Wednesday, February 15, 2017

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1 Wednesday, February 15, 2017 Topics for today Before and after assembly: Macros, Linkers Overview of Chapter 6 Branching Unconditional Status bits and branching If statements While statements The V and C bits Macro assemblers Many real assemblers allow users to declare macro instructions (macros). Here is an example.macro SWAP x,y Load r1 %x Move %y, %x Store r1 %x.mend Now the user can use SWAP as if it were a built-in instruction SWAP P,Q Implementation Processing of macros can be incorporated into Pass 1, however it is simpler to have macro definition and expansion handled separately by a Pass 0 thus Source Source Pass 0 - definitions Pass 1 etc. + expansions Macro definition table Pass 0 uses a table to store the macro definitions. Pass 0 outputs a copy of the source code from which macro definitions have been removed and into which macro calls have been expanded. Comp 162 Notes Page 1 of 13 February 15, 2017

2 Macros can be defined for IF, ENDIF, FOR, WHILE and so on, to give the assembly language a high-level flavor. After assembly: linking Source1 Assembler Object1 Source2 Assembler Object2 Linker Object Source3 Assembler Object3 Library, e.g. subroutines The linker program resolves external references, that is, references to symbols defined in another module 1. It may also adjust addresses to that the final object program is a single contiguous block. So far Warford has Chapter 6 (a) (b) introduced a high-level language (C). shown how the Pep/9 machine works (Chapter 4) and introduced an assembly language for programming it (Chapter 5). In Chapter 6 he brings the two threads together and shows how high-level language elements can be translated into assembly language. He covers both data structures (arrays and "structures") and control structures (if statements, while loops and function calls). Chapter 6 is a large one, containing much detailed information. Here is a top-level view 6.1 The run-time stack. Stack addressing mode. How the stack is used to implement local variables 6.2 Control Structures - 1: if and while 6.3 Control Structures - 2: function calls 6.4 Data structures - 1: arrays (also the switch statement) 6.5 Data structures - 2: "structures", pointers, dynamic memory 1 In C, for example, it is likely to be the linker that detects that you have misspelled the name of a library routine. Comp 162 Notes Page 2 of 13 February 15, 2017

3 We will skip consideration of the stack for the moment and look at how to implement programs with branches. Branching (Section 6.2) The unconditional branch instruction BR transfers control to the specified location. We have seen this informally when we branch round declarations at the beginning of a program as in br main A:.block 2 B:.block 2 main: deci A,d etc In general we also need conditional branch instructions that test some state of the machine to decide whether or not to branch. In Pep/9 many instructions (see Figure 5.2) set one or more of the status bits (NZVC) as a sideeffect of their operation. For example DECI sets N if input is negative, Z if it is zero and V if user enters a number too big for 16 bits LDWr sets N if a negative number is loaded and Z if zero is loaded CPWr The only purpose of the CPWr (Compare Words) instruction is to set status bits. It has no effect on any register or memory location. It sets status bits according to the value of the expression (register - operand). The conditional branch instructions test (but do not change) the status bits when deciding whether or not to branch. The 8 conditional branch instructions are Comp 162 Notes Page 3 of 13 February 15, 2017

4 opcode branch if branch if brle Less than or equal to N=1 or Z=1 brlt Less than N=1 breq Equal to Z=1 brne Not equal to Z=0 brge Greater than or equal to N=0 brgt Greater than N=0 and Z=0 brv Overflow V=1 brc Carry C=1 Pictorially DECI, LDWr, CPWr etc set N Z V C Status bits test BRGT, BREQ, BRV etc Suppose we need a program that reads N followed by N integers then outputs the sum of the N integers. The following partial implementation in Pep/9 assembly code illustrates use of branch instructions (some for error checking) how registers can be used in place of variables to speed up the program Comp 162 Notes Page 4 of 13 February 15, 2017

5 ldwa 0,i deci N,d brv error1 brlt error2 breq done ldwx N,d Register A holds total initially zero the number of numbers branch if N is out of range branch if user entered a negative count! branch if zero numbers to be read so just print total register X holds count of numbers still to be read top: deci Number,d input number to be summed use scratch location brv error3 branch if number out of range adda Number,d otherwise add it to the running total in register A brv error4 branch if sum is now out of range subx 1,i reduce count of numbers to be read brgt top branch if the count is still greater than zero done: stwa Total,d store the total in memory and deco Total,d output it stop N:.block 2 the number of numbers Number:.block 2 Total:.block 2.end one of the numbers the total of the numbers (for printing) IF statements Here are four pseudocode "if" statements and their translation to Pep/9 assembly code. If: Example 1 if (N<0) output (t) Generally, testing against 0 is easy. In Pep/9 this is simply ldwa N,d status bits set according to the value of N brge skip branch past the output if N >= 0 deco t,d skip: nop0 Note the use of the NOP instruction. A label must be attached to something. Comp 162 Notes Page 5 of 13 February 15, 2017

6 If: Example 2 if (count >= 39) output(1) else output(2) We need a CPW instruction in this case to compare count and 39. Note how we skip round the else part. ldwa count,d cpwa 39,i sets bits according to count-39 brlt else branch if count < 39 deco 1,i br endif skip round else code else: deco 2,i endif:nop0 If: Example 3 if (A>B) output(a) else output(b) We can put either the then or the else part first. The label else is never referenced, it is used to increase program readability. ldwa A,d cpwa B,d status bits depend on value of A-B brgt thenpart branch if A>B (A-B >0) else: deco B,d this is else part br skip remember to skip over the then part thenpart: deco A,d skip: nop0 If: Example 4 if (A>=14 && B<3 C!= 50) output (x) We can start with the simple second part of the OR ldwa C,d cpwa 50,i brne output ldwa A,d cpwa 14,i brlt skip ldwa B,d cpwa 3,i brge skip output: deco x,d skip: nop0 if C ne 50 condition is true so output otherwise test first part of condition first part of AND is false so no output second part of AND is false so skip output Comp 162 Notes Page 6 of 13 February 15, 2017

7 WHILE statements The implementation of conditional expressions in "while" statements is similar to their implementation in if statements. Here are three pseudocode examples translated into Pep/9 assemble code. While: Example 1 Our first example has a simple condition. input(n) while (N < 99) { output(n) input(n) } In Pep/9 this might be deci N,d top: ldwa N,d cpwa 99,i set bits according to N-99 brge skip skip loop if N >= 99 deco N,d deci N,d get next value for N br top back to test loop condition skip: nop0 While: Example 2 Our second example has two relational and one logical operator. In Pep/9 this might be input(n) while (N > 6 && N <= 10) { sum += N input(n) } top: deci N,d ldwa N,d sets bits according to value of N cpwa 6,i check first part of condition brle skip branch if N <= 6 no need to check N<=10 cpwa 10,i note - no need to reload N brgt skip skip the loop if N > 10 adda sum,d stwa sum,d sum += N br top skip: nop0 Comp 162 Notes Page 7 of 13 February 15, 2017

8 While: Example 3 Another example (this time with an or connector) is while (A > 2 B < 7) { input(a,b) output(a+b) } In Pep/9 this might be top: ldwa A,d cpwa 2,i brgt loop ldwa B,d cpwa 7,i brge skip loop: deci A,d get two numbers deci B,d ldwa A,d adda B,d skip: stwa -2,s deco -2,s br top nop0 first clause is true so don t need to check second first was false so check second if second clause is also false, don t do loop use stack for scratch space (see later) output sum of inputs Nested loops are of course possible. Label names and program text layout can help program clarity. To print a 3 by 5 table of stars * * * * * * * * * * * * * * * For (i=0 i<3, i++) { For (j=0 j<5 j++) Output( * ) Output( \n ) } In Pep/9 might be something like the following Comp 162 Notes Page 8 of 13 February 15, 2017

9 nested loops to print 3 by 5 stars ldwa 0,i stwa i,d outer: ldwa 0,i stwa j,d inner: ldbx '*',i stbx charout,d ldwa j,d adda 1,i stwa j,d j++ cpwa 5,i brlt inner branch if more to print on current line stop i:.block 2 j:.block 2.end ldbx '\n',i stbx charout,d ldwa i,d adda 1,i stwa i,d cpwa 3,i brlt outer i++ branch if more lines to print Structured programming theorem Warford refers to the "structured programming theorem" 2 which says that "if" and "while" are sufficient (along with sequence) to implement any algorithm. Thus, it must be possible to translate other constructs in C such as "do-while statements and "for" loops into if and while loops. E.g. and for (ABC) {action} is A while(b) { action C } do { action } while (C) is action while(c) {action} One strategy for developing assembly language programs is to write the algorithm first in a high-level notation using "if" and "while" then translate it. This avoids the temptation to use "branch" statements in an unstructured way (see Fig 6.16 for an example of unstructured branching). 2 Comp 162 Notes Page 9 of 13 February 15, 2017

10 The V and C bits. The V and C bits are tested by the brv and brc instructions respectively. The V bit The V bit is set on overflow and is useful for checking user input and arithmetic operations as in the earlier example program that summed N integers. The C bit The C bit is set by a carry out from arithmetic operations such as add and shift. The following code divides N by 2 (integer divide) and reports the result of the true division. For example if N is 13, it is replaced by 6 and the program outputs 6.5. ldwa N,d asra rightmost bit goes into C stwa N,d N/2 deco N,d brc halfout if C set, we need to output.5 br skip halfout: ldbx.,i stbx charout,d ldbx 5,i stbx charout,d skip: nop0 We can perform multilength arithmetic by allocating multiple storage locations to a variable then augmenting the arithmetic operations. For example, in Pep/9 we can use two words to hold a 32- bit number one holds the most significant half, the other the least significant half. We make use of the brc instruction in the code that adds two such 32-bit numbers and stores the result in a third. Our two numbers to be added are in variables A1, A2 and B1, B2. The result goes in C1 and C2. Pictorially: A1 B1 A2 B2 C1 C2 Comp 162 Notes Page 10 of 13 February 15, 2017

11 ldwa A2,d adda B2,d may set carry bit stwa C2,d ldwa A1,d brc carry branch if carry set in addition of least significant halves br nocarry carry: adda 1,i the extra 1 added to the addition of most significant halves nocarry: adda B1,d stwa C1,d Reading Read section 6.2 for material on ifs and whiles. Comp 162 Notes Page 11 of 13 February 15, 2017

12 Review Questions Consider the following Pep/9 program ldwx 0x0008,i L: stro S,d asrx shift X right arithmetically brne L stop S:.ascii Hello\n\x00.end 1. Predict the number of times that the program outputs the string before it stops. 2. What other constant(s) in the first line would result in the same number of strings being output? 3. What constant would we use in the first line if we want the program to output 10 strings? 4. What is the largest number of strings that we can have the program output in this way and then stop? 5. If the constant in the first line is changed to 0x8000 does it print strings and then stop? If so, how many strings? If not, why does it not stop? 6. If we change the third line in the original program above to aslx, predict the number of strings that it outputs before stopping. Comp 162 Notes Page 12 of 13 February 15, 2017

13 Review Answers 1. It prints a string then halves the number (truncated division). It will print 4 strings before the number reaches zero: 8, 4, 2, 1, 0 2. Any number that has the leftmost 1 in the same position as 8 so 9..F. 3. We could use 0x0200 or any number that has the leftmost 1 in the same position E.g., if the constant is 0x It does not stop because 0x8000 is a negative number and the right shift keeps the leftmost bit equal to 1 so the number never reaches zero. ( -1/2 = -1) 6. It stops after printing 13 strings when the sole 1 bit in the number is shifted out and the result is zero. Comp 162 Notes Page 13 of 13 February 15, 2017