-
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
No, that's primarily for SIMD
Most computers do not have special instructions that implement if, else, while, and for
-
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
No, jal puts the return address in $31 (aka $ra)
The result of a comparison for less than is a 32-bit integer value of 0 or -1
No, 0 or 1
There are no unconditional branch instructions, only conditional branches
Of course, beq $t,$t is always true
A single instruction can load a value from memory and add it to any register
No, lw is the only way to load a word, and you need a separate add
-
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 };
-
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; }
-
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.
-
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. :-)
-
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;
}
-
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.
-
Briefly explain: what MIPS instructions would you expect to
be generated by the assembly code: li $t0,0x12345678
-
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;
}
