1 / 23

MIPS Programming

MIPS Programming. Arrays Writing Procedures: Calling Convention. Memory Setup in C/Java. C++: int *intarray = new int[10]; Java: int[] intarray = new int[10]; What does this do? What does the memory look like? Where is intarray[5] located? intarray + 20

tudor
Download Presentation

MIPS Programming

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. MIPS Programming Arrays Writing Procedures: Calling Convention

  2. Memory Setup in C/Java • C++: int *intarray = new int[10]; • Java: int[] intarray = new int[10]; • What does this do? What does the memory look like? • Where is intarray[5] located? intarray + 20 • Where is intarray[i] located? intarray + 4*i intarray intarray &(intarray[0]) intarray + 20 &(intarray[0]) + 20

  3. Declaring & InitializingGlobal Arrays in HLL int GlobalA = 3; int GlobalB[] = {0x20040002, 0x20080001, 0x200b0001, 0x8b502a, 0x15400003, 0x01084020, 0x20840000, 0x800fffa, 0x10010200, 0x00000000}; int main(int argc, char *argv[]) { } public class MyClass{ public static int GlobalA = 3; public static int GlobalB[] = {0x20040002, 0x20080001, 0x200b0001, 0x8b502a, 0x15400003, 0x01084020, 0x20840000, 0x800fffa, 0x10010200, 0x00000000}; };

  4. Declaring & Initializing Global Arrays in MIPS .data GlobalA: .word 0x03; GlobalB: .word 0x20040002 0x20080001 0x200b0001 0x8b502a .word 0x15400003 0x01084020 0x20840000 0x800fffa .word 0x10010200 0x00000000 .text main:

  5. Declaring & InitializingLocal Arrays int main(int argc, char *argv[]) { int LocalA = 5; int LocalB[] = {1,2,3}; } Not possible in Java!!!!! In Java, arrays are references to Array objects. add $sp, $sp, -(24 + x + 4 + 12) # where x is space for preserved regs addi $t0, $0, 5 sw $t0, 12($sp) addi $t0, $0, 1 sw $t0, 0($sp) addi $t0, $0, 2 sw $t1, 4($sp) addi $t0, $0, 3 sw $t1, 8($sp) // and so forth

  6. Declaring & InitializingHeap Arrays in HLL int main(int argc, char *argv[]) { int *LocalA = (int *)malloc(4); *LocalA = 5; int *LocalB = (int *)malloc(12); LocalB[0] = 1; LocalB[1] = 2; LocalB[2] = 3; } public class MyClass{ public static void main(int argc, String argv) { int LocalA[] = new int[1]; LocalA[0] = 5; int LocalB[] = new int[3]; LocalB[0] = 1; LocalB[1] = 2; LocalB[2] = 3; } };

  7. Declaring & InitializingHeap Arrays in MIPS add $sp, $sp, -(24 + x + 8) # where x is space for preserved regs addi $a0, $0, 4 jal malloc sw $v0, 4($sp) // store the reference into the stack addi $t0, $0, 5 sw $t0, 0($v0) // initialize first elements as 5 (*LocalA = 5) addi $a0, $0, 12 jal malloc sw $v0, 0($sp) // store the reference into the stack addi $t0, $0, 1 sw $t0, 0($v0) // LocalB[0] = 1 addi $t0, $0, 2 sw $t0, 4($v0) // LocalB[1] = 2 addi $t0, $0, 3 sw $t0, 8($v0) // LocalB[2] = 3

  8. MIPS Example 5 Assumptions: A & B are global c is in the stack, 6 bytes from $sp Translate int A[100]; // ints are 4 bytes in C/Java char B[100]; // chars are 1 byte in C char c = B[50]; A[1] = A[5] + 7; Use LoadByte, not LoadWord, because char (in C) is 1 byte la $s1, B lb $t1, 50($s1) # $t1 = B[50]; sb $t1, 6($sp) # c = B[50]; la $s0, A lw $t0, 5 ($s0) ; x = A[5]; lw $t0, 20 ($s0) # $t0 = A[5]; addi $t0, $t0, 7 # $t0 = A[5] + 7; sw $t0, 4 ($s0) # A[1] = A[5] + 7;

  9. MIPS Example 6 Assumptions: &(A[0]) is in $s0, x is in $t0 i is in $s1 Translate int A[100]; int i; … x = A[i]; sll $t1, $s1, 2 # $t1 = i<<2 or i * 4 add $t1, $s0, $t1 # $t1 = (i*4+&A[0]) or &(A[i]) lw $t0, 0($t1) # $t0 = A[i];

  10. Procedure Calls Main Procedure • Procedure must work the same from any call • Procedure uses regs that main was using • We need a convention to • pass arguments • preserve registers, stack • pass return values Call Procedure Call Procedure

  11. MIPS-specific info Page 140, Figure 3.13

  12. MIPS-specific info – who cares? • Preserved – Value is same after call • Caller – excellent! no worries! • Procedure – may not destroy value • must store at beginning and restore at end to use • Not preserved – No guarantees • Caller – loses value in register • most store before call and restore after call • Procedure – may use without worries

  13. Steps for caller 0. Store any temp regs whose values we need • Pass function parameters to procedure • Transfer control to procedure • (then procedure executes) • Get return value • Restore any temp regs we saved away

  14. Steps for procedure • Allocate stack space • Store preserved regs we may use • Perform task • Place result in proper location for caller • Restore preserved regs we may have used • Transfer control back to caller

  15. Caller: 4. Restore regs int MyFunc(int g, int h) { return (g + h);} Caller: sw $t0, 0($sp) add $a0, $s2, $zero add $a1, $s3, $zero jal MyFunc add $s0, $v0, $zero lw $t0, 0($sp) Callee: Load from same location in the stack Assume: g,h are in $s2,$s3. We want return value in $s0. Caller wants to preserve $t0 across function call.

  16. Definitions • Leaf function • Makes no function calls • You need not store $ra into the stack • Non-leaf function • Contains a function call • You MUST store $ra in the stack and restore before returning to caller

  17. Allocating stack space Stack space for this function $sp $sp Minimum allocation: 24 bytes $sp Only allocate once per function!!!! $sp contains address of bottom of stack. What operation on $sp allocates space?

  18. Callee: 7. Return to caller int MyFunc(int g, int h) { return (g + h);} Caller: sw $t0, 0($sp) add $a0, $s2, $zero add $a1, $s3, $zero jal MyFunc add $s0, $v0, $zero lw $t0, 0($sp) Callee: addi $sp, $sp, -32 sw $s0, 0 ($sp) add $s0, $a0, $a1 add $v0, $s0, $zero lw $s0, 0 ($sp) addi $sp, $sp, 32 jr $ra Assume: g,h are in $s2,$s3. We want return value in $s0. Caller wants to preserve $t0 across function call.

  19. What actually goes in stack Extra Arguments $sp before call $fp during call $ra $fp P*4 bytes for preserved regs ($s0-$s7) preserved registers (and $a0-$a3) padding local data L*4 bytes for local data Extra outgoing arguments A*4 bytes for outgoing args $sp during call

  20. Assume it needs 2 saved registers Example Local 256-byte array int foo(int arg1, int arg2) { int myarray[64]; myarray[3] = 5; … bar(a1, a2, a3, a4, a5); … return (myarray[3]); } Non-leaf function

  21. Caller’s Stack Minimum Allocation addi $sp, $sp, - (1+64+1+2+6)*4 sw $ra, 73*4($sp) sw $fp, 72*4($sp) sw $s1, 67*4($sp) sw $s0, 66*4($sp) addi $t0, $zero,5 # $t0 = 5 sw $t0, (1+3)*4 ($sp)# myarray[3] = 5 … lw $ra, 73*4($sp) lw $fp, 72*4($sp) lw $s1, 67*4($sp) lw $s0, 66*4($sp) addi $sp, $sp, (1+64+1+2+6)*4 jr $ra $ra $sp before call $fp $a0 $a1 $a2 $a3 $s1 $s0 padding myarray $sp during call outgoing arg 5

  22. SumToN: add $v0, $0, $0 loop: slt $t0, $0, $a0 beq $t0, $0, end add $v0, $v0, $a0 addi $a0, $a0, -1 j loop end: jr $ra Another Example: int SumToN(int N) { int sum = 0; while (N > 0) { sum += N; N--; } return sum; }

  23. SumToNtotheM: add $sp, $sp, -32 sw $ra, 14($sp) add $t0, $0, $0 loop: slt $t0, $0, $a0 beq $t0, $0, end sw $a0, 28($sp) sw $a1, 24($sp) sw $t0, 0($sp) jal power lw $t0, 0($sp) lw $a0, 28($sp) lw $a1, 24($sp) add $t0, $v0, $t0 addi $a0, $a0, -1 j loop end: add $v0, $t0, $0 lw $ra, 14($sp) add $sp, $sp, 32 jr $ra Third Example: int SumNtotheM(int N, int M) { int sum = 0; while (N > 0) { sum += power(N,M); N--; } return sum; }

More Related