 Branch Instructions Unconditional –MAL uses: b label –TAL uses: j address MAL has many branch instructions, recall –Beq, bne, blt, bgt, ble, bge, bltz, bgtz, blez, bgez, bnez, beqz TAL only has –bltz, bgtz, blez, bgez, bne, beq

12 Branching In MAL, you would do something like –Blt $t1, $t2, again The assembler translates this to the following TAL code: –Sub $at, $t1, $t2 –Bltz $at, again The MAL beqz and bnez can be translated using $0 as an operand with beq and bne.

13 Problems Look at blt $t1, $t2, repeat, where –$t1 has close to the largest positive value and –$t2 has close to the most negative value When we do sub $at, $t1, $t2 we will get a value too big to be held -> overflow Could use subu which has no overflow. –But does not give the correct answer for branching

14 Comparison Solution Create a new instruction to do comparisons that will not have overflow problems (does not use subtraction). The following TAL instruction does this: –Slt $10, $11, $12 –Sets $10 to 1 if $11 is less than $12. –Sets $10 to 0 otherwise. –Also available in slti form Now use with bne, beq, etc. for branching

15 Load Address MAL has a load address (la) instruction. TAL does not. Let’s review what is being accomplished. –La $8, begin –Puts the 32 bit address of the label begin into register $8. Can’t have all 32 bits as an immediate operand, only 16. Use two immediate instructions each with 16 of the 32 bits.

16 Loading a register Need an instruction to load the higher 16 bits and another to load the lower 16 bits. Lui – load upper immediate – loads the upper half of the register with the value specified. Leaves the lower half with 0’s. Ori – or immediate. Logical oring of the rightmost 16 bits with the specified value.

17 Loading an address First you need the address to load. As a programmer, you should not deal with this (know how it is done by the assembler, but you should not hard code with fixed addresses). The assembler determines the address, say, 0X4083f3a4 (the 0X indicates base 16). –lui $8,0x4083 –Ori $8, $8, 0xf3a4

18 Machine Code Format The MIPS machine instructions are all 32 bits in length. Some bits are for the operation (opcode), and then some are for the operands. The immediate instructions have the following format: (R s – source register, R t – target reg.) 6-bits OpcodeR2R2 R1R1 Immediate operand

19 Immediate Example The instruction: –Addi $13, $7, 50 would be translated as

20 Arithmetic Logic Instructions Have the layout: The opcode is always The exended opcode specifies the function. Appendix C has opcodes and extended opcodes. OpcodeR2R2 R3R3 R1R1 Extended opcode

21 Arithmetic Instruction Example Let’s take the instruction: –Add $13, $7, $8 Converted to machine language would be:

22 The Assembly Process Need to be given where the data segment and text segment will start. The assembler will first go through the code creating a symbol table –This table has pairs (symbol name, address) as entries. –Just looks at labels. –Looks at commands and increments a counter While doing this, can replace MAL with TAL statements (and adjust the counter).

23 Pass 2 We needed to create a symbol table first since a branch instruction may refer to a label that we have not seen yet. Now we can go through the TAL program translating the instructions into machine language. Now we know the addresses, so we can fill them in.

24 Branching A branch statement has an address as an operand. This takes 32 bits. But we cannot use 32 bits for one operand of an instruction. The branch destination is calculated by the cpu as the contents of the PC+offset (immediate in the instruction – 16 bits). So the offset = address of destination – (address of branch instruction +4) Why +4???

25 More Scope This allows us to branch to destinations a distance from the PC between and Each instruction take 4 bytes. An instruction address is divisible by 4 Only branch to instruction addresses In binary, what do all values divisible by 4 have in common? Leave them out Now the 16 bits really represent values from to

26 Jump The jump instruction has only one operand So we have 26 bits for the address. Still not a 32 bit address  Can do the same thing as with branch –Don’t store the last 2 zero bits since all destination addresses are divisible by 4 Now we can represent 28 bits with just 26 bits. –Still not a 32 bit address 

27 Jump address We need 4 more bits. This will be the high order 4 bits of the PC So the cpu will create the jump address by taking the 26 bits in the operand, adding two 0’s to the right and copying the left 4 bits of the PC into the left 4 bits of the jump address. Reverse this process to create the jump address

28 Long Branches Recall that we can only branch a distance of about 2 17 from the current instruction. What if we need a conditional branch for a longer distance. Bgtz $5, long_away Change to Blez $5, around J long_away Around: ….

29 Program Relocation Each time a program is loaded into memory, it may be loaded into a different location. –We can have several programs running on a machine at any time. Addresses need to be adjusted based on where the program starts to load. This is the job of the loader

30 Absolute vs. Relative Address Not all addresses need to be adjusted. Some addresses are absolute addresses –Those used in a la instruction –Those used in a jump –Etc Some addresses are relative addresses –Those used in a branch instruction (relative to the PC). Relative addresses do not need to be adjusted (relocated).


Download ppt "Chapter 10 The Assembly Process. What Assemblers Do Translates assembly language into machine code. Assigns addresses to all symbolic labels (variables."

Similar presentations


Ads by Google