Translate the following code into MIPS assembly language: 1. if(C==0){ A = B + D[12]; }else{ A = B++; } 2. for(I=0; I < 100; I++){ A[ I ] += A[ I ]; }
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A: The equivalent MIPS instruction of this high-level language.
Translate the following code into MIPS assembly language:
1.
if(C==0){
A = B + D[12];
}else{
A = B++;
}
2.
for(I=0; I < 100; I++){
A[ I ] += A[ I ];
}
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- Convert this following C++ code below to easy68k assembly language: int sum = 0; int c =1; for (unsigned short i = 0; i <= 10; i++) { if (c == 1 ) { sum += i; c = 0; } else { c = 1; } }Write each of the following pseudocode in assembly language and explain:(a) if( var1 >= var2 )var3 = +127;else{var3 = -128;}(b) if ( val1 > ecx ) And ( ecx > edx ) then X = 7Fh else X = 80h; (c) while( eax > ebx)eax = eax - 1;MIPS Programming Assignment Part I Rewrite C statements int i = 11; int j = 10; int A[] = { 0x11, 0x22, 0x33 }; int B[] = { 0x0, 0x1, 0x2, 0x3, 0x4, 0x5, 0x6, 0x7, 0x8 }; B[ 8 ] = A[ i - j ]; in MIPS assembly. Assume that addresses of variables i, j, A, and B are loaded into the registers $s3, $s4, $s6, and $s7, respectively: .data i: .word 11 j: .word 10 A: .word 0x11, 0x22, 0x33 B: .word 0x0, 0x1, 0x2, 0x3, 0x4, 0x5, 0x6, 0x7, 0x8 .text la $s3, i # load address of i la $s4, j # load address of j la $s6, A # load address of A la $s7, B # load address of B When you finished programming, add a comment at the end of your code which specifies what value is assigned to the location of B[8] after the program executes. Part IIWrite MIPS Assembly program that allows the user to enter a string of text. Determine whether the entered string is a palindrome (a word or a phrase that reads the same backward as forward, like "kayak" and "level") and print the result on the screen.
- Convert the C function below to MIPS assembly language. Make sure you follow the MIPS calling conventions. unsigned int sum(unsigned int n) { if (n == 0) return %; else return n+ sum(n-1); %3D%3Dplease modify the c language program into a pep 9 assembly language #include <stdio.h> int num, answer; int divide ( int numer, int denom ) { int quotient, remain; remain = numer; quotient = 0; while ( remain >= denom ) { remain -= denom; quotient++; } // end for return quotient; } // end of divide( ) int modulus ( int n, int d ) { int quot, rem; rem = n; quot = 0; while ( rem >= d ) { rem -= d; quot++; } // end for return rem; } // end of modulus( ) int main() { printf("? "); scanf("%d", &num); answer = divide(num, 20); printf("%d / 20 = %d\n", num, answer); answer = modulus(num, 20); printf("%d mod 20 = %d\n", num, answer); answer = divide(num, 35); printf("%d / 35 = %d\n", num, answer); answer = modulus(num, 35); printf("%d mod 35 = %d\n", num, answer); return 0; } // end of mainPlease can you convert this part of a c program to pep 9 assembly language void findLog ( int *result, int num, int b ) { int k; num--; *result = 0; while (num > 0) { *result = *result + 1; num = divide(num, b); } // end while } // end of findLog( )
- Convert the following into Pep/9 Assembler: #include <iostream>using namespace std;int square(int n){ int i; int sq; sq = 0; for (i = 0; i < n; i++){ sq = sq + n; } return sq;}int main (){ int num; cout << "Enter a number: "; cin >> num; cout << num << " squared = " << square(num) << endl; return 0;} Submit: Source file along with screen captures showing the program running in the Pep simulator.ASSEMBLY Write assembly code for: int var1 = 1; for(int i = 0; i < 10; i++) var1++;Convert the following into Pep/9 Assembler: #include using namespace std; int square(int n){ int i; int sq; sq = 0; for (i = 0; i < n; i++){ sq = sq + n; } return sq; } int main (){ int num; cout << "Enter a number: "; cin >> num; cout << num << " squared = " << square(num) << endl; return 0; }
- Translate the following C program to Pep/9 assembly language. #include int inpNum, sum, count, min; int main() { sum = 0; count = 0; min = 32767; printf("? "); scanf("%d", &inpNum); while (inpNum != 0 ) { sum += inpNum; count++; if (inpNum < min) { min = inpNum; } printf("? "); scanf("%d", &inpNum); } // end while printf("Sum printf("Count printf("Min } // end of main = = %d\n", sum); = %d\n", count); %d\n", min);Define the concept of pointer arithmetic.Please convert to C language #include <bits/stdc++.h>using namespace std; //folding method int main(){ string s; cout << "enter number\n"; cin >> s; //folding and summing int sum = 0; for (int i = 0; i < s.length(); i += 2) { if (i + 1 < s.length()) sum += stoi(s.substr(i, 2)); else //when only one digit is left for folding sum += stoi(s.substr(i, 1)); } cout << s << "->" << sum % 10; return 0;} Output:
