Spring 2021 CPE380 Assignment 3

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  1. For this question, check all that apply. Which of the following are correct statements about assembly languages for most modern computers?
    a+b is an add producing an rval, but a[b] is an add that produces an lval
    The same add instruction could compute an integer result or an address
    Local variables for a function are usually allocated in registers or in the stack frame
    Enable masking is a key aspect of MIMD supercomputers and Multi-core processors
    Most computers do not have special instructions that implement if, else, while, and for
  2. For this question, check all that apply. MIPS is not a CISC architecture, but a RISC with a simple and highly regular instruction set. Which of the following attributes correctly describe the MIPS instruction set?
    Each instruction is encoded in precisely one 32-bit word
    There is a call instruction that pushes the return address on the stack
    The result of a comparison for less than is a 32-bit integer value of 0 or -1
    There are no unconditional branch instructions, only conditional branches
    A single instruction can load a value from memory and add it to any register
  3. An array a of 32-bit integers is specified as shown below. On the MIPS architecture, suppose &(a[0]) is 4000. What is the rval of a[3]? 
    int a[3] = { 601, 86, 42, 22 };

  4. Which of the following segments of C code best describes what the following MIPS assembly code does? (Note: oddly, unlike nearly every programming language, MIPS add detects integer overflow -- so you really should always use addu as this code does.) 
    	la	$t0, x
	lw	$t1, 0($t0)
	la	$t2, y
	lw	$t3, 0($t2)
l1:	bne	$0, $t3, l2
	addu	$t1, $t1, $t3
	beq	$t1, $t1, l1
l2:	sw	$t1, 0($t0)

    x=x+y;
    if (y==0) { x=x+y; }
    if (y!=0) { x=x+y; }
    while (y==0) { x=x+y; }
    while (y!=0) { x=x+y; }
  5. What is the 32-bit hexadecimal value used to represent the MIPS assembly language instruction ori $t0,$0,1? Hint: you can use SPIM to find out.
  6. This MIPS/SPIM program includes a subroutine called myadd that performs x=(y+z);. In the space below, replace the myadd subroutine with one named negz that will make x have the value which is the 2's complement negation of z (i.e., -z). You should test your routine using SPIM before you submit it, which will require merging it with a test framework like the one used in this MIPS/SPIM program -- but only submit the negz routine here. Remember that you can and should comment your code, especially if there are any known bugs. Half off for documented bugs. :-)
  7. This (now familiar) MIPS/SPIM program includes a subroutine called myadd that performs x=y+z;. In the space below, replace the myadd subroutine with one named popcount that will compute x=popcount(y);, the Population Count of y, the number of 1 bits in y's value. and the following C code gives a simple algorithm to compute it. This uses a little trick credited to Brian Kernighan (and described here) to count the "population" of 1s; t0 & (t0 - 1) removes the least-significant 1 bit from the value of t0. 
    extern int x, y, z;

void
popcount(void)
{
    int t0 = y;
    int t1 = 0;

    while (t0) {
        t1 = t1 + 1;
        t2 = t0 - 1;
        t0 = t0 & t2;
    }
    x = t1;
}

  8. This MIPS/SPIM program includes a subroutine called myadd that performs x=(y+z);. In the space below, replace the myadd subroutine with one named sort that will re-order the values of x, y, and z to be in increasing order: the smallest value should end up in x and the largest in z. Your code should take advantage of the fact that x, y, and z are consecutive words in memory; you should treat x as an array of three elements. You should test your routine using SPIM before you submit it, which will require merging it with a test framework like the one used in this MIPS/SPIM program -- but only submit the sort routine here.
  9. Briefly explain: what MIPS instructions would you expect to be generated by the assembly code: li $t0,0x12345678
  10. What does the following C++ program print? Why? 
    #include <iostream>
using namespace std;
int a=380, b=480, c=42;
void f(int *x, int &y, int z) { ++*x; ++y; ++z; }
int main() {
  cout <<"a="<<a <<" b="<<b <<" c="<<c <<endl;
  f(&a,b,c);
  cout <<"a="<<a <<" b="<<b <<" c="<<c <<endl;
}

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CPE380 Computer Organization and Design.