Problem with Scanning an Infix Expression

Transcription

1 Operator Notation Consider the infix expression (X Y) + (W U), with parentheses added to make the evaluation order perfectly obvious. This is an arithmetic expression written in standard form, called infix form. There are two other forms, prefix and postfix, which are occasionally used in software. Infix form has the operator in between two operands, as in X + Y. It is what we are used to, but it is hard for computers to process. The two other forms are prefix and postfix. I give the LISP variant of prefix. Prefix: (+ ( X Y) ( W U) ) Postfix: XY WU + Implicit in each of these examples is the assumption that single letter variables are used. Each of the prefix and postfix is easier for a computer to interpret than is infix. The assumption is that the expression is scanned either left to right or right to left. Here, we arbitrarily scan left to right. We return to this expression, but first consider a much simpler expression.

2 Problem with Scanning an Infix Expression Suppose that we have a five symbol expression. We are scanning left to right. We assume: all variables are single alphabetic characters Each alphabetic characters is taken singly and represents a variable At the beginning we have five unknown symbols: We read the first character and find it to be a variable: We read the next character and find it to be an operator: Something is being added. X X + We read the next character and find it to be a variable: X + Y. Can we do the addition now? It depends on what the next character is. There are two basic possibilities: X + Y + Z X + Y Z In the first option, we can do the addition immediately, forming (X + Y) to which Z is later added. In the second option, we must wait and first do the multiplication to get (Y Z) to which X is added.

3 Scanning a Postfix Expression With the same assumptions, we scan a five symbol postfix expression. At the beginning we have five unknown symbols: We read the first character and find it to be a variable: Y We read the next character and find it also to be a variable: Y Z At this point we need an operator. We read the next character and find it to be an operator: Because this is postfix, the evaluator can immediately form the product term (Y Z) We read the next character and find it to be a variable: At this point, we need another operator. We have two operands: (Y Z) and X. Y Z Y Z X We read the last character and find it to be an operator: Y Z X + We do the addition and have the term (Y Z) + X.

4 The Stack Data Type Stacks are LIFO (Last In, First Out) data structures. Stacks are the data structure most naturally fit to evaluate expressions in postfix notation. The stack is best viewed as an abstract data type with specific methods, each of which has a well defined semantic (we can say exactly what it does). We can also consider the stack as having a top. Items are added to the stack top and then removed from the stack top. The position of the stack top is indicated by a stack pointer (SP). More on this later; it has two different definitions, each of which is valid. The stack data type has three necessary methods, and two optional methods. The mandatory methods are Push, Pop, and IsEmpty. The standard optional methods are Top, and IsFull. Note: When studying computer networks, we speak of a protocol stack, which has nothing to do with the ADT (Abstract Data Type) stack that we study today. We shall discuss the TCP/IP protocol stack later in this course.

5 Pushing Items onto the Stack Consider the sequence: Push (X), Push (Y), and Push (Z). This sequence adds three items to the stack, in the specified order. Push X Push Y Push Z After the third operation, Z is at the top of the stack and is the first to be removed when a Pop operation is performed. The SP (Stack Pointer) is a data item internal to the stack that indicates the location of the top of the stack. It is never made public by any proper stack implementation.

6 Evaluation of the Postfix Expression Y Z X + Postfix notation is also called RPN for Reverse Polish Notation. Rules: 1. Scan left to right. 2. Upon reading a variable, push it onto the stack. 3. Upon reading a dyadic operator, pop two variables from the stack, perform the operation, and push the result. We read the first character and find it to be a variable: Y We read the next character and find it also to be a variable: Y Z We read the next character and find it to be an operator: Y Z

7 Evaluation of the Postfix Expression Y Z X + Continues We read the next character and find it to be a variable: Y Z X We read the last character and find it to be an operator: Y Z X + After this, the stack will be popped and the value placed into some variable. This example just looked at the expression, without considering the assignment.

8 Stack Based (Zero Operand) Machine Evaluate the expression Z = (X Y) + (W U) which must be evaluated in its postfix (RPN) version: XY WU + Push Push Mult Push Push Mult Add Pop X Y W U Z X X Y X Y X Y W X Y W U X Y W U X Y +W U

9 More Explicit Evaluation: W = Y Z It might be good at this point to make explicit a point that many consider obvious. Suppose that before the evaluation above, W = 0, Y = 10, and Z = 15. The push pop code to execute this instruction will go as follows: PUSH Y // Value of Y is placed on the stack. PUSH Z MULT // Value of Z is placed onto the stack. // Pop the top two values, multiply them // and push the result onto the stack. POP W // Pop the value at stack top and place into W. At the start, here is what we have.

10 PUSH Y More Explicit Evaluation: Part 2 // Value of Y is placed on the stack. PUSH Z // Value of Z is placed onto the stack.

11 MULT More Explicit Evaluation: Part 3 // Pop the top two values, multiply them // and push the result onto the stack. POP W // Pop the value at stack top and place into W.

12 Practical Evaluation of Postfix Operators Here are some steps for manual evaluation. 1. Scan the postfix expression left to right. Find the first operator. 2. Determine how many operands that operator requires. Call it N. Select the previous N operands, apply to the operator. 3. Remove the operator and its operands from the expression. Replace these with the operands results. 4. Repeat Here is an example, using single digit numbers to start with * (6 5 +) 4 2 * (11) 4 2 * (11) (4 2 ) * (11) (2) * 22

13 Practical Evaluation of Prefix Operators Note: We assume Lisp notation, with full use of parentheses. 1. Scan the expression left to right. If the expression does not contain another parenthesized expression, evaluate it. Otherwise, attempt to evaluate its subexpression. Formally, this is viewed as a tree traversal. 2. Keep moving up until the last is evaluated. Here is an example. Again, it will start with single digit numbers in order to be more easily read. Evaluate the prefix expression: ( + ( * 4 2 ) ( 5 2 ) ( * ( + 6 4) (+ 1 1 ) ) ) (* 4 2) can be evaluated, so: ( + 8 ( 5 2 ) ( * ( + 6 4) (+ 1 1 ) ) ) ( 5 2) is defined as 5 2, so: ( ( * ( + 6 4) (+ 1 1 ) ) ) (+ 6 4) can be evaluated, so: ( ( * 10 (+ 1 1 ) ) ) (+ 1 1) can be evaluated, so: ( ( * 10 2 ) ) (* 10 2) can be evaluated, so: ( ) ( ) = 31, so: 31

14 Expression Tree for the Prefix Evaluation The expression is ( + ( * 4 2 ) ( 5 2 ) ( * ( + 6 4) (+ 1 1 ) ) ). Software to evaluate this prefix expression would first build this tree (not very hard to do) and then evaluate it. For example, the node (+ 6 4) would be replaced by 10, etc.

15 Expression Tree Evaluation

16 Single Accumulator Machine Evaluate the expression Z = (X Y) + (W U) This requires an extra storage location, which I call T for Temp. Load X // AC now has X Mult Y // AC now has (X Y) Store T // Now T = (X Y). AC still has (X Y) Load W // AC now has W Mult U // AC now has (W U) Add T // AC now has (X Y) + (W U) Store Z // And so does Z.

17 Multiple Register Machines Evaluate the expression Z = (X Y) + (W U) Standard CISC Approach (There are many variants of this one.) Load R1, X // R1 has X Mult R1, Y // R1 has (X Y) Load R2, W // R2 has W Mult R2, U // R2 has (W U) Add R1, R2 // R1 has (X Y) + (W U) Store R1, Z // And so does Z Load Store RISC Load R1, X // R1 has X Load R2, Y // R2 has Y Mult R1, R2 // R1 has (X Y) Load R2, W // R2 now has W Load R3, U // R3 has U Mult R2, R3 // R2 has (W U) Add R1, R2 // R1 has (X Y) + (W U) Store R1, Z // And so does Z.

18 Instruction Types There are basic instruction types that are commonly supported. 1. Data Movement (Better called Data Copy ) These copy bytes of data from a source to a destination. If X and Y are 32 bit real numbers, then the instruction Y = X makes a copy of the four bytes associated with X and places them in the address for Y. 2. Arithmetic The standard arithmetic operators are usually supported. If division is supported, one usually also has the mod and rem functions. On business oriented machines, decimal arithmetic is always supported. Graphics oriented machines usually support saturation arithmetic. Real number arithmetic is often not handled directly in the CPU, but by a coprocessor attached to it. Early on (Intel / 80487) this was a cost consideration. RISC machines follow this approach in order to keep the CPU simple. 3. Boolean Logic Here I differ from the book s description. Boolean instructions are often used for non Boolean purposes, such as bit manipulation. The real Boolean instructions are the conditional branches.

19 More Instruction Types 4. Bit Manipulation These use instructions that appear to be Boolean, but in fact operate differently. This is a distinction lost on many students, perhaps because it is rarely important. More on this in a moment. 5. Input / Output The computer must communicate with external devices, including permanent storage, printers and keyboards, process control devices (through dedicated I/O registers), etc. A few architectures have a dedicated input device and a dedicated output device. All commercial machines have addressable I/O devices; i.e., the CPU issues a device identifier that appears to be an address to select the device. From the CPU viewpoint, each I/O device is nothing more than a set of registers (control, status, and data) and some timing constraints. 6. Transfer of Control These alter the normal sequential execution of code. At the primitive machine language level, all we have is unconditional jumps, conditional jumps, and subroutine invocations. Higher level languages elaborate these features with structured constructs such as conditional statements, counted loops, and conditional loops.

20 More on Logical Instructions vs. Bitwise Instructions This is quite important in C, C++, and Java. Consider the standard implementation of Boolean values as 8 bit bytes. This is done for convenience in addressing by the CPU, as single bits are not addressable. Let A = C = B = D = Logical operators in C++ AND && Expression && Expression OR Expression Expression Bitwise operators in C++ AND & Expression & Expression OR Expression Expression XOR ^ Expression ^ Expression Logical Bitwise A && B = 0 (FALSE) A & B = A B = 1 (TRUE) A B = C && D = 1 (TRUE) C & D = C D = 1 (TRUE) C D = Source: The Annotated C++ Reference Manual (Sections ) Margaret Ellis and Bjarne Stroustrup, Addison Wesley, 1990.

21 A Context for Bitwise Operators For simplicity I consider a very old (late 1960 s) Line Printer, a predecessor to today s laser printer. We examine the Status/Control register for the LP 11. This register is called LPS for Line Printer Status in the literature. We have here two status bits and a control bit. Status bits Bit 15 Error If Error = 1, then there is a device error, such as power off, no paper in the printer, etc. Bit 7 Done If Done = 1, the printer is ready for the next line. Control bit Bit 6 IE If IE = 1, the printer is enabled to raise an interrupt whenever Done becomes 1 or Error becomes 1.

22 More on the LPS Register Why this arrangement of bits? The PDP 11, for which the LP 11 was used, did not support 8 bit arithmetic. A 16 bit integer was the smallest that the CPU would handle. Viewed as a 16 bit signed integer, we note that the error bit (Bit 15) is the sign bit. To test for an error, we just read the LPS into a register and test if it is negative. Testing the Done Bit Recall that the Done Bit is bit 7 and that is 0x0080. LPS E DI x LPS & 0x D If ( 0 = = (LPS & 0x0080) ) then the Line Printer is Not Done

23 Still More on the LPS Register Testing and Setting the Interrupt Enable Bit Recall that the Done Bit is bit 6 and that is 0x is 0xFFBF. Testing the Interrupt Enable Bit LPS E DI x LPS & 0x I If ( 0 = = (LPS & 0x0040) ) then the Line Printer Interrupt is disabled.

24 Yet More on the LPS Register Enabling Interrupts (Setting the I Bit) LPS E DI x LPS 0x0040 E D Setting LPS = LPS 0x0040 enables the interrupt and leaves the other bits unchanged. Disabling Interrupts (Clearing the I Bit) LPS E DI xFFBF LPS & 0xFFBF E D Setting LPS = LPS & 0xFFBF disables the interrupt and leaves the other bits unchanged.

25 Overview of Addressing Modes Many computer instructions involve memory, and must generate memory addresses. In this introduction, we shall discuss addressing modes in general terms. Later, we shall investigate examples specifically based on IA 32 code. All modes of addressing memory are based on an EA (Effective Address), which is the actual memory address issued by the control unit. First, we study some modes that do not address memory. Immediate Mode In this mode, there is no reference to memory. The argument is part of the instruction. The simplest example is the Boz 7 instruction HALT. There is no argument. In the IA 32 trap, int 21h, the hexadecimal value is part of the machine code. The machine language would have two bytes: 0xCC 21; the first byte is the opcode. Here is another example with decimal notation. MOV EAX, This sets the value in the EAX register to decimal Register Direct Here is a simple IA 32 example: MOV EAX, EBX. It moves the value in the EBX register into the EAX register. Memory is not referenced.

26 Calculating the Effective Address By definition of the term Effective Address, all memory references use that address when accessing memory. M[EA] denotes the contents of the effective address. Consider two basic operators: Load from memory and Store to memory Load Register from Memory: R M[EA] Store Register to Memory: M[EA] (R) Understanding addressing modes then becomes equivalent to understanding how to calculate the Effective Address. Each instruction referencing memory has fields (collections of bits) indicating how to compute the EA. In general, the only fields used are: the Address Field the Index Register this contains an address, which may be modified. this contains the register used as array index. If this field is 0, indexing is not used. the Indirect Bit this is used for Indirect and Indexed Indirect addressing. If this is 1, indirect addressing is used; otherwise not. Instructions that do not use an address field conventionally have that field set to 0.

27 Direct and Indirect Addressing Opcode Registers Mode Flags Address Field In each of direct and indirect addressing, the computation is simple: Just copy the address field itself. In direct addressing, the address field contains the address of the argument. In indirect addressing, the address field contains the address of a pointer to the argument. Put another way, the contents of the memory indicated by the address field contain the address of the argument. In subroutine linkage, argument passing by preference basically uses indirect addressing.

28 Direct and Indirect Addressing (Example) Consider the following address map as an example for our addressing modes. Z represents address 0x0A. Z == 0x0A and M[Z] = 5. Address A B C D E F Contents C 1E A 10 3 F E D 8 Direct Addressing LDR %R1, Z Effective Address: EA = Z Here EA = 0x0A. Effect: R1 M[Z] Here R1 M[0x0A] or R1 = 5. Indirect Addressing LDR %R1, *Z Effective Address: EA = M[Z] Here EA = M[0x0A] = 5 Effect: R1 M[EA] = M [ M[Z] ] Here R1 M[0x05] or R1 = 11. NOTE: These examples are based on the Boz 7 architecture, rather simpler than the IA 32. The Boz 7 is a didactic design by the author of these notes.

29 Indexed Addressing This is used to support array addressing in all computers. Languages such as C and C++ implement this directly using zero based arrays. In this mode, the contents of a general purpose register are used as an index. The contents of the register are added to the address field to form the effective address. EA = (Address Field) + (Register) Although we expect the index to be smaller than the value of the address, this may not always (or often) be the case. The contents of the address field should be considered as having a different data type from those of the register, which holds a signed integer.

30 Indexed Addressing (Example) LDR %R1, Z, 3 // Z indexed by the general purpose register R3. Z represents address 0x0A. Z == 0x0A and M[Z] = 5. (R3) = = 4 Address A B C D E F Contents C 1E A 10 3 F E D 8 Effective address: EA = Z + (%R3) Here EA = 0x0A + 4 = 0x0E Effect: R1 M[EA] = M[0x0E] or R1 = F In this, the most common, variant of indexed addressing we see Z as an array. The first five elements of the array are placed in memory as follows: Memory Map 0x0A 0x0B 0x0C 0x0D 0x0E Contents: Z[0] Z[1] Z[2] Z[3] Z[4]

31 Based Addressing This is quite similar to indexed addressing, so much so that it seems redundant. This mode arises from an entirely different consideration. Indexed addressing basically supports the use of arrays. Based addressing supports the Operating System in partitioning the user address space. User addresses are mapped into physical addresses that cannot conflict.

32 Indexed Indirect Addressing This is a combination of both modes. Question: Which is done first? Pre Indexed Indirect: The indexing is done first. We have an array of pointers. Post Indexed Indirect: The indirect is done first. We have a pointer to an array.

33 Indexed Indirect Addressing (Example) Let Z refer to address 0xA, so that M[Z] = = 0x05. Let (R3) = 0x07. Address A B C D E F Contents C 1E A 9 3 F E D 8 Pre Indexed Indirect: LDR %R1, *Z, 3 Effective address EA = M[ Z + (R3) ] = M[0xA + 0x7] = M[ 0x11] = 0x08. Effect Post Indexed Indirect: LDR %R1, *Z, 3 R1 M[ M[Z + (R3)] ] or R1 M[0x08] or R1 1E Effective address EA = M[Z] + (R3) = M[0x0A] + 0x07 = 0x05 + 0x07 = 0x0C. Effect R1 M[0x0C] or R1 0x09.

34 Double Indirect Addressing: Use 1 Suppose a common data structure that is used by a number of programs. In modern systems, this could be a DLL (Dynamic Linked Library) module. Each program contains a pointer to this structure, used to access that structure. Singly indirect addressing, in which the pointer is stored in every program, presents problems when the data block is moved by the memory manager. Every reference to that block must be found and adjusted. In doubly indirect addressing, all programs have a pointer to a data structure which itself is a pointer. When the data block is moved, only this one pointer must be adjusted. As an extra benefit, this intermediate pointer can be expanded to a data structure containing an ACL (Access Control List) and other descriptors for the data block.

35 Double Indirect Addressing: Use 2 The Operating System uses double indirection to activate a device driver associated with a given I/O device. For each type of I/O device, the O/S stores the address of a device vector, which contains the actual address of the software to handle the device. This arrangement allows the operating system and the device drivers to be developed independently. The only fixed location is the address of this device vector. When the driver software is loaded, the O/S places it in the memory location that is most convenient for the current configuration and copies that start address into the address part of the device vector. Double indirection is then used to call the driver. This is really a vectored interrupt mechanism, described more fully in a later chapter.