Accessing Memory Chapter 5 Lecture notes for SPARC Architecture, Assembly Language Programming and C, Richard P. Paul by Anu G. Bourgeois

2 Memory Addresses are 32 bits wide Therefore, 2 32 bytes in memory Each location is numbered consecutively Memory data types –Byte = 1 byteHalfword = 2 bytes –Word = 4 bytesDoubleword = 8 bytes

3 Data types C TypeSPARCBitsUnsignedSigned charbyte80, , 127 shorthalf160, 65,535-32,768, 32,767 int, longword320, 4.294x10 9  x 10 9 All memory references must be aligned x byte quantities must begin in an address divisible by x

Memory Allocation.section “.data” input:.word 0x limit:.half 0x9a prompt:.asciz “Hello!” Address (decimal) Data (hex) a ‘H’ 307‘e’ 308‘l’ 309‘l’ 310‘o’ 311‘!’ input limit prompt

Memory Allocation.section “.data” input:.word 0x limit:.half 0x9a prompt:.asciz “Hello!” Address (decimal) Data (hex) a ‘H’ 307‘e’ 308‘l’ 309‘l’ 310‘o’ 311‘!’ – divisible by 4 input takes up 4 byte locations

Memory Allocation.section “.data” input:.word 0x limit:.half 0x9a prompt:.asciz “Hello!” Address (decimal) Data (hex) a ‘H’ 307‘e’ 308‘l’ 309‘l’ 310‘o’ 311‘!’ – divisible by 2 limit takes up 2 byte locations Note: location 305 will have leading zeroes fill in the extra byte not specified in the data section

Memory Allocation.section “.data” input:.word 0x limit:.half 0x9a prompt:.asciz “Hello!” Address (decimal) Data (hex) a ‘H’ 307‘e’ 308‘l’ 309‘l’ 310‘o’ 311‘!’ – divisible by 1 Each element of prompt takes up 1 byte location

8 Addressing Variables Load and Store operations are the only instructions that reference memory Both instructions take two operands –One memory, and one register Can access memory using different data types (byte, half word, word, double word)

9 Load Instructions MnemonicOperation ldsb Load signed byte, propagate sign left in register ldub Load unsigned byte, clear high 24 bits of register ldsh Load signed halfword, propogate sign left in register lduh Load unsigned halfword, clear high 16 bits of register ld Load word ldd Load double, reg. # even, first 4 bytes into reg. n, next 4 into reg. n + 1

10 set input, %o0 ld [%o0], %o1 %o1 = 0x ldub [%o0 + 7], %o2 %o2 = ‘e’ ldsh [%o0 + 3], %o3 error ldsh [%o0 + 4], %o4 %o4 = 0x9a ldsb [%o0 + 4], %o5 %o5 = 0xffffff9a

11 Store Instructions MnemonicOperation stb Store low byte of register, bits 0-7, into memory sth Store low two bytes of register, bits 0-15 into memory st Store register std Store double, reg. # even, first 4 bytes from reg. n, next 4 from reg. n + 1 Why don’t we have all the same options as we did for the load instructions?

Address (decimal) Data (hex) mov 300, %o0 mov 0x , %o1 st %o1, [%o0] sth %o1, [%o0 + 6] sth %o1, [%o0 + 9] stb %o1, [%o0 + 13] error 78

Address (decimal) Data (hex) mov 0x9abcdef0, %o2 mov 0x , %o3 std %o2, [%o0 + 4] f0 de bc 9a

Rules to Remember 1.Lower byte Lower address 2.Reference is to the lowest address 3.Memory references must be byte aligned 14

Data Section.section “.data” first:.word 0x03, 0x0ef321 second:.byte 0x5c.align 2 third:.half 0x987, 0x7e string:.asciz “done” PointerAddressdata first f3 4060e second4085c 409undef third e string414‘d’ 415‘o’ 416‘n’ 417‘e’

Allocating Space.skip is a psedo-op that will provide space without any initialization my_array:.skip 4*100 Provides space for a 100-word unintialized array 16