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3. Convert the following C code to assembly. Assume a, b, and c are in main...
Convert the following c code into instructions MIPS Assembly language. Assume a associated with $s0, i with $s1, and j with $s2. if( i < 10) { if( j < 10) { a = 0; } else { a = 1;} } else { a = 2; }
DO 2 AND 3.
Question #1 Convert the following assembly program into machine code. LDAA WSA1 STAA $1000 INCA LDAB $1000 DECB ABA Here: Bra Here Question # 2 (a) in Question # 1, determine the values of PC, A, B, N, Z, V, C after each instruction (b) Verify your answers by using CodeWarrior Question # 3 Assume that the program in Question 1 is saved in memory starting at location $COOA. Drawa memory diagram showing the contents (in...
5. Assume DS-1000, CS- 800, Ax-3412H, and BX-0200H. For the following x86 assembly code: MOV [BX +1000H], AX a) Translate the assembly code to machine code (in Hex). The opcode of MOV instruction is 100010. b) Show the values of AX, BX, and new values in memory, if there are, in the following figure c) (10 points) AFTER executing the instruction. (5 points) What is the memory address accessed by the instruction, assuming real mode operation? Show it in the...
Given the following C Code segment convert it to ARM assembly. Assume the following register assignment have been made before your section of code begins. C Variable Register assignment r1 y r2 r3 r10 j r11 int x=0, y=0, z=0; int main() { for (int i = 0; i<10; i++) for (int j 0; j < 20; j++) if (i* j > 100) X++; if (i j >= 15) y++; + فہه Z = X + y; }
Assume that we are using the x86-64 architecture Convert the following code fragment (in the C language) to assembly code. Explain your steps. [25 marks] long test ( long a, long b, long c ) { long t; if (a>b) { if (a>c) t=a; else t=c; } else { t=b; } return t; }
Convert the following C fragment to equivalent MIPS assembly language. Assume that the variables a, b, c, d, i and x are assigened to registers $t1, $t2, $t3, $t4, $s0 and $s1 respectively. Assume that the base address of the array A and B is in register $a0 and $a1 respectively. if (a > 0) b = a + 10; else b = a - 10;
C2. Convert the following c-code to MIPS code. The base memory address of the array a is 0x8000_400C, which must be loaded in the base register Ss0. Register assignments: int a[10]; while (i !- 10) 1 if ( i%2 == 0) a[i] = i * 2; else ail - i* 3 i++i
C2. Convert the following c-code to MIPS code. The base memory address of the array a is 0x8000_400C, which must be loaded in the base register Ss0. Register...
Convert the high-level code into a MIPS-assembly program. Assume a is in $s0, b is in $s1, and c is in $s2. c = 0; while (a>0) { a = a-b; c++; }
its
brr[8]
(40%) Convert the following C-pseudo code into MIPS assembly code as a standalone program (including main and all the required directives). You can use any register. You must comply, however, with the convention of register usage. Before writing your code perform an explicit register allocation phase. Note that the C snippet is int arr[8]; int brr[4]-{1, 2, 3, 4, 5, 6, 7, 8) int i-8; while (i>-0) arrli]-brr[i-); (40%) Convert the following C-pseudo code into MIPS assembly code...
Translate the following C code to MIPS assembly code. Assume that the value of i is in register $t0, and $s0 holds the base address of the integer MemArray if (i > 10) MemArray[i] = 0; else MemArray[i] = -MemArray[i];