Memory Hierarchy Locality of Reference Cache Memory Organization Virtual Memory Today's Agends.
Memory Organization II
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Memory Organization II Microprocessor and Interfacing 261214
description
Memory Organization II. Microprocessor and Interfacing 261214. PIC Flash Memory Segments. 12. 10. 0. 2K. 2K. 2K. 2K. PIC’s RAM Banking. Instruction Format. f = 7 bit Maximum memory = 2^7 = 128 Bytes. PIC 16F877 has 512 bytes of RAM: How do we access all of it?. 7. 0. - PowerPoint PPT Presentation
Transcript of Memory Organization II
Memory Organization IIMicroprocessor and Interfacing
261214
PIC Flash Memory Segments
2K2K2K2K
01012
PIC’s RAM Banking
Instruction Format
f = 7 bitMaximum memory = 2^7 = 128 Bytes
PIC 16F877 has 512 bytes of RAM: How do we access all of it?
STATUS (F# 0x03)
Bit 5-6 (RP0,RP1) in STATUS are used for RAM Page selection
07
Total Memory becomes 2^9 = 512 Bytes
Creating a 9 bit RAM address
STATUS
07
068
f (File#)
06713
9 Bit RAM Address
OPCODE
void main(){
int i;i = 1;i++;
}
.................... int i;
....................
.................... i=1; 000D: MOVLW 01000E: BCF 03.5000F: MOVWF 21.................... i++; 0010: INCF 21,F
This code should now make sense
RAM location
First 32 bytes of every pageare reserved. Except on pages 3 and 4
User RAM space.Last 16 Bytes are mirrored
The big picture: where the bits are
Bit 5-6 (RP0,RP1) in STATUS are used for RAM Page selection
07
Bit 3-4 in PCLATH are usedfor memory access
07 4 3
6 5
2. เร่�ง Memory Operations ด้�วย Indirect Addressing
ทำ��ไมต้�องม� Indirect Addressing?สมม�ติ�ว่�าเราม� Array ในภาษา C ดั�งน��
int value[8];
หลั�งจากที่��ใช้�งานม�นไปส�กพั�กเราอยาก Clear ค่�าใน array ที่��งหมดัม�ค่�าเป#น 0
Code จะเป#นอย�างไรถ้�าเราไม�สามารถ้ใช้�ติ�ว่แปรอ�างติ'าแหน�งใน Array เช้�นห�ามใช้� value[i] = 0;
With indirect addressing
For (i=0 ; i<8 ; i++) {value[i] = 0;
}
Without indirect addressing
Value[0] = 0;Value[1] = 0;Value[2] = 0;Value[3] = 0;Value[4] = 0;Value[5] = 0;Value[6] = 0;Value[7] = 0;
PIC’s Indirect Addressing
Opcode File
Creating a 9 bit indirect RAM address
STATUS
07
078
8-bit address
07
9 Bit RAM Address
FSR (File #4)
Example: clear RAM locations 0x20 – 0x27 (8 Bytes)
Instruction Comment
BCF 0x03, 7 ;select page
MOVLW 0x20
MOVWF 0x04 ;Set FSR
Next CLRF 0x00 ;clear RAM
INCF FSR,F ;increase counter
BTFSS FSR,3 ;all done?
GOTO Next ; no, clear next
0x20 = 0b0010 00000x27 = 0b0010 0111
The big picture: where the bits are
Bit 5-6 (RP0,RP1) in STATUS are used for RAM Page selection
07
Bit 3-4 in PCLATH are usedfor memory access
07 4 3
6 5
Bit 7 (IRP) in STATUS is used for indirect Addressing
Indirect Addressing ExFind the sum of values in RAM address 0x22-0x25 and storeat address 0x21
Address Function
0x21 Sum
0x22 Value1
0x23 Value2
0x24 Value3
0x25 Value4
End Result: Sum = Value1+Value2+Value3+Value4
; 0x21 = sum result variableclrf 0x21; 0x22 - 0x25 = input numbers to be summedmovlw 4movwf 0x22movlw 3movwf 0x23movlw 2movwf 0x24movlw 1movwf 0x25
