My closest friend is going through an EE course (I'm his last hope : /), I have knowledge of Java from about 7 years ago, but his (outline) latest EE programming assignment is to use the MIPS Assembly to do the following:
Write a program that takes two positive integers (m and n) and computes:
x= (m^n) - (1+2+3+…+n) * min(m,n)!
Both the integers should be greater than zero. I'm not allowed to use any R-type arithmetic instructions (add, mult, sub). Instead I'm to write the code for their functions using other instructions???? "Your program should continue getting new values for m and n after each computation until the user enters zero which would be the end for your program."
I do not have access to any of his previous assignments and trying to dive head 1st into assembly language WITHOUT using (add, mult, sub) isn't working out for me too well.
ece.ucdavis.edu/~vojin/CLASSES/EEC70/W2001/pr4.pdf Prof seemed to be using an ols assignment from when he was teaching at UC Davis.
//edit Here is a c++ version of the problem, it does not cover all of the assignment's bases, but it's a starting point:
#include <iostream.h>
//x = (m^n) - (1+2+3+...+n) * ((min(m,n))!)
int m; //User Input
int n; //User Input
double answer; //Answer yo.
int findMin(int, int); //Takes 2 int inputs and outputs the smallest int.
int minFound; //Function output
double factorial(int); //Do eet.
double factOutput; //Function output
double sumN(int); //1+2+3+...+n
double sumFound; //Function output
double expMtoN(int, int); //m^n, float for number size,
double expFound; //Function output, float for number size,
int main(void)
{
cout << "Please enter a positive integer (m): ";
cin >> m;
//Escape if zero.
if ( m == 0)
{
cout << "User input for \"m\" is equal to zero; escape on zero." << endl;
return 0;
}
cout << "Please enter a positive integer (n): ";
cin >> n;
//Escape if zero.
if ( n == 0)
{
cout << "User input for \"n\" is equal to zero; escape on zero." << endl;
return 0;
}
expFound = expMtoN(m, n); //m^n
sumFound = sumN(n); //1+2+3+...+n
minFound = findMin(m, n); //Takes 2 int inputs and outputs the smallest int.
factOutput = factorial(minFound); //Factorial math for minFound (z!)
answer = expFound - sumFound * factOutput; //x = (m^n) - (1+2+3+...+n) * ((min(m,n))!)
cout << endl;
cout << m << " raised to the power of " << n << " is: " << expFound << endl;
cout << "Sum of " << n << " is: " << sumFound << endl;
cout << "Lowest number out of " << m << " and " << n << " is: " << minFound << endl;
cout << minFound << " factorial is: " << factOutput << endl;
cout << endl << "x = (m^n) - (1+2+3+...+n) * ((min(m,n))!)" << endl;
cout << "x = " << answer << endl;
}
//all temp variables below are confined to their respective functions.
//return functions output temp into variable from main.
double expMtoN(int userBase, int userExp)
{
double temp = 1; //Must establish 1 so you are not multiplying by zero.
for ( int i = 1; i <= userExp; i++ )
temp *= userBase;
return temp;
}
double sumN(int userN)
{
double temp = 0;
for ( int i = 1; i <= userN; i++ )
temp = temp + i;
return temp;
}
int findMin(int userM, int userN)
{
if( userM <= userN )
return userM;
else
return userN;
}
double factorial(int minFound)
{
double temp;
if ( minFound <= 1 )
return 1;
temp = minFound * factorial(minFound - 1);
return temp;
}
Input.s
;-----------------------------------------------------------------------------
;Subprogram call by symbol "InputUnsigned"
;expect the address of a zero-terminated prompt string in R1
;returns the read value in R1
;changes the contents of registers R1,R13,R14
;-----------------------------------------------------------------------------
.data
;*** Data for Read-Trap
ReadBuffer: .space 80
ReadPar: .word 0,ReadBuffer,80
;*** Data for Printf-Trap
PrintfPar: .space 4
SaveR2: .space 4
SaveR3: .space 4
SaveR4: .space 4
SaveR5: .space 4
.text
.global InputUnsigned
InputUnsigned:
;*** save register contents
sw SaveR2,r2
sw SaveR3,r3
sw SaveR4,r4
sw SaveR5,r5
;*** Prompt
sw PrintfPar,r1
addi r14,r0,PrintfPar
trap 5
;*** call Trap-3 to read line
addi r14,r0,ReadPar
trap 3
;*** determine value
addi r2,r0,ReadBuffer
addi r1,r0,0
addi r4,r0,10 ;Decimal system
Loop: ;*** reads digits to end of line
lbu r3,0(r2)
seqi r5,r3,10 ;LF -> Exit
bnez r5,Finish
subi r3,r3,48 ;´0´
multu r1,r1,r4 ;Shift decimal
add r1,r1,r3
addi r2,r2,1 ;increment pointer
j Loop
Finish: ;*** restore old register contents
lw r2,SaveR2
lw r3,SaveR3
lw r4,SaveR4
lw r5,SaveR5
jr r31 ; Return