Exam 1 Solutions CS 213 Spring 2009 ********* Problem 1 ********* Description Decimal Binary Bias 3 ------ Smallest Positive positive 1/16 000001 Lowest finite -14 111011 Smallest positive normalized 1/4 000100 ------ -7/16 100111 ------ 5/4 001101 ------ -5/8 101001 ------ 13(12) 011010 ********* Problem 2 ********* a) 32 bytes b) 6 bytes c) fun2 fun4 fun3 fun1 d)0x000f 0xbfb2ffdc 0x0000000c 0x0012 0x000000f3 0x000000d5 0x000000c 0x01 ********* Problem 3 ********* int switchfn(inta, longb) { int y=0,x=)xdeadbeef; switch(A*b) { case 1: return 24; case 6: a= x + b*4; return a; case 0: return a+b; case 4: x=a; y*=b; break; case 2: a= y==x; case 3: b = y0) { a = 0; for(b=len-1; b>0; b--) { if(array[b]>array[a]) { a=b; } } len--; tmp = array[len]; array[len] = array[a]; array[a] = tmp; } ********* Problem 5 ********* a a b b a d d b b a c Update 10/6/11 by Taerim Kim: Although C was accepted as the solution in Spring 2009, none of the choices for Problem 5 Question 11 are correct. ********* Problem 6 ********* a) Enter new stack frame by saving old base pointer and allocating space by subtracting from stack pointer b)Fill into dissasembly of foo 0x8, 0xc, 0x10, 0x14 or 8, 12, 16, 20 c) The new foo is not saving the old ebp. This makes the stack frame 4 bytes shorter old new ------ ------ | args | | args | | ret | | ret | | ebp | ebp | | | | | | | | | | | ... | | ... | | | | | |______| esp |______| esp d) 0x44, 0x48, 0x4c, 0x50 or 68, 72, 74, 80 e) Advantages: save space (4 bytes per stack frame) save time (don't execute enter leave instructions) %ebp can now be used as GPR Drawbacks: makes debugging "impossible" as you can no longer easily get a stack trace by climbing the stack with the base pointer