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Published byDenisse Wingate Modified over 3 years ago

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Cosc 2150 Arrays in assembly code

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Variables and addresses Uncompiled ld [a], %r1 addcc %r1, 2, %r3 ARC has three addressing modes —immediate, direct, and indexed In the following code segment. compiled load [2064], %r1 addcc %r1, 2, %r3 immediate direct

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Variables and addresses (2) Direct addressing —Like IAS, ARC loads the value at a memory location —So when, load [a], %r1 is “compiled”, a symbol table is built. A is located at memory location 2064 –Given us, the final code with load [2064], %r1 –The [] tells us this is a direct addressing mode. So given the following variables and memory a: 5 2064 b: 6 2068 c: 7 2072 d: 8 2076 e: 9 2080 We can find each variable in a sequence “down” memory. —So load [2072], %r1 ! load c’s value into %r1

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Arrays vs variables Using the following table variable c points to memory location 2072 (direct) arr[2] points to memory location 2072, but we used index addressing. —the arr’s base address is 2064 plus 8, given us the address of 2072

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Array addressing ElementAddress = Base + (Index – Start)*size Base = base address of the array Start = array [-10..10] of integer —The start would be –10 –Note, In c/c++ starting index is normally 0 Index = element in the array you want —array[5], index =5 Size = size of individual element in bytes

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Example array [-10..10] of integer; array[5] = 2; EA = array + (5 – -10)*4 = array + 60 add %r0, 5, %r1 ! load index sub %r1, -10, %r1 ! subtract start sll %r1, 2, %r1 ! multi by size =4 add %r0, 2, %r2 ! load 2 st %r2, array + %r1 !store 2, in array[5]

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C/C++, normally starts arrays at 0, so the formula can be simplified to EA = array + (index)*size In ARC, the size will always be 4, so EA = array + index*4

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Visual example of an array in memory

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Example 1 int c[5]; for (I=0; I<5; ++I) { c[I] = I+1; } Assembly shown in simulator.begin.org 2048 ! for (I=0; I<5; main: ld [I], %r2 ! load I into register 2 top:subcc %r2, 5, %r0 bpos done ! figure out array address in memory sll %r2, 2, %r1 !multi index by 4 ! Need I+1 add %r2,1, %r3 ! add I +1 ! store I +1 in c[I] st %r3, %r1 + c ! Increment I=I +1 add %r2, 1, %r2 st %r2, [I] ba top done:halt I:0 c: 0 0.end

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Example 2 int a[5] ={1,2,3,4,5}; int b[5] ={0,0,0,0,0}; int I; for (I=0; I <5; ++I) { b[I] = a[I] + 2; }.begin.org 2048 ! for (I=0; I<5; main: ld [I], %r2 ! load I into register 2 top:subcc %r2, 5, %r0 bpos done ! figure out array address in memory sll %r2, 2, %r1 !multi index by 4 ! load a[I] and add 2 ld %r1+a, %r3 add %r3, 2, %r3 ! store a[I] + 2 in b[I] st %r3, %r1 + b ! I=I +1 add %r2,1, %r2 !add I +1 st %r2, [I] ba top done:halt I:0 a:1 2 3 4 5 b: 0 0.end

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Example 3 int a[5] ={1,2,3,4,5}; int b[5] ={5,4,3,2,1}; int I; for (I=0; I <5; ++I) { a[I] = a[I] + b[I]; } again shown in simulator

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ARC code, example 3.begin.org 2048 ! for (I=0; I<5; main: ld [I], %r4 ! load I into register 4 top:subcc %r4, 5, %r0 bpos done sll %r4,2, %r1 ! load a[I] into register 2 ld %r1+a, %r2 ! load b[I] into register 3 ld %r1+b, %r3 add %r2, %r3, %r5 ! store a[I] + b[I] in a[I] st %r5, %r1 +a add %r4, 1, %r4 st %r4, [I] ba top done: halt I:0 a:1 2 3 4 5 b: 5 4 3 2 1.end

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Example 4 How would we write the following code a[i] = b[i-1];

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Q A &

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