CS1021 Tutorial 2
Q1 The NXP LPC2468 has 512KiB of flash memory starting at address 0x00000000. What is the last address of the memory area (in hexadecimal)?
512KiB = 512 x 1024 = 524,288 bytes Convert 524,288 to hexadecimal
16 524,288
16 32,768 r0 16 2,048 r0 16 128 r0 16 8r0 16 0r8
524,288 = 0x00080000
Last address of memory area is 0x00000000 + 0x00080000 – 1 = 0x0007FFFF Need to subtract 1 because first byte at offset 0 (not 1).
Q2 The NXP LPC2468 has 64KiB of read write memory starting at address 0x40000000. What is the last address of the memory area (in hexadecimal)?
Q3 One hundred 8 bit unsigned integers are stored consecutively in memory starting at address 0x00002000. What is the address of the byte containing (i) the first integer (ii) the 22nd integer (iii) the 75th integer and (iv) the last integer?
Q4 One hundred 32 bit signed integers are stored consecutively in memory starting at address 0x004420C0. What is the address of the word containing (i) the first integer (ii) the 22nd integer (iii) the 75th integer and (iv) the last integer?
0x0044224C 0x00442248
0x004420CC 0x004420C8 0x004420C4 0x004420C0
The first integer is stored at 0x004420c0. As each integer is 32 bits or 4 bytes, the address of the 2nd integer is 0x004420c0 + 4 = 0x004420c4, and so on. The nth integer is at address 0x004420c0 + 4(n-1). If n = 100, the 100th integer is at address 0x004420c0 +
CS1021 Tutorial 2 2018 jones@scss.tcd.ie
last integer
99th integer
4th integer
3rd integer
2nd integer
first integer
1
…
CS1021 Tutorial 2 2018 jones@scss.tcd.ie 4*99 = 0x004420c0 + 0x018c = 0x0044224C. The addresses of the 22nd (0x00442114)
and 75th (0x004421E8) integers can be computed in a similar way.
Q5 Assuming that x is stored in R1, y in R2, z in R3 and the result in R0:
(i) Write ARM assembly language instructions to compute x + y + z.
ADD R0, R1, R2 ; R0 = x + y ADD R0, R0, R3 ; R0 = x + y + z
(ii) Write ARM assembly language instructions to compute y – x – z.
(iii) Write ARM assembly language instructions to compute x2 + y2 + z2.
(iv) Write ARM assembly language instructions to compute 5(x + y).
MOV R0,#5 ;R0=5
ADD R4, R1, R2 ; R4 = x + y MUL R0, R4, R0 ; R0 = 5*(x + y)
Need to work around the limitations of the MUL instruction – dst and src1 registers must not be the same and src2 cannot be an immediate value.
(v) Write ARM assembly language instructions to compute (x + y)(y – z).
(vi) Write ARM assembly language instructions to compute 3×4 – 5x – 16y4z4.
MOV R0,#3 ;R0=3
;R1=x2 ;R4=3×2 ;R0=3×4
;R4=5x
;R0=3×4 -5x ;R4=2y
;R4=2yz
;R5=4y2z2
; R5 = 16y4z4 ;R0=3×4 -5x-16y4z4
MUL R4, R1, R1 MUL R0, R4, R0 MUL R0, R4, R0 MOV R4,#5 ;R4=5
MUL R4, R1, R4 SUB R0, R0, R4 ADD R4, R2, R2 MUL R4, R3, R4 MUL R5, R4, R4 MUL R4, R5, R5 SUB R0, R0, R4
the limitations of the MUL instruction – dst and src1 registers
Need to work around
must not be the same and src2 cannot be an immediate value.
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