Wednesday, September 21, 2016

Transcription

1 Wednesday, September 21, 2016 Topics for today More high-level to translations Compilers and Assemblers How assemblers work Symbol tables ILC Pass 1 algorithm, Error checking Pass 2 Immediate mode and equate Assembler variants: Disassembler, Cross assembler High-level language and To further reinforce ideas about mapping high-level constructs into, here are nine more examples of programs and their possible translations 1 int a,b,c; input(a,b) c=a+b; output (c); deci a,d deci b,d lda a,d adda b,d ; a + b sta c,d deco c,d a: block 2 b: block 2 c: block 2 2 int a,c; input(a); c=a+5; output(c); deci a,d lda a,d adda five,i ; a + 5 sta c,d deco c,d a: block 2 c: block 2 five:equate 5 Comp 162 Notes Page 1 of 17 September 21, 2016

2 3 int a,b,c; input (a,b); c=(a+b)/2; output (c); deci b,d lda a,d adda b,d asra sta c,d deco c,d a: block 2 b: block 2 c: block 2 ; (a+b)/2 4 int a; input(a); output (3*a+27); deci a,d lda a,d asla ; 2a adda a,d ; 3a adda constant,i ; 3a+27 sta a,d ; used for result deco a,d a: block 2 constant:equate 27 5 int a,b; input(a,b); output("the sum is "); output(a+b); deci a,d deci b,d lda a,d adda b,d sta a,d ; a now a+b stro label,d deco a,d a: block 2 b: block 2 label:ascii "The sum is \x00" Comp 162 Notes Page 2 of 17 September 21, 2016

3 6 int a,b; input (a,b); output(a); output(" + " ); output(b); output(" = " ); output(a+b); deci a,d deci b,d lda a,d adda b,d sta temp,d deco a,d stro plus,d deco b,d stro equals,d deco temp,d a: block 2 b: block 2 temp: block 2 plus: ascii " + \x00" equals:ascii " = \x00" ; for a+b 7 int a; input(a); a = a + a%4; output(a); deci a,d lda a,d anda 3,i ; a%4 adda a,d ; a+a%4 sta a,d deco a,d a: block 2 Comp 162 Notes Page 3 of 17 September 21, 2016

4 8 int score, exam1, exam2; const int bonus = 10; int main() { output ( Enter exam scores: ) input (exam1); input (exam2); score = (exam1+exam2)/2 + bonus; output( Your semester score is: ); output(score); } br main score: block 2 exam1: block 2 exam2: block 2 bonus: equate 10 main: stro prompt,d deci exam1,d deci exam2,d lda exam1,d adda exam2,d asra adda bonus,i sta score,d stro label,d deco score,d prompt:ascii Enter exam scores: \x00 label:ascii Your semester score is: \x00 Comp 162 Notes Page 4 of 17 September 21, 2016

5 9 int A, B, C, result; int main() { output( Enter three integers: ); ; } input(a); input(b); input(c); result = (A/4) + (B%32) + (C*3) 24 output( Result is: ); output(result); stro prompt,d deci a,d deci b,d deci c,d lda a,d asra asra sta result,d ; a/4 lda b,d anda 31,i adda result,d sta result,d ; a/4 + b%32 lda c,d asla adda c,d adda result,d suba 24,i sta result,d ; a/4 + b%32 + 3*c - 24 stro label,d deco result,d a: block 2 b: block 2 c: block 2 result:block 2 prompt:ascii Enter three integers:\x00 label: ascii Result is: \x00 Compilers and Assemblers Both compilers and assemblers are translation programs Both use symbol tables that contain information about identifiers in the source program but the compiler symbol table has to record information such as type and scope of the identifiers A compiler does type checking that an assembler does not do (in assembly code everything is bits ) Comp 162 Notes Page 5 of 17 September 21, 2016

6 How assemblers work In a sentence: An assembler translates symbols to numbers There are various symbols in an assembly language program but they fall into two groups: (1) those that are part of the assembly language itself eg deco stro block (2) those that a user is free to make up (within length limits) the labels on instructions or data locations The forward reference problem In general, an assembler cannot read a program just once and output a translation as it goes along This is because a symbol can be used before it is defined Consider the following deci N,d ; input count of lines N : block 2 ; variable for line count When the assembler reads the first line, it does not (yet) know the value that N represents (the address of the line it labels) Even if the language requires users to declare all data at the top of the program we may still have forward references in branch instructions as in the first line of the following br main A: block 2 B: block 2 main: deci A,d So a typical assembler reads the source program twice (makes two passes over the text): Pass 1: get values of symbols Pass 2: do the actual translation 1 An assembler makes use of tables in translating the program and there will be two types of tables corresponding to the two types of symbol Fixed Tables - contain those symbols that are part of the language such as the opcodes and directives The structure of these tables can be optimized for fast access, eg entries ordered alphabetically 1 It is possible to write an assembler that only requires one pass but it is tricky Reading a source program twice is rarely a problem so two passes are preferred Comp 162 Notes Page 6 of 17 September 21, 2016

7 Entries in the opcode table would associate, with each symbolic opcode, information such as its numeric translation, number of operands, valid addressing modes and so on For example, Symbol Template Operand Bytes Modes Allowed ROLr r No 1 STr 1110raaa Yes 3 d,s,x,n,sf,sx,sxf STOP No 1 BR a Yes 3 i,x The fixed tables are initialized before the user program is read and do not change Variable Table - contains those symbols that the user is free to choose such as labels on instructions and data locations (think about the identifiers in a high-level language program) This table is empty before the user program is read and entries are made during Pass 1 We can depict the operation of a two-pass assembler thus: Source Program Pass 1 Pass 2 Object Program Fixed Tables Variable Table Organization of tables In theory, it does not matter how fast the assembler performs the translation as long as it is correct But in practice we would like it to run as fast as possible Much of what an assembler does is look up symbols in tables so if we can make look-ups fast, the assembly operation as a whole will be quick Comp 162 Notes Page 7 of 17 September 21, 2016

8 Fixed tables In a fixed table we know what the entries are Some possibilities for organizing such a table for fast access are (a) order by expected frequency common symbols will be found quicker (b) order by actual frequency Start with some ordering then when an entry is used move it closer to the beginning of the table (c) order alphabetically and do a binary/logarithmic search (d) use a hash function to map each symbol to a different address in a table To look up a symbol from the program, apply the hash function and look at the contents of the corresponding address (more on hashing below) Variable tables The problem with the user-defined symbols is that when we being pass one we don t know how many symbols there will be or what they will be Thus options (a), (b) and (c) above are not good It is common to use (d) - a hash-table approach We need to make provision for dealing with two symbols that hash to the same address Hashing: a small example Hashing involves devising a function that maps the objects to be stored (in our case user-defined identifiers) onto addresses (probably indexes of a table) If the function is simple, hashing can be very fast A simple function is: ( rank(1 st character) + rank(last character) ) % table-size The following shows the value of this function for some typical identifiers assuming table addresses ranging from 0 to 9 Symbol Rank(1 st ) Rank(Last) Rank(1st)+rank(last) Rank(1 st )+Rank(last))%10 TOP N COUNT LAB LOOP When a symbol hashes to an address that is already full we can just use the next free one After entering the examples above, our 10-entry table looks like Comp 162 Notes Page 8 of 17 September 21, 2016

9 Index Contents Value COUNT 4 LAB 5 6 TOP 7 8 N 9 LOOP Assembler operation: Pass 1 Typically, Pass 1 does the majority of the error checking of the source program In the case of, Pass 1 does all the checking Pass 1 is responsible for determining the value of each user-defined symbol It constructs a table of symbols and corresponding values It uses a variable called the Instruction Location Counter (ILC) to keep track of the amount of space required in memory by a translation of the program read so far If we are translating the following program by hand charo A,d charo B,d A: byte H B: byte I We can figure the translation of the first line is 3 bytes (and will occupy bytes 0,1,2 in memory), the translation of the second line is also three bytes (bytes 3, 4 and 5) the translation of the third line occupies one byte (byte 6) So the symbol A is a label on byte 7 and therefore the value of the symbol A is 7 Similarly the value of B is 8 We could replace the first line of the program by charo 7,d and it would translate the same Next we will see how Pass 1 of the assembler uses and updates the value of the ILC Comp 162 Notes Page 9 of 17 September 21, 2016

10 Pass 1: ILC The Instruction Location Counter (ILC) is a variable used by the assembler to assign appropriate values to labels it is initialized to zero it is incremented by number of bytes required by translation of the current line When a label is encountered, its value is (usually) the current value of the ILC (See later for an exception) Example Label Program Line Bytes occupied ILC before ILC after deci N,d lda N,d asra next: deci M,d adda N,d brv end br next end: M: block N: block n/a 24 The final value of the ILC is the size of the program in bytes After Pass 1 we have the following values (see highlighted entries in the trace above) Symbol Value N 22 next 7 M 20 end 19 limitations? Large block declarations work The following program assembles lda 1,i x: block y: block 2 Comp 162 Notes Page 10 of 17 September 21, 2016

11 A program with 10,000 different labels also appears to assemble correctly Pass 1: Algorithm Here is a pseudocode algorithm for Pass 1 ILC = 0 read line while ( line!= END) { if errors(line) output errormessage } else if ( notallcomment(line)) { extractsymbols (line,symbollist) updatetable (symbollist,ilc) ILC = ILC + spaceneeds(line) // see below } if (end-of-file) { output ("missing "); exit } else read line Computing the space needs of a source line Pass 1 has to figure out how many bytes will be generated by the translation of a line (symbolic instruction or pseudo operation) so it can update the ILC correctly This can be more or less complicated In it is relatively straightforward Here are some examples of lines and the number of bytes its translation takes up in the object program Line Bytes byte 4 1 ascii "*" 1 asra 1 word 0 2 ascii "Hi" 2 loada 10,i 3 ascii "ABCDE" 5 (in general, the number of chars) ascii X=\t \n Result\x00 14 (might be tricky to count them!) block N N Comp 162 Notes Page 11 of 17 September 21, 2016

12 Here is another example of processing by Pass 1 (verify the ILC values for yourself) Program ILC before ILC after a: block b: byte c: word d: word 0x e: ascii f: ascii 0x g: block 0x h: byte '\x12' Error checking Here is a list of typical errors that might be detected by an assembler during Pass 1 * symbol defined more than once * instruction has incorrect number of operands * misspelled directive/opcode * invalid addressing mode * missing * invalid constant (eg, a non-digit character in a decimal number) * missing symbol definition (detected at the end of Pass 1) * invalid string delimiter Note that while the assembler will detect an error in a line, the error message it gives may not be the best indicator of what the error is For example, when erroneous operand is present main: asra 2;ERROR: Comment expected Assembly operation: Pass 2 If Pass 1 does not find any errors in the source program then Pass 2 uses the symbol table constructed in Pass 1 and the table of opcode information to translate the program For example if M has a value of 15 after Pass 1 then LDA M,d Translates to = C0 00 0F LDr A d <value of symbol M> Comp 162 Notes Page 12 of 17 September 21, 2016

13 We assume that in order to do the error checking of Pass 1 and the translation of Pass 2, the table of opcodes contains information such as the following (see Figure 52) Symbol Template Operand Bytes Modes Allowed ROLr r Yes 1 STr 1110raaa Yes 3 d,s,x,n,sf,sx,sxf STOP No 1 Error detection in Pass 2 In there are no errors that are detectable only in Pass 2 Here is an example of an error in another assembly languages that can be detected only in Pass 2 The fragment includes a branch instruction to a label further down the program X: brgt X ; jump to X if result greater than 0 Suppose that the format of the two-byte brgt instruction is as follows opcode Relative address of destination The (signed) relative address (one byte) will be added to the PC if the branch is to be taken We can thus jump to locations approximately ± 127 bytes away from the branch instruction During Pass 1 we do not know where X is so no error can be signaled If during Pass 2 we find that X is too far away for us to put the distance in the single byte available then we indicate an error The programmer will have to find a different way to implement this branch, for example brle skip jump X skip: X: ; jump, like instructions in can address all memory Comp 162 Notes Page 13 of 17 September 21, 2016

14 The equate directive The equate directive lets us associate a symbolic name with a value It is processed in Pass 1 and ignored in Pass 2 (it doesn t generate any executable code) so it does not matter where it appears in the program We have seen how to figure out the values that Pass 1 will give to symbols If equate is used, just remember that it does not occupy any space in the object program Consider the following a: block 25 maxu: equate 19 count: word 0 total: word 0 tab : equate 0x09 n: block 8 t: ascii "END" w: byte 'x' Verify for yourself that after Pass 1 the symbol table with contain the following symbols and values Symbol Value a 0 maxu 19 count 25 total 27 tab 9 n 29 t 37 w 40 Assembler variant 1: Disassembler (see p 207) We cannot in general perform a reverse translation of an object program because data and instructions are indistinguishable in memory For example, in and subsp 2,i translates to byte h word 2 translates to Comp 162 Notes Page 14 of 17 September 21, 2016

15 A disassembler can offer two or more translations of a memory block and let the user sort out which is correct If it has access to the symbol table used in assembly, it may be able to make the translation a little more readable In, a memory dump shows the (printable) ASCII characters For example, a memory dump of ; test program lda 48,i stro mess1,d deco 51,i stro mess2,d mess1: ascii "Adding 34 and 17 gives: \x00" mess2: ascii " Correct! \n\x00" gives the following 0000 C D À0A A&Add E ing 34 a E nd 17 gi A ves: C F orrect! A Assembler variant 2: Cross assembler A cross assembler is just an assembler that produces an object program for a system different from the one on which it itself runs An example would be using an assembler on an Intel-based Windows system to produce code for an appliance (eg, microwave or BMW or cellphone) that uses a non-intel CPU Often used when the target machine does not have a memory large enough to hold an assembler or perhaps where the target machine only exists on paper Reading Unfortunately, Warford does not have much on how the assembler works Section 51 has what little there is Comp 162 Notes Page 15 of 17 September 21, 2016

16 Review questions 1 How many bytes does the following directive occupy in the object program ascii \t\tthis is the result\n\n\x00 2 If we now hash ERROR into our hash table, which entry does it occupy? 3 Consider example program 4 If we replace adda constant,i by adda 27,i and remove the line constant: equate 27 Does the object program get bigger, get smaller or stay the same? 4 Suppose we want to have synonyms for opcodes so that, if we wish we could write LA instead of LDBYTEA Which symbol table would have to change and what could that change be? 5 Consider the following program A: byte 4 B: block 4 C: ascii 4 D: byte 4 E: equate 4 Do any of the symbols have value 4? If so, which? Comp 162 Notes Page 16 of 17 September 21, 2016

17 Review answers 1 23 bytes 2 Hash (ERROR) is (5+18)%10 = 3 Entry 3 is full, entry 4 is full so it is inserted into entry 5 3 Stays the same the object programs are identical 4 We would add a line to the opcode table The new entry for LA would be the same as the entry for LDBYTEA in all fields except the mnemonic 5 Only E A has value 0, B has value 1, C has value 5 and D has value 6 Comp 162 Notes Page 17 of 17 September 21, 2016