2011 ICS Middle Examination

2011 ICS Middle Examination

PROBLEM-1

Problem-1 • Ref: section 2.2, 2.3 • Conditions • 6-bits machine • Two’s complement arithmetic • Signed integer • Right shift is performed arithmetically • Short is encoded using 3 bits

short sy = -4; unsigned short usy = sy; int y = sy; int x = -30; unsigned int ux = x; Problem-1 9 = 001 001 -9 = 110 110 + 1 = 110 111

short sy = -4; unsigned short usy = sy; int y = sy; int x = -30; unsigned int ux = x; Problem-1 x = -30 30 = 011 110 -30 = 100 001 + 1 = 100 010 ux = x = 100 010

short sy = -4; unsigned short usy = sy; int y = sy; int x = -30; unsigned int ux = x; Problem-1 sy = -4 short 4 = 100 -4 = 011 + 1 = 100 y = sy = 111 100 sign extension

short sy = -4; unsigned short usy = sy; int y = sy; int x = -30; unsigned int ux = x; Problem-1 sy = -4 short 4 = 100 -4 = 011 + 1 = 100 usy = sy = 100

short sy = -4; unsigned short usy = sy; int y = sy; int x = -30; unsigned int ux = x; Problem-1 x = 100 010 x>>2 = 11100 010 = 111 000arithmetically (x>>2)<<1 = 111 0000 = 110 000

short sy = -4; unsigned short usy = sy; int y = sy; int x = -30; unsigned int ux = x; Problem-1 x = 100 010 y = 111 100 x + y = 100 010 + 111 100 = 1011 110overflow

short sy = -4; unsigned short usy = sy; int y = sy; int x = -30; unsigned int ux = x; Problem-1 x = 100 010 y = 111 100 !y = 000 000 x & !y = 100 010 & 000 000 = 000 000

short sy = -4; unsigned short usy = sy; int y = sy; int x = -30; unsigned int ux = x; Problem-1 UMax = 2^6 – 1 = 111 111 TMax = 2^5 – 1 = 011 111 -TMin = -2^5 = -(100 000) = 011 111 + 1 = 100 000

PROBLEM-2

Problem-2 • Ref: lab-1 • Conditions • The coding rules are same as lab1 • Legal ops: ! ~ & ^ | + << >> • Constants: <= 8 bits (0x0 ~ 0xFF)

Problem-2 /* [absValue] * calculate the abs value of x * ex: absValue(5) = 5 * absValue(-29) = 29 * legal ops: ~ & ^ | - << >> * #ops: 8 */ int absValue (int x) { /* fill your code */ return }

Problem-2 /* [absValue] * calculate the abs value of x * ex: absValue(5) = 5 * absValue(-29) = 29 * legal ops: ~ & ^ | - << >> * #ops: 8 */ int absValue (int x) { /* fill your code */ return } if x >= 0 abs = x = x^(0)-(0) if x < 0 abs = x^0xF.. + 1 = x^(-1)+1 = x^(-1)–(-1) abs = x^mask-mask mask = x>>31

Problem-2 /* [absValue] * calculate the abs value of x * ex: absValue(5) = 5 * absValue(-29) = 29 * legal ops: ~ & ^ | - << >> * #ops: 8 */ int absValue (int x) { int mask = x >> 31; abs = x ^ mask – mask; return abs; } ex: x=5 mask = 5>>31 = 0 abs = 5^0 – 0 = 5 ex: x=-5= 0xF..FB mask = 0xF..FF abs = 0xF..FB ^ 0xF..FF - (-1) = 0x0..04 + 0x0..01 = 0x5

Problem-2 /* [swapBits] * swap n bits started from i * with n bits started from j * ex: swapBits(0x2F,1,5,3) = 0xE3 * legal ops: ~ & ^ | - << >> */ intswapBits (int x, int i, int j, int n) { int mask = (1 << ?) - 1; intxorTemp = ((x >> ? ) ^ (x >> ? ))&? ; return x ? ((xorTemp << i) ? (xorTemp << j) }

Problem-2 x = 0x2F, i = 1, j = 5, n = 3 0x2F = 0010 1111 swapBits(0x2F) = 1110 0011 = E3 x = 0x..A..B.. swapBits(x) = 0x..B..A.. A = A ^ B ^ B = (A ^ B) ^ B B = B ^ A ^ A = (B ^ A) ^ A x = 0x..A..B.. ^ 0x..T..T.. (T = A ^ B) swapBits(x) = 0x..B..A..

Problem-2 x = 0x..A..B.. T = A ^ B = 0x0..0..A ^ 0x0..A..B & 0x0....0M {M = (1<<n)-1} = ((x >> j) ^ (x >> i)) & M swapBits(x) = 0x..A..B.. ^ 0x..T..T.. {0x..T.. | 0x....T..} = x ^ ((T << j) | (T << i))

Problem-2 /* [swapBits] * swap n bits started from i * with n bits started from j * ex: swapBits(0x2F,1,5,3) = 0xE3 * legal ops: ~ & ^ | - << >> */ intswapBits (int x, int i, int j, int n) { int mask = (1 << n) - 1; intxorTemp = ((x >> i) ^ (x >> j))& mask; return x^((xorTemp << i)|(xorTemp << j) }

PROBLEM-3

Problem-3 • Ref: section 3.3, 3.4, 3.5 • Conditions • 32-bit little endian machine • 4 byte size and hex • Each operation take effect on the memory and register

Problem-3

PROBLEM-4

Problem-4 • Ref: section 3.3~3.6, 3.7 • Conditions • 32-bit little endian machine

int mystery(int i) { if(i != 0) return i + mystery(i-1) return i; } int main(void) { return mystery(10); } Problem-4 08048324 <mystery> L1 8048324: 55 L2 : 89 e5 L3 : 83 ec 08 L4 ? : 83 7d 08 00 00

int mystery(int i) { if(i != 0) return i + mystery(i-1) return i; } int main(void) { return mystery(10); } Problem-4 08048324 <mystery> L1 8048324: 55 L2 8048325: 89 e5 L3 8048327: 83 ec 08 L4 804832a: 83 7d 08 00 00

2011 ICS Middle Examination