9/29: Lecture Topics

1. 9/29: Lecture Topics • Conditional branch instructions • Unconditional jump instructions • Hexadecimal/Binary/Decimal • Instruction encoding

2. Mailing List and Computer Labs • CSE410 mailing list • Subscribe if you weren’t automatically or you use a different account • Send email to majordomo@cs.washington.edu with ‘subscribe cse410’ in the body • Computer accounts • MSCC Lab in basement of Communications building • Username = passwd =

3. Conditional Branch A change of the flow of control of the program that depends on a condition ... ... ? ... yes no ...

4. Control flow in C • Conditional branch constructs: • while, do while loops • for loops • Unconditional jump constructs: • goto • switch statements

5. Branching Instructions • beq • bne • other real instructions: b, bgez, bgtz, more... • other pseudoinstructions: beqz, bge, bgt, ble, blt, more...

6. Pseudoinstructions • One-to-one mapping between assembly and machine language not strictly true • Assembler can do some of the work for you • Example: slt is a real instruction; blt is a pseudoinstruction • move is a pseudoinstruction

7. Labels in MIPS • MIPS allows text tags • a name for a place to branch to • Under the covers, the assembler converts the text tag into an address loop: add $t0, $t0, $t0 # bne $t0, $t1, loop # sw $t2, 0($t3) #

8. Building a while loop in MIPS Idea: translate i=0; while(i<100) { i++; } into assembly.

9. How about a for loop? • One answer: translate from for to while, then use while loop conventions • This is typically how compilers handle loops i=0; while(i<N) { do_stuff(i); i++; } for(i=0; i<N; i++) { do_stuff(i); }

10. Example C Program int array[100]; void main() { int i; for(i=0; i<100; i++) array[i] = i; }

11. An Assembly Version main: move $t0, $0 # la $t1, array # start: bge $t0, 100, exit # sw $t0, 0($t1) # addi $t0, $t0, 1 # addi $t1, $t1, 4 # j start # exit: j $ra #

12. Unconditional Jump • (Mostly) the same as branch • No choice about it; just go to the label • How are branch and jump different? • we’ll see when we get to instruction encoding

13. Binary Numbers • Instead of 0-9 we can only use 0 and 1 • Also known as base-2 • Counting to 10ten in binary 0, 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010 • Binary arithmetic 1 1011 +1010 10101 11001001 - 1010110

14. Binary -> Decimal • Converting from Base-2 to Base-10 12 = 20 = 110 102 = 21 = 210 1002 = 22 = 410 10002 = 23 = 810 11102 = 1*23 + 1*22 + 1*21 + 0*20 = 8+4+2+0 = 1410 1010112 =

15. Decimal -> Binary • Repeatedly divide by 2 until you reach 0 • Reverse the order of the remainders 5010 50/2 = 25 remainder 0 25/2 = 12 remainder 1 12/2 = 6 remainder 0 6/2 = 3 remainder 0 3/2 = 1 remainder 1 1/2 = 0 remainder 1 1100102 7510 75/2 = remainder /2 = remainder /2 = remainder /2 = remainder /2 = remainder /2 = remainder /2 = remainder

16. Hexadecimal Numbers • Instead of 0-9 we use 0-15 • 0-15 is confusing so we use 0-9, A-F • Also known as base-16 • Counting to 18ten in hex 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, 10, 11, 12 • Hex arithmetic 1 1 2E97 +A2B3 D14A A3AD38B993CA93C930B -A3AD38B993CA93C930A

17. Hexadecimal <-> Binary • Each four binary digits represents one hexadecimal digit • Can quickly convert either way • 0x2A07 = 0010 1010 0000 0111 • 1111 0011 0101 1011 = 00002 = 016 00012 = 116 00102 = 216 00112 = 316 01002 = 416 01012 = 516 01102 = 616 01112 = 716 10002 = 816 10012 = 916 10102 = A16 10112 = B16 11002 = C16 11012 = D16 11102 = E16 11112 = F16

18. Hexadecimal -> Decimal • Converting from Base-16 to Base-10 116 = 160 = 110 1016 = 161 = 1610 10016 = 162 = 25610 100016 = 163 = 153610 A32 = A*161 + 3*160 = 10*161 + 3*160= 160 + 3 = 163

19. Decimal -> Hexadecimal • Repeatedly divide by 16 until you reach 0 • Reverse the order of the remainders 50010 500/16 = 31 remainder 4 31/16 = 1 remainder 15 (F) 1/16 = 0 remainder 1 1F416

20. What you need to know • Be able to convert to/from small decimal numbers from/to binary or hex • Be able to/from convert any binary number from/to a hex number

21. Stored Program • So far, we’ve seen that data comes from memory • It turns out that the program comes from memory also • Each instruction in the program has an address • Von Neumann computer (p. 33)

22. Program Layout (p. 160) Address 0 Reserved 0x00400000 Program instructions Text 0x10000000 Static data Global variables 0x10008000 Dynamic data and stack 0x7fffffff

23. The Text Segment • Let’s zoom in on the text segment: 0x00400000 Text 0x10000000 0x00400000 add $t0, $t1, $t2 0x00400004 sub $t0, $t0, $t3 lw $s1, 4($t0) 0x00400008

24. Jump Register • Now we’re ready for the jr instruction • Syntax: • Effect: jump to the instruction at the address in $t0 jr $t0

25. Another Look at Labels • Labels are text tags • The assembler translates them into addresses and does search-and-replace loop: add $t0, $t0, $t0 bne $t0, $t1, loop sw $t2, 0($t3)

26. Instruction Encoding • How does the assembler translate each instruction into bits? add $t0, $s1, $s2 assembler 00000010001100100100000000100000

27. Fields • Split the 32-bit instruction into fields: • op+funct: Which instruction? • rs, rt: Register source operands • rd: Register destination operand 32 bits op rs rt rd shamt funct 6 bits 5 bits 5 bits 5 bits 5 bits 6 bits

28. Encoding an add instruction • Let’s translate: • Opcode for add is 000000 (p. A-55) • The funct for add is 100000 • The code for $t0 is 01000 (p. A-23) • The code for $s1 is 10001 • The code for $s2 is 10010 add $t0, $s1, $s2

29. Instruction Formats • The 32 bits can be split up more than one way • For add, it was 6-5-5-5-5-6 • This is called R-format • For lw, we use 6-5-5-16 instead • This is called I-format 32 bits op rs rt address 6 bits 5 bits 5 bits 16 bits

30. Translating into I-format • The opcode for lw is 100011 • The code for $t0 is 01000 • The code for $t1 is 01001 • Binary for 100 is lw $t0, 100($t1)

31. Branch vs. Jump • Branch instructions use the I-format • The destination address is 16 bits • You can only branch ±32KB away • Jump instructions use another format called the J-format • The destination address is 26 bits • You can jump up to ±32MB away • What’s the tradeoff?

32. What to Take Away • I don’t care if you: • Know how many bits are in each field • Know what the opcodes are • Know what the codes for the registers are • I do care that you: • Know what fields are • Know why we need more than one instruction format