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C Stack Frames / Pointer variables. Stack: Local Variables Pass & Return values Frame Ptr linkage (R5) and PC linkage (R7) Pointer Variables: Defining & using Pass by value Pass by reference (pointer). Compiling C. Allocating Space for Variables. x0000. Trap Vectors. x0100.
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C Stack Frames / Pointer variables Stack: Local Variables Pass & Return values Frame Ptr linkage (R5) and PC linkage (R7) Pointer Variables: Defining & using Pass by value Pass by reference (pointer)
Allocating Space for Variables x0000 Trap Vectors x0100 Intr Vectors • Global data section • All global variables stored here(actually all static variables) • R4 points to Global Variables • Run-time stack • Used for local variables (among other things) • R6 points to top of stack • R5 points to top frame on stack • New frame created for each “block”or scope (goes away when block exited) • Accessing a variable: • Global: LDR R1, R4, #x • Local: LDR R2, R5, #-y Offset = distance from beginning of storage area x0200 Op Sys x3000 instructions PC R4 global data Heap R6 run-time stack R5 xFE00 Device Registers xFFFF
Example C program #include <stdio.h> int main() { /* Declare local variables */ int amount; /* The number of bytes to be transferred */ int rate; /* The average network transfer rate */ int time; /* The time, in seconds, for the transfer */ int hours; /* The number of hours for the transfer */ int minutes; /* The number of mins for the transfer */ int seconds; /* The number of secs for the transfer */ /* Get input: number of bytes and network transfer rate */ printf("How many bytes of data to be transferred? "); scanf("%d", &amount); printf("What is the transfer rate (in bytes/sec)? "); scanf("%d", &rate); /* Calculate total time in seconds */ time = amount / rate; /* Convert time into hours, minutes, seconds */ hours = time / 3600; /* 3600 seconds in an hour */ minutes = (time % 3600) / 60; /* 60 seconds in a minute */ seconds = ((time % 3600) % 60); /* remainder is seconds */ /* Output results */ printf("Transfer Time : %d h %d m %d s \n", hours, minutes, seconds); return 0 }
Symbol Table • Like assembler, compiler needs to know information associated with identifiers • in assembler, all identifiers were labels and information is address • Compiler keeps more information - Name (identifier) - Type - Location in memory - Scope Where are local variables stored? Why?
Local Variable Storage • Local variables are stored in anstack frame. (also known as anactivation record) • Symbol table “offset” gives thedistance from the base of the frame. • R5is theframe pointer– holds addressof the base of the current frame. • Because stack grows downward,base is the highest address of the frame,and variable offsets are <= 0. seconds minutes hours time rate amount R5
Context Frame (Activation Record) Format(Note: you will see that there is some inconsistency as to where the Frame begins) R6 Function stacked stuff …….. …….. Local Variables Caller’s Frame Pointer (R5) Caller’s R7 (contains caller’s caller’s PC) Function Return Value Function Pass Value 1 …….. Function Pass Value n R5 - - - - - - - - Frame …….. Local Variables “Previous R5” - - - - - - - -
Function Call Implementation Caller pushes arguments (last to first). Caller invokes subroutine (JSR). Callee allocates return value, pushes R7 and R5. Callee allocates space for local variables. Callee executes function code. Callee stores result into return value slot. Callee pops local variables, pops R5, pops R7. Callee returns RET (or JMP R7). Caller loads return value and pops arguments. Caller resumes computation…
Function Call Implementation (with nested calls) Caller pushes arguments (last to first). Caller invokes subroutine (JSR). Callee allocates return value, pushes R7 and R5. Callee allocates space for local variables. Callee executes function code. Caller pushes arguments (last to first). Caller invokes subroutine (JSR). Caller loads return value and pops arguments. Caller resumes computation… Callee stores result into return value slot. Callee pops local variables, pops R5, pops R7. Callee returns RET (or JMP R7). Caller loads return value and pops arguments. Caller resumes computation…
Context Frame or Activation Record Format (Stack PTR) R6 Function stacked stuff …….. …….. Local Variables Caller’s Frame Pointer (R5) Caller’s R7(contains ITS caller’s PC) Function Return Value Function Pass Value 1 …….. Function Pass Value n Called Program Called Program (Frame PTR) R5 Called Program Called Program Calling program …….. Local Variables “PUSHED” on Stack By:
Declaring Pointer variables int *ptr; ptr is a pointer variable that points to an int type variable char *cp; cp is a pointer variable that points to a character type variable double *dp; dp is a pointer variable that points to a double type variable *is referred to as theindirection operator, or dereference operator *ptr returns the value of the variable pointed to by pointer variable ptr & is referred to as the unaryor monadicoperator &variable returns the address of the variable variable
Using Pointer Variables • A pointer is a variable which contains the address in memory of another variable. • We can have a pointer to any variable type. • The unary or monadic operator & provides the “address of a variable”. • The indirection or dereference operator * gives the “contents of an object pointed • to by a pointer variable”. • To declare a pointer to a variable:int *pointer; int x=1, y=2; /* let x be at x3000 and y at 3001 */ int *ip; /* ip is an int type pointer */ ip=&x; /* ip=x3000 x= 1 y=2 */ y=*ip /* ip=x3000 x= 1 y=1 */ x=ip; /* ip=x3000 x=x3000 y=1 */ *ip=3; /* ip=x3000 x=3 y=1 */ Note: ip = ip + 1 actually increments ip by 4. Why?
Example Define local variables: int object; int *ptr; Now, let’s assign values to them: object = 4; ptr = &object *ptr = *ptr + 1; The last statement above is equivalent to: object = object + 1 What are the final values of object and ptr ?
Example: Parameter Passing by Value #include <stdio.h> void Swap(int firstVal, int secondVal); int main() { int valueA = 3; int valueB = 4; Swap(valueA, valueB); return 0; } void Swap(int firstVal, int secondVal) { int tempVal; /* Holds firstVal when swapping */ tempVal = firstVal; firstVal = secondVal; secondVal = tempVal; return; } Snapshot before return from subroutine
Example: Parameter Passing by Reference #include <stdio.h> void NewSwap(int *firstVal, int *secondVal); int main() { int valueA = 3; int valueB = 4; NewSwap(&valueA, &valueB); return 0; } void NewSwap(int *firstVal, int *secondVal) { int tempVal; /* Holds firstVal when swapping */ tempVal = *firstVal; *firstVal = *secondVal; *secondVal = tempVal; return; } Snapshots During the Exchange What happened differently using pass by reference, rather than pass by value ?
Scanf( ) function • Recall reading from the keyboard in C: scanf(“%d”, &input); • Why do we use &input ?
Pointer Example #include <stdio.h> int IntDivide(int x, int y, int *quoPtr, int *remPtr); int main() { int dividend; /* The number to be divided */ int divisor; /* The number to divide by */ int quotient; /* Integer result of division */ int remainder; /* Integer remainder of division */ int error; /* Did something go wrong? */ printf("Input dividend: "); scanf("%d", ÷nd); printf("Input divisor: "); scanf("%d", &divisor); error = IntDivide(dividend,divisor,"ient,&remainder); if (!error) /* !error indicates no error */ printf("Answer: %d remainder %d\n", quotient, remainder); else printf("IntDivide failed.\n"); } int IntDivide(int x, int y, int *quoPtr, int *remPtr) { if (y != 0) { *quoPtr = x / y; /* Modify *quoPtr */ *remPtr = x % y; /* Modify *remPtr */ return 0; } else return -1; }