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15-213 Recitation 2 – 2/11/02. Outline Stacks & Procedures Homogenous Data Arrays Nested Arrays Structured Data struct s / union s Arrays of structs. James Wilson e-mail: [email protected] Office Hours: Friday 1:30 – 3:00 Wean Cluster 52xx. Reminders Lab 2: Tuesday, 11:59. - PowerPoint PPT Presentation
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15-213 Recitation 2 – 2/11/02
Outline• Stacks & Procedures• Homogenous Data
– Arrays– Nested Arrays
• Structured Data– structs / unions– Arrays of structs
James Wilson
e-mail:
Office Hours:
Friday 1:30 – 3:00
Wean Cluster 52xx
Reminders• Lab 2: Tuesday,
11:59
Stacks• Grows down• Stores local variables that can’t fit in
registers• Stores arguments and return addresses• %esp Stack Pointer
– Points to the top value on the stack
• %ebp Base Pointer– Points to a function’s stack frame
• pushl– Decrements, then places value
• popl– ‘Returns’ value, then increments
Stack Frames
• Abstract partitioning of the stack
• Each Frame contains the state for a single function instant
Stack Pointer(%esp)
Frame Pointer(%ebp)
Return Addr
SavedRegisters
ArgumentBuild
Old %ebp
LocalVariables
Arguments
CallerFrame
Procedures
call: Caller Responsibilities• Arguments (pushl)
– In what order?
• Return Address (done by call)ret: Callee Responsibilities
• Save Registers (especially %ebp)• Set up Stack Frame• Return value in %eax
Problem 1: Call Chain
void absdiff(int *result, int x, int y){ int z;
if (x >= y) z = x - y; else z = y - x;
*result = z; return;}
int main(){ int result; int x,y;
x = 5; y = -3; absdiff(&result, x, y); printf("|(%d) - (%d)| = %d\n", x, y, result);
return 0;}
Problem 1: Answer
<main>: push %ebp mov %esp,%ebp sub $0x18,%esp movl $0x5,-8(%ebp) movl $0xfffffffd, -12(%ebp) add $0xfffffffc,%esp mov -12(%ebp),%eax push %eax mov -8(%ebp),%eax push %eax lea -4(%ebp),%eax push %eax call <absdiff>
Old %ebp
result
5
-3
%esp
-3
5
&result
Rtn Address %esp
%ebp
Problem 1: Answer<absdiff>:
push %ebpmov %esp,%ebpsub $0x18,%espmov 0xc(%ebp),%eaxcmp 0x10(%ebp),%eaxjl .L1 mov 0xc(%ebp),%eaxmov 0x10(%ebp),%edxmov %eax,%ecxsub %edx,%ecxjmp .L2
.L1mov 0x10(%ebp),%eaxmov 0xc(%ebp),%edxmov %eax,%ecxsub %edx,%ecx
.L2mov 0x8(%ebp),%eaxmov %ecx,(%eax)mov %ebp,%esppop %ebpret
%esp
-3
5
&result
Rtn Address
%ebpOld %ebp
%esp
*
*
*
Problem 1: Answer
<main>:….. add $0x10, %esp mov -4(%ebp),%eax push %eax mov -12(%ebp),%eax push %eax mov -8(%ebp),%eax push %eax push $0x80484d8 call <printf>
mov %ebp,%esp pop %ebp ret
Old %ebp
result
5
-3
%esp
result
-3
5
%ebp
$0x80484d8
Rtn Address
%esp
Problem 2: Recursion
With the following code, what does the stack look like if we call fib(1, 1, 0) and reach the point where if(n==0) holds true?
int fib(int n, int next, int result){ if(n == 0) return result;
return fib(n - 1, next + result, next);}
Problem 2: Answer0 ; third argument to fib1 ; second1 ; firstret ; call fiboldebp ; <--- ebp of fib’s caller0 ; <--- push result1 ; <--- next + result0 ; <--- n - 1ret ; call fiboldebp
Homogenous Data: Arrays
• Allocated as contiguous blocks of memory
Address Computation Examples
• int cmu[5] = {…}• cmu begins at memory address 40
cmu[0] 40 + 4*0 = 40cmu[3] 40 + 4*3 = 52cmu[-1] 40 + 4*-1 = 36cmu[15] 40 + 4*15 = 100
Problem 3: Arrays
get_sum: pushl %ebp movl %esp,%ebp pushl %ebx
movl 8(%ebp),%ebx # ebx = 1st arg movl 12(%ebp),%ecx # ecx = 2nd arg xorl %eax,%eax # eax = 0 movl %eax,%edx # edx = 0 cmpl %ecx,%eax # jge .L4 # if (ecx >= 0) goto L4.L6: addl (%ebx,%edx,4),%eax # eax += Mem[ebx+edx*4] incl %edx # edx ++ cmpl %ecx,%edx # jl .L6 # if (edx < ecx) goto L6
.L4: popl %ebx movl %ebp,%esp popl %ebp ret
Problem 3: Answer
int get_sum(int * array, int size){ int sum = 0; int i=0; for (i=0; i<size; i++) sum += array[i]; return sum;}
get_sum: pushl %ebp movl %esp,%ebp pushl %ebx
movl 8(%ebp),%ebx movl 12(%ebp),%ecx xorl %eax,%eax movl %eax,%edx cmpl %ecx,%eax jge .L4.L6: addl (%ebx,%edx,4),%eax incl %edx cmpl %ecx,%edx jl .L6
.L4: popl %ebx movl %ebp,%esp popl %ebp ret
Problem 4: Nested arrays
int main(int argc, char **argv){ int i,j,r=0; for (i=0; i<argc; i++) { j=0; while(argv[i][j] != '\0') { r ^= argv[i][j]; j++; } } return r;}
Problem 4: Answer
main: pushl %ebp movl %esp,%ebp pushl %edi pushl %esi pushl %ebx movl 12(%ebp),%edi xorl %esi,%esi xorl %ebx,%ebx cmpl 8(%ebp),%esi jge .L4.L6: xorl %ecx,%ecx movl (%edi,%ebx,4),%eax cmpb $0,(%eax) je .L5 movl %eax,%edx
.L9: movsbl (%ecx,%edx),%eax xorl %eax,%esi incl %ecx cmpb $0,(%ecx,%edx) jne .L9.L5: incl %ebx cmpl 8(%ebp),%ebx jl .L6.L4: movl %esi,%eax popl %ebx popl %esi popl %edi movl %ebp,%esp popl %ebp ret
structs and unions
• Organize data• structs store multiple elements,
unions store a single element at a time
• Members of a union change how you look at data
• unions used for mutually exclusive data
Alignment
• Contiguous areas of memory• Each block is aligned
– Size is a multiple of a base value– “Base value” is the largest alignment of
data types in structure
• Why?– Efficient load/store from memory– Virtual Memory paging
• This applies to any variable type
Structure of a struct
• Find largest alignment– Size of structure must be a multiple of this
• For each element e (top to bottom):– Find alignment of e
• Starting offset must be a multiple of this
– Pad previous element with empty space until alignment matches
– Allocate alignment worth of space to e
• Pad last element with empty space until alignment of structure matches
• Note this isn’t optimal!
Structure of a union
• Find largest alignment– Size of structure must be a multiple of this
• Allocate this much space
Examplesstruct one { union two { int i; int i; double d; double d; char c[2]; char c[2];} }
Problem 5: Structs#define MAX_STRING 20
#struct student #{# char first [ MAX_STRING ];# char last [ MAX_STRING ];# char id [ MAX_STRING ];# int age;#};
# struct student s1;# struct student *s2;
# strncpy(s1.first, fName,# sizeof(fName));
# strncpy(s1.last, lName,# sizeof(lName));
.LC0: .string "Jack".LC1: .string "Black".LC2: .string "12345ZXY" .align 32.LC3: .string "Name : %s %s\nID : %s\nAge : %i\n“
main: pushl %ebp movl %esp, %ebp
subl $88, %esp subl $4, %esp pushl $5 pushl $.LC0 leal -72(%ebp), %eax pushl %eax call strncpy
addl $16, %esp subl $4, %esp pushl $6 pushl $.LC1 leal -72(%ebp), %eax addl $20, %eax pushl %eax call strncpy
Problem 5: Structs addl $16, %esp
subl $4, %esppushl $9pushl $.LC2leal -72(%ebp), %eaxaddl $40, %eaxpushl %eaxcall strncpy
addl $16, %espmovl $19, -12(%ebp)
leal -72(%ebp), %eaxmovl %eax, -76(%ebp)
subl $12, %espmovl -76(%ebp), %eaxpushl 60(%eax)movl -76(%ebp), %eaxaddl $40, %eaxpushl %eaxmovl -76(%ebp), %eaxaddl $20, %eaxpushl %eaxpushl -76(%ebp)pushl $.LC3call printf
addl $32, %espmovl $0, %eaxleaveret
strncpy(s1.id, ID, sizeof(ID));
# s1.age = AGE;
# s2 = &s1;
# printf("Name : %s %s\nID : %s\nAge : %i\n",# s2->first, s2->last, s2->id, s2 ->age);
# clean up and exit
Problem 5: Answer
int main ( int argc, char** argv ){ struct student s1; struct student *s2;
strncpy ( s1.first, fName, sizeof ( fName ) ); strncpy ( s1.last , lName, sizeof ( lName ) ); strncpy ( s1.id , ID , sizeof ( ID ) ); s1.age = AGE;
s2 = &s1; printf ( "Name : %s %s\nID : %s\nAge : %i\n",
s2 -> first, s2 -> last, s2 -> id, s2 -> age ); return 0;}
Problem 6: Arrays of structs.LC0: .string "%i : %s - “.LC1: .string "Has no partner\n“.LC2: .string "Partnered with : %s \n“ .align 32.LC3: .string "Claims parter: %s, but not mutual\n“
partner_check: pushl %ebp movl %esp,%ebp subl $20,%esp pushl %ebx nop movl $0,-4(%ebp) .p2align 4,,7.L3: cmpl $5,-4(%ebp) jle .L6 jmp .L4 .p2align 4,,7.L6: addl $-4,%esp movl -4(%ebp),%eax movl %eax,%ecx movl %ecx,%edx sall $4,%edx addl %eax,%edx
#define MAX_STRING 20#define MAX_STUDENTS 6
#struct student #{# char first [ MAX_STRING ];# char last [ MAX_STRING ];# char id [ MAX_STRING ];# int age;# struct student *partner; #};
# Lines with < symbols are moving# data for the printf command at# the end of the < block. left# in for informational purposes
<<<<<<
leal 0(,%edx,4),%eaxmovl %eax,%edxaddl 8(%ebp),%edxleal 40(%edx),%eaxpushl %eaxmovl -4(%ebp),%eaxpushl %eaxpushl $.LC0call printfaddl $16,%espmovl -4(%ebp),%eaxmovl %eax,%ecxmovl %ecx,%edxsall $4,%edxaddl %eax,%edxleal 0(,%edx,4),%eaxmovl 8(%ebp),%edxcmpl $0,64(%edx,%eax)jne .L7addl $-12,%esppushl $.LC1call printfaddl $16,%espjmp .L5.p2align 4,,7
<<<<<<<<<< printf ( "%i : %s - ", j, class[j].ID );
<< printf ( "Has no partner\n" );
Problem 6: Arrays of structs
.L7:movl -4(%ebp),%eaxmovl %eax,%ecxmovl %ecx,%edxsall $4,%edxaddl %eax,%edxleal 0(,%edx,4),%eaxmovl 8(%ebp),%edxmovl 64(%edx,%eax),%eaxmovl -4(%ebp),%edxmovl %edx,%ebxmovl %ebx,%ecxsall $4,%ecxaddl %edx,%ecxleal 0(,%ecx,4),%edxmovl %edx,%ecxaddl 8(%ebp),%ecxcmpl %ecx,64(%eax)jne .L9addl $-8,%espmovl -4(%ebp),%eaxmovl %eax,%ecxmovl %ecx,%edxsall $4,%edxaddl %eax,%edxleal 0(,%edx,4),%eaxmovl 8(%ebp),%edxmovl 64(%edx,%eax),%eaxaddl $40,%eaxpushl %eaxpushl $.LC2call printfaddl $16,%espjmp .L5.p2align 4,,7
<<<<<<<<<<<<< printf ( "Partnered with : %s \n",< class[j].partner->ID );
Problem 6: Arrays of structs
.L9:addl $-8,%espmovl -4(%ebp),%eaxmovl %eax,%ecxmovl %ecx,%edxsall $4,%edxaddl %eax,%edxleal 0(,%edx,4),%eaxmovl 8(%ebp),%edxmovl 64(%edx,%eax),%eaxaddl $40,%eaxpushl %eaxpushl $.LC3call printfaddl $16,%esp
.L10:
.L8:
.L5:incl -4(%ebp)jmp .L3.p2align 4,,7
.L4:
.L2:movl -24(%ebp),%ebxmovl %ebp,%esppopl %ebpret
<<<<<<<<<<<<<<<< printf ( "Claims parter: %s, but not mutual\n",< class[j].partner->ID );
Problem 6: Arrays of structs
void partner_check ( struct student * class ){ int j; for (j = 0; j < MAX_STUDENTS; j++ ) { printf ( "%i : %s - ", j, class[j].ID ); if ( (class+j)->partner == NULL )
printf ( "Has no partner\n" );else if ( (class+j)->partner->partner == (class+j) ) printf ( "Partnered with : %s \n", class[j].partner->ID );else printf ( "Claims parter: %s, but not mutual\n",
class[j].partner->ID ); }}
Problem 6: Answer