Memory address mapping of u-boot

summary

1. This article mainly analyzes Memory address mapping
2. ARM The length in the segmented mapping of is 1MB, So we're all 4G Need... Within the scope 4G/1MB=4096 A mapping unit

Code

1. Memory address mapping 1

.macro FL_SECTION_ENTRY base,ap,d,c,b
	.word (\base << 20) | (\ap << 10) | \
	      (\d << 5) | (1<<4) | (\c << 3) | (\b << 2) | (1<<1)
.endm
mmu_table:
	.set __base,0
	// Access for iRAM
	.rept 0x100
	FL_SECTION_ENTRY __base,3,0,0,0
	.set __base,__base+1
	.endr

• .rept and .endr It's a pair of features 0x100 It's ten years ago 256, So the cycle 256 Time , Then this paragraph is 256MB
• FL_SECTION_ENTRY Before macro 20 Bit sets the base address
• In the loop body ,__base For the initial 0, Then add 1 Enter again FL_SECTION_ENTRY, Reason by analogy .
• .rept 0x100 Can be seen as .rept 0x100 - 0x0, Then the mapped virtual address is 0 To 0x10000000
• Physical address from 0 Start , until 0x1000000, Therefore, the mapped virtual address is equal to the real address
• Every time you move left 20 position （10 Is it K,20 Is it M）, So from 0 To 0x100,base The correspondence between and address is as follows

address	base
0	0
0x00100000	1
0x00200000	2
0x00300000	3
0x00400000	4
……	……
0x10000000	256(0x100)

2. Memory address mapping 2

// Not Allowed
.rept 0x200 - 0x100
.word 0x00000000
.endr

• Circulated 0x200-0x100=0x100（256） Time , Is mapped 256MB
• The mapping of 256MB The virtual address of corresponds to the physical address 0 Address

3. Memory address mapping 3

.set __base,0x200
// should be accessed
.rept 0x600 - 0x200
FL_SECTION_ENTRY __base,3,0,1,1
.set __base,__base+1
.endr

• Mapping 1G(0x400) Of memory space
• Both physical and virtual addresses are from 0x20000000 From to 0x60000000

4. Memory address mapping 4

.rept 0x800 - 0x600
.word 0x00000000
.endr

• Mapping 512MB（0x200） Space to 0 Address

5. Memory address mapping 5

.set __base,0x800
.rept 0xb00 - 0x800
FL_SECTION_ENTRY __base,3,0,0,0
.set __base,__base+1
.endr

• Mapping 768MB(0x300) Of memory space
• Both physical and virtual addresses are from 0x80000000 to 0xb0000000

6. Memory address mapping 6

.set __base,0xB00
.rept 0xc00 - 0xb00
FL_SECTION_ENTRY __base,3,0,0,0
.set __base,__base+1
.endr

• Mapping 256MB(0x100) Memory space
• Both physical and virtual addresses are from 0xb0000000 From to 0xc0000000

7. Memory address mapping 7

// 0xC000_0000 Mapping to 0x2000_0000
.set __base,0x300
//.set __base,0x200
// 256MB for SDRAM with cacheable
.rept 0xD00 - 0xC00
FL_SECTION_ENTRY __base,3,0,1,1
.set __base,__base+1
.endr

• Mapping 256MB(0x100) Memory space
• Physical address from 0x30000000 To 0x40000000
• Virtual address from 0xc0000000 To 0xd0000000

8. Memory address mapping 8

set __base,0xD00
.rept 0x1000 - 0xD00
FL_SECTION_ENTRY __base,3,0,0,0
.set __base,__base+1
.endr

• Mapping 768MB（0x300） Memory space
• Both physical and virtual addresses are from 0xd0000000 To 0x1000000000

summary

Virtual address （16 Base number ）	Physical address （16 Base number ）	length
0-10000000	0-10000000	256MB
10000000-20000000	0	256MB
20000000-60000000	20000000-60000000	1GB
60000000-80000000	0	512MB
80000000-b0000000	80000000-b0000000	768MB
b0000000-c0000000	b0000000-c0000000	256MB
c0000000-d0000000	30000000-40000000	256MB
d0000000-100000000	d0000000-0x100000000	768MB

• So this MMU The function of is to put the virtual address 0xc0000000-0xd0000000 Mapping to 0x30000000-0x40000000
• It is the same as the physical address except for two unreachable virtual addresses
• In the physical address 30000000-40000000 Did Twice mapping , So here uboot in , Virtual address 30000000-40000000 And c0000000-d0000000 yes The same address , It's just that it may be in different on-board chips .