CPSC 213 Midterm 2019W Term 1 (Oct 21, 2019) – Solution 1. Numbers and Memory [8 marks]
a) Considerthefollowingcoderunningonalittle-endianmachinewheretheaddressofiis0x1000. int i = 0x1234;
What is the hex value of the byte at address 0x1000? 0x34
b) Considerthefollowingcodewheretheaddressofsis0x1000.
struct SS {
char c;
int i;
char d[2]; };
struct SS s;
What is the value of the expression &s.i ?
0x1004
What is the value of the expression sizeof(s) ? 12
c) Consider the following code
int i = 0x56789a;
char c = (char) i;
int j = c;
What is the hex value of c ? 0x9a
What is the hex value of j ?
0xffffff9a
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2. Global Variables and Arrays [9 marks]
Consider the following code where the value of the variable i is already loaded into r0 and where the assembly- codelabelsa,b,andcrepresenttheaddressofthevariablesa,b,andc,respectively. Treateachsubquestion separately; do not use values computed in prior subquestions. No comments needed.
int* a;
int b[2];
int c;
a) Give assembly code for the statement c = *a;
ld $a, r1 ld(r1),r1 ld(r1),r1 ld $c, r2 str1,(r2)
# r1 = &a #r1=a #r1=*a # r2 = &c #c=*a;
b) Give assembly code for the statement a = b;
ld $b, r1 # r1 = b ld $a, r2 # r2 = &a str1,(r2) #a=b
c) Give assembly code for the statement b[i] = *(a + i) + i;
ld $a, r1
ld (r1), r1 ld(r1,r0,4),r2 add r0, r2
ld $b, r3 str2,(r3,r0,4)
# r1 = &a
# r1 = a #r2=*(a+i)=a[i] # r2 = *(a + i) + i
# r3 = b #b[i]=*(a+1)+i
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3. Structs and Instance Variables [9 marks]
Consider the following code, where f is a global variable, assuming whatever assembly labels make sense, and
where the size of a pointer is 8-bytes. Use r0 for the variable i. No comments needed.
struct X {
int* a;
int b[4];
struct Y c;
}
struct X* f;
struct Y {
int d;
struct X* e; }
a) Give assembly code for i = f->c.e->a[1]; (i.e., store the result in r0)
ld $f, r0 # r0 = &f
ld(r0),r0 #r0=f
ld 32(r0), r0 # r0 = f->c.e
ld (r0), r0 # r0 = f->c.e->a
ld 4(r0), r0 # r0 = i’= f->c.e->a[1]
b) Give assembly code for f->b[1] = i; (i.e., just use r0 for i)
ld $f, r1 # r1 = &f ld(r1),r1 #r1=f
st r0, 12(r1) # f->b[1] = i’
c) What is the minimum number of memory references required to compute the value of the following expression (not including fetching the instructions themselves from memory)?
f->c.e->c.e[0].a[0]
01234567
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4. Static Control Flow [7 marks] XXXX THIS TOPIC NOT COVERED on 2020 T1 MIDTERM Assumingx,yandzareglobalvariablesthathavealreadybeendeclaredandinitialized. Nocommentsneeded.
a) Giveassemblycodefortheloop(youmustimplementaloop)
for (int x=0; x
# last two lines optional (since x in declared in for statement) ld $x r1 # r1 = &x
str0,(r1) #x=x’
b) Giveassemblycodefortheprocedurecall(ignoreanythinghavingtodowiththestack).
foo();
c) Give assembly code for the return statement (ignore anything having to do with the stack).
return;
gpc $6, r6
j foo
j (r6)
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5. C Pointers [8 marks]
Give the values of the listed variables following the execution of this block of code.
int i = 1;
int j = 10;
int* p = &i;
int* q = &j;
*p = *p + *q;
q = p;
*p = *p + *q;
*q = *q + *p;
p = (int*) 0x1000;
int k = &p[3] – &p[1];
q = p + 2;
What is the value of
What is the value of
What is the value of
What is the value of
i ?
j ?
k ?
q ?
44
10
2
0x1008
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6. Dynamic Allocation [10 marks]
Consider each of the following code blocks where add, doAdd, and doSomethingElse are in three different modules of a large program. For each block indicate whether the code has a memory leak or dangling pointer and which technique (if any) is the single best way (only one) to fix the bug or improve the code (even if it doesn’t have a bug). Select none if the code does not need to be improved or if the needed change isn’t listed.
a)
int* add(int *a, int *b, n) {
int* c = malloc(n * sizeof(int));
for (int i = 0; i
rc_free_ref(a->s);
free(a);
}
a = malloc(sizeof (struct A)); a->s = s; rc_keep_ref(a->s->i); rc_keep_ref(a->s);
}
c = c + 1;
*s->i = c;
}
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//////////
// Module B
int bar() {
struct S* s = rc_malloc(sizeof(struct S)); s->i = rc_malloc(sizeof(int));
foo(s);
int t = *s->i;
rc_free_ref(s->i);
rc_free_ref(s);
return t;
}
//////////
// Shared between models A and B
struct S {
int* i;
};
//////////
// Module C
int main() {
int s = 0;
for (int i=0; i<100; i++)
s = s + bar();
printf("%d\n", s);
}
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8. Reading Assembly Code [10 marks]
a) Add high-level, C-like comments to the following assembly program. Be sure to clearly indicate the location of high-level structure such as loops, if-statements, reading from or writing to variables or arrays etc. Partial marks on this question will be determined by the amount of high-level structure you identify.
ld $a, r0
ld (r0), r0
ld $b, r1
ld (r1), r1
not r1
inc r1
ld $0, r2
L0: mov r2, r3
add r1, r3
beq r3, L4
mov r2, r3
inc r3
L1: mov r3, r4
add r1, r4
beq r4, L3
ld (r0, r2, 4), r5 ld(r0,r3,4),r6 mov r5, r7
not r7
inc r7
add r6, r7
bgt r7, L2
st r6, (r0, r2, 4) st r5, (r0, r3, 4)
L2: inc r3
br L1
L3: inc r2
br L0
L4: halt
#r0=&a
#r0=a
#r1=&b #r1=b=b’ #r1=~b’
#r1=-b’ #r2=i’=0 #r3=i’
#r3=i’ -b’
# goto L4 if i’ == b’ #r3=i’
#r3=j’ =i’+1 #r4=j’ #r4=j’ -b’
<= top of outer for loop (i’)
<= top of inner for loop (j’)
# goto L3 if
#r5=a[i’]
#r6=a[j’]
#r7=a[i’]
# a7 = ~a[i’]
# a7 = -a[i’]
#a7=a[j’]-a[i’]
# goto L2 if a[j’] > a[i’]
# a[i’] = a[j’]’ if a[j’] <= a[i’] <= THEN: swap two # a[j’] = a[i’]’ if a[j’] <= a[i’] <= array elements # j’++
# goto top of inner loop
# i’++
# goto top of outer loop
<= IF statement
<= end of inner loop
<= end of outer loop
j’==b’
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<= load two values from array
b) Give C code that does what this assembly code does. For full credit, your code must be straight-forward C, like you would write if you were writing this from scratch and not just translating it from the assembly code.
for (int i=0; i