Run Time Environment. Activation Records Procedure Linkage Name Translation and Variable Access

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Run Time Environment Activation Records Procedure Linkage Name Translation and Variable Access Copyright 2015, Pedro C. Diniz, all rights reserved. Students enrolled in the Compilers class at the University of Southern California have explicit permission to make copies of these materials for their personal use.

Procedure Abstraction What is a Procedure? Basic Unit of Abstraction and Program Reasoning Why do We use Them? To allow us to build (very) large programs Conceptually allows us to abstracts from all the details How to Generate Code? Storage Allocation Scoping, i.e., what is visible and where? Control Transfer

The Procedure as a Control Abstraction Procedures have well-defined control-flow The Algol-60 procedure call Invoked at a call site, with some set of actual parameters Control returns to call site, immediately after invocation s = p(10,t,u); int p(a,b,c) int a, b, c; { int d; d = q(c,b);... } caller callee

The Procedure as a Control Abstraction Procedures have well-defined control-flow The Algol-60 procedure call Invoked at a call site, with some set of actual parameters Control returns to call site, immediately after invocation s = p(10,t,u); int p(a,b,c) int a, b, c; { int d; d = q(c,b);... } int q(x,y) int x,y; { return x + y; }

The Procedure as a Control Abstraction Procedures have well-defined control-flow The Algol-60 procedure call Invoked at a call site, with some set of actual parameters Control returns to call site, immediately after invocation s = p(10,t,u); int p(a,b,c) int a, b, c; { int d; d = q(c,b);... } int q(x,y) int x,y; { return x + y; }

The Procedure as a Control Abstraction Procedures have well-defined control-flow The Algol-60 procedure call Invoked at a call site, with some set of actual parameters Control returns to call site, immediately after invocation s = p(10,t,u); int p(a,b,c) int a, b, c; { int d; d = q(c,b);... } int q(x,y) int x,y; { return x + y; }

The Procedure as a Control Abstraction Procedures have well-defined control-flow The Algol-60 procedure call Invoked at a call site, with some set of actual parameters Control returns to call site, immediately after invocation s = p(10,t,u); int p(a,b,c) int a, b, c; { int d; d = q(c,b);... } int q(x,y) int x,y; { return x + y; } Most Languages Allow Recursion

Compilation Issues How to Generate Code Storage for Local Variables Transfer of Arguments and Return Results How to Execute a Procedure How to Access Local and Non-Local Variables How to Communicate between Caller and Callee How to Transfer Control between Caller and Callee The Role of the Symbol Table Keep track of where names are defined and declared Scope and Lifetime

Run-Time Storage Organization Code Static data & Global data stack Classical Organization Code, Static, & Global Data have known size Use symbolic labels in the code Heap & Stack grow and shrink over time Stack used for Activation Records (AR) Heap for data (including AR) whose lifetime extends beyond activation. heap This is a virtual address space

Activation Record Basics ARP parameters register save area return value return address caller s ARP local variables Space for parameters to the current routine Saved register contents If function, space for return value Address to resume caller Help with non-local access To restore caller s AR on a return Space for local values & variables (including spills) One AR for each invocation of a procedure

Activation Records on the Stack What Happens on a Call? Passing of Arguments Transfer of Control What Happens on a Return? Recovery of Results (if any) Transfer of Control ARP Need to Save/Restore Execution ARP Context? ARP, PC, Access Link, Register Values Stack parameters register save area return value return address caller s ARP local variables parameters register save area return value return address callee s ARP local variables Caller s Activation Record Callee s Activation Record Call Return

Procedure Linkages Standard procedure linkage procedure p prolog pre-call post-return epilog procedure q prolog epilog Procedure has standard prolog standard epilog Each call involves a pre-call sequence post-return sequence These are completely predictable from the call site Þ depend on the number & type of the actual parameters

Procedure Linkages Pre-Call Sequence Sets up Callee s basic AR Helps preserve its own environment The Details Allocate Space for the Callee s AR except space for local variables Evaluates each parameter & stores value or address Saves return address, caller s ARP into callee s AR If s are used Find appropriate lexical ancestor & copy into callee s AR Save any caller-save registers Save into space in caller s AR Jump to address of callee s prolog code

Procedure Linkages Post-Return Sequence Finish restoring caller s environment Place any value back where it belongs The Details Copy return value from callee s AR, if necessary Free the callee s AR Restore any caller-save registers Restore any call-by-reference parameters to registers, if needed Also copy back call-by-value/result parameters Continue execution after the call

Procedure Linkages Prolog Code Finish setting up the callee s environment Preserve parts of the caller s environment that will be disturbed The Details Preserve any callee-save registers If display is being used Save display entry for current lexical level Store current ARP into display for current lexical level Allocate space for local data Easiest scenario is to extend the AR Find any static data areas referenced in the callee Handle any local variable initializations With heap allocated AR, may need to use a separate heap object for local variables

Procedure Linkages Epilog Code Wind up the business of the callee Start restoring the caller s environment If ARs are stack allocated, this may not be necessary. (Caller can reset stacktop to its pre-call value.) The Details Store return value? No, this happens on the return statement Restore callee-save registers Free space for local data, if necessary (on the heap) Load return address from AR Restore caller s ARP Jump to the return address

Simplified Example 1. program sort(input, output);; 2. var a: array [0..10] of integer;; 3. x, i: integer;; 4. procedure readarray;; 5. var i : integer;; 6. begin a end { readarray } ;; 7. procedure exchange(i, j: integer);; 8. begin 9. x := a[i];; a[i] := a[j];; a[j] := x;; 10. end { exchange } ;; 11. procedure quicksort(m, n : integer);; 12. var k, v: integer;; 13. function partition(y, z : integer) : integer;; 14. var i, j : integer;; 15. begin a 16. v 17. exchange(i,j);; 18. end { partition } 19. begin end { quicksort } 20. begin end { sort } ARP ARP sort ret address arp link a,x quicksort(1,9) ret address arp link quicksort(1,3) ret address arp link partition(1,3) ret address arp link i,j

Activation Record Details Where do activation records live? If lifetime of AR matches lifetime of invocation, AND If code normally executes a return Keep ARs on a stack If a procedure can outlive its caller, OR Yes! This stack. If it can return an object that can reference its execution state ARs must be kept in the heap If a procedure makes no calls AR can be allocated statically Efficiency prefers static, stack, then heap C o d e S G t l a & o t b i a c l H e a p S t a c k

Activation Record Details How does the Compiler find the Variables? They are at known offsets from the AR pointer Code nesting level and AR offset within a procedure Level specifies an ARP, offset is the constant Variable-length data If AR can be extended, put it after local variables Leave a pointer at a known offset from ARP Otherwise, put variable-length data on the heap Initializing local variables Must generate explicit code to store the values Among the procedure s first actions

Storage for Blocks within a Single Procedure B0: { int a, b, c B1: { int v, b, x, w B2: { int x, y, z. } B3: { int x, a, v } } } Fixed length data can always be at a constant offset from the beginning of a procedure In our example, the a declared at level 0 will always be the first data element, stored at byte 0 in the fixed-length data area The x declared at level 1 will always be the sixth data item, stored at byte 20 in the fixed data area The x declared at level 2 will always be the eighth data item, stored at byte 28 in the fixed data area But what about the a declared in the second block at level 2?

B0: { int a, b assign value to a B1: { int v(a), b, x B2: { int x, y(8). } } } Variable-length Data Arrays If size is fixed at compile time, store in fixedlength data area If size is variable, store descriptor in fixed length area, with pointer to variable length area Variable-length data area is assigned at the end of the fixed length area for block in which it is allocated a b v b x x y(8) v(a) Includes variable length data for all blocks in the procedure Variable-length data

Translating Local Names How does the compiler represent a specific instance of x? Name is translated into a static coordinate < level,offset > pair level is lexical nesting level of the procedure offset is unique within that scope Subsequent code will use the static coordinate to generate addresses and references level is a function of the table in which x is found Stored in the entry for each x offset must be assigned and stored in the symbol table Assigned at compile time Known at compile time Used to generate code that executes at run-time

Scoping Scoping Rules Define which instance each name refers to Lexical Scoping Look at the source text of the code Determine the closest (nesting structure) name Ex. FORTRAN, C, Pascal. Dynamic Scoping Check at Run-Time the closest variable with the same name Ex. Scheme, Lisp, Miranda, etc.

Lexical Scoping Example 1. program sort(input, output);; 2. var a: array [0..10] of integer;; 3. x, i: integer;; 4. procedure readarray;; 5. var i : integer;; 6. begin a end { readarray } ;; 7. procedure exchange(i, j: integer);; 8. begin 9. x := a[i];; a[i] := a[j];; a[j] := x;; 10. end { exchange } ;; 11. procedure quicksort(m, n : integer);; 12. var k, v: integer;; 13. function partition(y, z : integer) : integer;; 14. var i, j : integer;; 15. begin a 16. v 17. exchange(i,j);; 18. end { partition } 19. begin end { quicksort } 20. begin end { sort } readarray Call Tree Sort a[],x,i quicksort m,n, partition exchange quicksort m,n, partition y,z,i,j exchange i,j

Nested Procedures & Symbol Tables sort 1. program sort(input, output);; 2. var a: array [0..10] of integer;; nil var a array 44 integer 0 var x scalar 4 integer 3. x, i: integer;; 4. procedure readarray;; 44 var i scalar 4 48 integer 5. var i : integer;; 6. begin a end { readarray } ;; 7. procedure exchange(i, j: integer);; 8. begin readarray exchange quicksort 9. x := a[i];; a[i] := a[j];; a[j] := x;; 10. end { exchange } ;; 11. procedure quicksort(m, n : integer);; 12. var k, v: integer;; 13. function partition(y, z : integer) : integer;; var i scalar 4 integer 0 par i scalar 4 integer 0 par j scalar 4 integer 4 par i scalar 4 integer 0 par j scalar 4 integer 4 var k scalar 4 integer 8 var v scalar 4 integer 12 14. var i, j : integer;; 15. begin a 16. v partition 17. exchange(i,j);; 18. end { partition } 19. begin end { quicksort } par y scalar 4 integer 0 par z scalar 4 integer 4 var i scalar 4 integer 8 20. begin end { sort } var j scalar 4 integer 12

Static Allocation subroutine A() integer i real a(100), b(10,10) do 100 i=1, 10 a(i*10) = b(i,10) 100: continue end Code for A Code for B integer i real a(100), b(10,10) subroutine B() integer i, c(10) do 200 i=1, 10 c(i) = 0 200: continue end Activation record for A Local variables are bound to fixed location in storage Values can be retained across procedure call (static) Save PC in AR but no need for stack Limitations: Fixed size variables only Does not support recursion No dynamic memory allocation Advantages: Simplified code generation integer I, c(10) Activation record for B

Lexical Scopes Without Nested Procedures 1. program sort(input, output);; 2. var a: array [0..10] of integer;; 3. x: integer;; 4. procedure readarray;; 5. var i : integer;; 6. begin a end { readarray } ;; 7. procedure exchange(i, j: integer);; 8. begin Easy location of variables Either local, i.e., in the AR Global, i.e. at specified global offset Why Do with Need a Stack? 9. x := a[i];; a[i] := a[j];; a[j] := x;; 10. end { exchange } ;; 11. function partition(y, z : integer) : integer;; 12. var i, j : integer;; 13. begin a 14. exchange(i,j);; 15. end { partition } 16. procedure quicksort(m, n : integer);; 17. var k, v: integer;; 18. begin end { quicksort } 19. begin end { sort }

Lexical Scope With Nested Procedures 1. program sort(input, output);; 2. var a: array [0..10] of integer;; 3. x: integer;; 4. procedure readarray;; 5. var i : integer;; 6. begin a end { readarray } ;; 7. procedure exchange(i, j: integer);; 8. begin 9. x := a[i];; a[i] := a[j];; a[j] := x;; 10. end { exchange } ;; 11. procedure quicksort(m, n : integer);; 12. var k, v: integer;; 13. function partition(y, z : integer) : integer;; 14. var i, j : integer;; 15. begin a 16. v 17. exchange(i,j);; 18. end { partition } 19. begin end { quicksort } 20. begin end { sort } Problem! Now quicksort might have to access a, x at sort Also, partition needs to access k, v at quicksort But which one? Need to keep Track of Depth (static) Nesting Depth sort at depth 1 readarray, quicksort at depth 2 partition at depth 3 Implementation Link Chasing in AR Reflect Nesting Structure During Calls Display indexed by depth

Activation Records on the Stack sort sort sort sort a,x a,x a,x a,x quicksort(1,9) quicksort(1,9) quicksort(1,9) quicksort(1,9) quicksort(1,3) quicksort(1,3) quicksort(1,3) partition(1,3) partition(1,3) i,j i,j exchange(1,3)

Lexical Scoping Example 1. program sort(input, output);; 2. var a: array [0..10] of integer;; 3. var x, i: integer;; 4. procedure readarray;; 5. var i : integer;; 6. begin a end { readarray } ;; 7. procedure exchange(i, j: integer);; 8. begin 9. x := a[i];; a[i] := a[j];; a[j] := x;; 10. end { exchange } ;; 11. procedure quicksort(m, n : integer);; 12. var k, v: integer;; 13. function partition(y, z : integer) : integer;; 14. var i, j : integer;; 15. begin a 16. v 17. exchange(i,j);; 18. end { partition } 19. begin end { quicksort } 20. begin end { sort }

Access Links and How to Use Them Suppose procedure p at lexical nesting depth n p refers to non-local variable a at depth n q n p, them a can be found: 1. Follow n p - n q s from AR of p 2. Access the variable at offset a in that AR Example: partition code at depth = 3 refers to v and a at depth 2 and 1 for which the code should traverse 1 and 2 s respectively. Since (n p - n q ) can be computed at compile-time this tracing method is always feasible. What happens if n q > n p?

Access Links and How to Use Them Suppose procedure p at lexical nesting depth n p refers to non-local variable a at depth n q n p, them a can be found: 1. Follow n p - n q s from AR of p 2. Access the variable at offset a in that AR Example: partition code at depth = 3 refers to v and a at depth 2 and 1 for which the code should traverse 1 and 2 s respectively. Since (n p - n q ) can be computed at compile-time this tracing method is always feasible. What happens if n q > n p? Variable a is not visible! The compiler will never allow this access.

How to Set Up Access Links? Procedure p at depth n p calls q at depth n q Code generated as part of the calling sequence: Case n q > n p : procedure q is nested more deeply than p; it must be declared within p, i.e. n q = n p + 1; Why? Action: copy ARP pointer of the caller s to the callee s as this creates an additional indirection (another level) Case n q n p : all the ARs of the procedures up to p are the same, simply need to access the link of the most recent invocation of p; Action: Follow n q - n p +1 s you reach the correct AR of procedure r that encloses p to set the in the AR of q.

How to Set Up Access Links? Code generated as part of the calling sequence: Case n q > n p : procedure q is nested more deeply than p; it must be declared within p, i.e. n q = n p + 1; Why? Action: copy ARP pointer of the caller s to the callee s Example: quicksort (n p =2) calls partition(n p =3) sort a,x quicksort(1,9) quicksort(1,3) sort a,x quicksort(1,9) quicksort(1,3) partition(1,3) i,j

How to Set Up Access Links? Code generated as part of the calling sequence: Case n p > n p : all the ARs of the procedures up to p are the same, simply need to access the link of the most recent invocation of p; Action: Follow n q - n p +1 s you reach the correct AR of procedure r that encloses p to set the in the AR of q. Example: quicksort (n p =2) calls quicksort (n p =2) sort a,x quicksort(1,9) sort a,x quicksort(1,9) quicksort(1,3)

Lexical Scope with Nested Procedures sort sort sort sort a,x a,x a,x a,x quicksort(1,9) quicksort(1,9) quicksort(1,9) quicksort(1,9) quicksort(1,3) quicksort(1,3) quicksort(1,3) partition(1,3) partition(1,3) i,j i,j exchange(1,3) sort calls quicksort quicksort calls quicksort quicksort calls partition partition calls exchange n q = n p + 1 n q = n p n q = n p + 1 n p = n q + 1

Display Following Access Links can take a long Time Solution? Keep an auxiliary array of pointers to AR on the stack Storage for a non-local at depth i is in the activation record pointed to by d[i] called display. Faster because you need to follow a single pointer How to Maintain the display? When AR of procedure at depth i is set up: Save the value of d[i] in the new AR Set d[i] to point to the new AR. Just before an activation ends, d[i] is reset to the saved value Values in saved at a specific offset on the AR like ARP and return

Display: Example d[1] d[2] d[3] sort a,x P at depth n p calls Q at depth n q Case n q > n p : then n q = n p + 1 First n p elements of the display do not change; just set d[n q ] to new AR. Case n q n p : Enclosing procedures at levels 1,..., n q -1 are the same; save old d[n q ] in new AR and make d[n q ] point to new AR.

Display: Example d[1] d[2] d[3] sort a,x quicksort(1,9) P at depth n p calls Q at depth n q Case n q > n p : then n q = n p + 1 First n p elements of the display do not change; just set d[n q ] to new AR. Case n q n p : Enclosing procedures at levels 1,..., n q -1 are the same; save old d[n q ] in new AR and make d[n q ] point to new AR.

Display: Example d[1] d[2] d[3] sort a,x quicksort(1,9) quicksort(1,3) saved d[2] P at depth n p calls Q at depth n q Case n q > n p : then n q = n p + 1 First n p elements of the display do not change; just set d[n q ] to new AR. Case n q n p : Enclosing procedures at levels 1,..., n q -1 are the same; save old d[n q ] in new AR and make d[n q ] point to new AR.

Display: Example d[1] d[2] d[3] sort a,x quicksort(1,9) quicksort(1,3) saved d[2] partition(1,3) P at depth n p calls Q at depth n q Case n q > n p : then n q = n p + 1 First n p elements of the display do not change; just set d[n q ] to new AR. Case n q n p : Enclosing procedures at levels 1,..., n q -1 are the same; save old d[n q ] in new AR and make d[n q ] point to new AR. i,j

Display: Example d[1] d[2] d[3] sort a,x quicksort(1,9) quicksort(1,3) saved d[2] partition(1,3) P at depth n p calls Q at depth n q Case n q > n p : then n q = n p + 1 First n p elements of the display do not change; just set d[n q ] to new AR. Case n q n p : Enclosing procedures at levels 1,..., n q -1 are the same; save old d[n q ] in new AR and make d[n q ] point to new AR. i,j exchange(1,3) saved d[2]

Display: Example d[1] d[2] d[3] sort a,x quicksort(1,9) quicksort(1,3) saved d[2] partition(1,3) P at depth n p calls Q at depth n q Case n q > n p : then n q = n p + 1 First n p elements of the display do not change; just set d[n q ] to new AR. Case n q n p : Enclosing procedures at levels 1,..., n q -1 are the same; save old d[n q ] in new AR and make d[n q ] point to new AR. i,j exchange(1,3) saved d[2]

Display: Example d[1] d[2] d[3] sort a,x quicksort(1,9) quicksort(1,3) saved d[2] partition(1,3) P at depth n p calls Q at depth n q Case n q > n p : then n q = n p + 1 First n p elements of the display do not change; just set d[n q ] to new AR. Case n q n p : Enclosing procedures at levels 1,..., n q -1 are the same; save old d[n q ] in new AR and make d[n q ] point to new AR. i,j

Display: Example d[1] d[2] d[3] sort a,x quicksort(1,9) quicksort(1,3) saved d[2] P at depth n p calls Q at depth n q Case n q > n p : then n q = n p + 1 First n p elements of the display do not change; just set d[n q ] to new AR. Case n q n p : Enclosing procedures at levels 1,..., n q -1 are the same; save old d[n q ] in new AR and make d[n q ] point to new AR.

Display: Example d[1] d[2] d[3] sort a,x quicksort(1,9) P at depth n p calls Q at depth n q Case n q > n p : then n q = n p + 1 First n p elements of the display do not change; just set d[n q ] to new AR. Case n q n p : Enclosing procedures at levels 1,..., n q -1 are the same; save old d[n q ] in new AR and make d[n q ] point to new AR.

Display: Example d[1] d[2] d[3] sort a,x P at depth n p calls Q at depth n q Case n q > n p : then n q = n p + 1 First n p elements of the display do not change; just set d[n q ] to new AR. Case n q n p : Enclosing procedures at levels 1,..., n q -1 are the same; save old d[n q ] in new AR and make d[n q ] point to new AR.

Communicating Between Procedures Most languages provide a parameter passing mechanism Expression used at call site becomes variable in callee Two common binding mechanisms Call-by-reference passes a pointer to actual parameter Requires slot in the AR (for address of parameter) Multiple names with the same address? Call-by-value passes a copy of its value at time of call Requires slot in the AR Each name gets a unique location (may have same value) Arrays are mostly passed by reference, not value Can always use global variables

Complications Passing Functions as Arguments? Just an address of the code, not that hard to implement Need to verify at run-time the number of arguments. Access Link needs to be passed to understand how to follow it What if AR outlives Execution of Procedure? When is this possible? What to do? Dynamically Linked Libraries What are the issues with regular libraries? Position-Independent Code Why is this important?

Code Sharing Traditionally Link all Libraries with your code Drawbacks: Space as each executable includes the code of all libraries it uses (big as every function needs to be included at link time) Bugs in libraries require recompilation and linking Solution: Dynamically Linked Libraries Loaded and linked on-demand during execution Advantages: Single Copy in the system rather than replicated. Executable has only what is really needs. Bugs can be fixed later not requiring re-linking

Shared Libraries Make it Look Like a Statically Linked Linking? Name Resolution: finding bindings for symbols Determine before hand if linking will succeed Check for undefined or multiply defined symbols Create a table of symbols for each shared library Pre-execution linking checks the tables Run-time dynamic linker is guaranteed to fail if and only if the preexecution static linker would.

Summary What Have We Learned? AR is a Run-time Structure to hold State regarding the Execution of a Procedure AR can be allocated in Static, Stack or even Heap Links allow Call-Return and Access to Non-local Variables Symbol-Table plays Important Role Linkage Conventions Saving Context before Call and restoring after Call Need to understand how to generate code for body