CSE 351: Week 4. Tom Bergan, TA

Transcription

1 CSE 35 Week 4 Tom Bergan, TA

2 Does this code look okay? int binarysearch(int a[], int length, int key) { int low = 0; int high = length - ; while (low <= high) { int mid = (low + high) / ; int midval = a[mid]; if (midval < key) low = mid + ; else if (midval > key) high = mid - ; else return mid; // key found return -; // key not found

3 Does this code look okay? int binarysearch(int a[], int length, int key) { int low = 0; int high = length - ; while (low <= high) { int mid = (low + high) / ; int midval = a[mid]; if (midval < key) low = What mid + if ; length > 30? else if (midval > key) high = mid - ; else return mid; // key found return -; // key not found 3

4 Does this code look ok? int mid = (low + high) / ; What if length > 30?... then we could have low = 30 = 0x high = 30 + = 0x low + high = 3 + = 0x int midval = a[mid]; Oops, in two s complement, this is a negative number! (low + high) / = 0xC = Crashes because mid < 0 4

5 How can we fix the bug? int mid = (low + high) / ; int mid = low + ((high - low) / ); (There are other ways, but I think this is the simplest to understand) 5

6 This was an actual bug in Java java.util.arrays.binarysearch This bug went unnoticed for years. See http//googleresearch.blogspot.com/006/06/extra-extra-read-all-about-it-nearly.html Understanding binary number representations is important! 6

7 Check your textbook Don t use the international edition! The homework problems are different. 7

8 Today Questions on Hw or Lab? Procedure calls 8

9 Procedure Call Example Caller int z = sum(, ); Callee int sum(int x, int y) { return x + y; 9

10 Procedure Call Example (IA3/Linux) Caller int z = sum(, ); Caller in assembly 0x800 pushl $ 0x8005 pushl $ 0x8009 call sum addl $8, *note these instruction addresses are completely made up for this example 0

11 Procedure Call Example (IA3/Linux) Caller int z = sum(, ); Caller in assembly 0x800 pushl $ 0x8005 pushl $ 0x8009 call sum addl $8, *note these instruction addresses are completely made up for this example

12 Procedure Call Example (IA3/Linux) Caller int z = sum(, ); Caller in assembly 0x800 pushl $ 0x8005 pushl $ 0x8009 call sum addl $8, *note these instruction addresses are completely made up for this example

13 Procedure Call Example (IA3/Linux) Caller int z = sum(, ); Caller in assembly 0x800 pushl $ 0x8005 pushl $ 0x8009 call sum addl $8, *note these instruction addresses are completely made up for this example 3

14 Procedure Call Example (IA3/Linux) Callee int sum(int x, int y) { return x + y; Callee in assembly (simple version) movl 8(), %edi movl 4(), %eax addl %edi, %eax ret 8() 4() y x Registers %edi 4

15 Procedure Call Example (IA3/Linux) Callee int sum(int x, int y) { return x + y; Callee in assembly (simple version) movl 8(), %edi movl 4(), %eax addl %edi, %eax ret 8() 4() y x Registers %eax %edi 5

16 Procedure Call Example (IA3/Linux) Callee int sum(int x, int y) { return x + y; Callee in assembly (simple version) movl 8(), %edi movl 4(), %eax addl %edi, %eax ret 8() 4() y x %eax has the return value! Registers %eax 3 %edi 6

17 Procedure Call Example (IA3/Linux) Callee int sum(int x, int y) { return x + y; Callee in assembly (simple version) movl 8(), %edi movl 4(), %eax addl %edi, %eax ret %eax has the return value! Registers %eax 3 y x %edi 7 %eip

18 Procedure Call Example (IA3/Linux) Caller int z = sum(, ); Caller in assembly 0x800 pushl $ 0x8005 pushl $ 0x8009 call sum addl $8, *note these instruction addresses are completely made up for this example 8 Registers %eax %edi %eip 3

19 Procedure Call Example (IA3/Linux) Caller int z = sum(, ); Caller in assembly 0x800 pushl $ 0x8005 pushl $ 0x8009 call sum addl $8, *note these instruction addresses are completely made up for this example 9 Registers %eax %edi %eip 3

20 Procedure Call Example (IA3/Linux) Caller int z = sum(, ); Problem - What if Caller used %edi before making the call? Caller in assembly 0x800 pushl $ 0x8005 pushl $ 0x8009 call sum addl $8, *note these instruction addresses are completely made up for this example 0 Registers %eax %edi %eip 3

21 Procedure Call Example (IA3/Linux) Caller int d = 5; int z = sum(, ); Problem - What if Caller used %edi before making the call? Caller in assembly 0x7fff movl $5, %edi 0x800 pushl $ 0x8005 pushl $ 0x8009 call sum addl $8, *note these instruction addresses are completely made up for this example sum() overwrote %edi! Need to save... Registers %eax %edi %eip 3

22 Saving Registers Some are caller save - IA3 %eax, %edx, %ecx - These are very commonly used (caller should expect they will be clobbered) Some are callee save - IA3 %ebx, %edi, %esi - These are less commonly used from prior example

23 Callee Procedure Call Example (IA3/Linux) Callee in assembly (better version) setup body int sum(int x, int y) { return x + y; pushl %ebp movl, %ebp pushl %edi movl (%ebp), %edi movl 8(%ebp), %eax addl %edi, %eax %ebp y x cleanup movl (), %edi movl %ebp, popl %ebp ret 3

24 Callee Procedure Call Example (IA3/Linux) int sum(int x, int y) { return x + y; Callee in assembly (better version) setup body pushl %ebp movl, %ebp pushl %edi movl (%ebp), %edi movl 8(%ebp), %eax addl %edi, %eax %ebp old %ebp y x cleanup movl (), %edi movl %ebp, popl %ebp ret 4

25 Callee Procedure Call Example (IA3/Linux) int sum(int x, int y) { return x + y; Callee in assembly (better version) setup body cleanup pushl %ebp movl, %ebp pushl %edi movl (%ebp), %edi movl 8(%ebp), %eax addl %edi, %eax movl (), %edi movl %ebp, popl %ebp ret 5 %ebp old %ebp y x

26 Callee Procedure Call Example (IA3/Linux) int sum(int x, int y) { return x + y; Callee in assembly (better version) setup body cleanup pushl %ebp movl, %ebp pushl %edi movl (%ebp), %edi movl 8(%ebp), %eax addl %edi, %eax movl (), %edi movl %ebp, popl %ebp ret %ebp old %ebp old %edi saved %edi y x 6

27 Callee Procedure Call Example (IA3/Linux) int sum(int x, int y) { return x + y; Callee in assembly (better version) setup body cleanup pushl %ebp movl, %ebp pushl %edi movl (%ebp), %edi movl 8(%ebp), %eax addl %edi, %eax movl (), %edi movl %ebp, popl %ebp ret 7 (%ebp) 8(%ebp) %ebp old %ebp old %edi Key %ebp is fixed for the entire function y x

28 Callee Procedure Call Example (IA3/Linux) int sum(int x, int y) { return x + y; Callee in assembly (better version) setup body cleanup pushl %ebp movl, %ebp pushl %edi movl (%ebp), %edi movl 8(%ebp), %eax addl %edi, %eax movl (), %edi movl %ebp, popl %ebp ret (%ebp) 8(%ebp) %ebp restoring %edi old %ebp old %edi y x 8

29 Callee Procedure Call Example (IA3/Linux) int sum(int x, int y) { return x + y; Callee in assembly (better version) setup body cleanup pushl %ebp movl, %ebp pushl %edi movl (%ebp), %edi movl 8(%ebp), %eax addl %edi, %eax movl (), %edi movl %ebp, popl %ebp ret %ebp restoring %ebp old %ebp old %edi y x 9

30 Callee Why use a frame pointer? (%ebp) int sum(int x, int y) { return x + y; To make debugging easier - may move - %ebp is fixed Your compiler emits a symbol map y (%ebp) x 8(%ebp) gdb uses this map when you write print x (%ebp) 8(%ebp) %ebp old %ebp old %edi y x 30

31 Aside how does gdb s backtrace work? Follow return addresses! - use old %ebp to find prior frame Pseudocode while (pc is not in main ) { pc = 4(%ebp) %ebp = (%ebp) (%ebp) 8(%ebp) 4(%ebp) %ebp old %ebp old %edi y x 3

32 How is x86-64 different? Pass the first six arguments in registers - In this order %rdi,%rsi,%rdx,%rcx,%r8,%r9 New register save convention - Callee save %rbx,%rbp,%r,%r3,%r4,%r5 - Others are caller save By default, gcc omits the frame pointer - It has to emit more complex debug info (e.g., the location of argument x relative to can change) 3

33 Procedure Call Example (x86-64/linux) Caller int z = sum(, ); Caller in assembly movl $, %edi movl $, %esi call sum edi not rdi because int is 3-bits Callee int sum(int x, int y) { return x + y; Callee in assembly addl %esi, %edi movl %edi, %eax ret x86-64 with gcc does not use a frame pointer 33 Tip you can force gcc to emit code with a frame pointer using gcc -fno-omit-frame-pointer