1. 火雨辰
homeWork1

2. Q1 binary calculation
It’s a pair of unsigned Multiplication, and each value is represented by 4-bits unsigned integer. Please solve the equation step by step! Or you can’t get marks of this question.
0011 * x * y = (a)
0100 * x * y = (b)

3. Q1 binary calculation
3x+4y=9
4x+3y=8
x=5/7,y=12/7

4. Q1 binary calculation Solution1： (b)–(a) 0001*x+1111*y=1111 (c)
(c)* *x+1110*y= (d)
(c)+(d) 0011*x+1101*y=1101 (e)
(e) –(a) 1001*y=  y=0100
0011*x=  x=0011

5. Q1 binary calculation
Solution2: x=abcd(2) y=efgh(2) (a)  (a+b+e)(b+c+h)(a+d)d = 1001 (b)  (e+f+c)(f+g+d)(g+h)h = 1000 From right to left solve the function

6. Q1 binary calculation
Solution3: (3x+4y) mod 16= 9 (4x+3y) mod 16= 8 3x + 4y = a 4x + 3y = b x,y ϵ [0,15] x = 3, y = 4

7. Q2 Operations
Given A and B with hexadecimal expression 0xDC and 0xE3 respectively.Calculate the values of the following expressions:
A & B 0xC0
A && B 0x01
A | B 0xFF
A || B 0x01
~ A |~ B 0x3F
! A || B 0x01
A&&~ B 0x01

8. X>>3 (Arithmetic)
Q2 Operations X
X<<4
X>>3 (Logical)
X>>3 (Arithmetic)
Hex
0xF3
0x30
0x1E
0xFE
0x0E
0xE0
0x01
0xCD
0xD0
0x19
0xF9
0xA9
0x90
0x15
0xF5

9. Q3 Endian
Consider the following calls to show_bytes() from Figure2.4 page 76 (English text book)
int val = 0x1234ABCD;
byte_pointer valp = (byte_pointer) &val;
show_bytes(valp,1); /* A. */
show_bytes(valp,2) /* B. */
show_bytes(valp,3) /* C. */

10. Q3 Endian
Indicate which of the values would be printed by each call on a little-endian and on a big-endian machine:
A.Little endian: _CD_____; Big endian: __12____;
B.Little endian: _CDAB___; Big endian: __1234_;
C.Little endian: _CDAB34_; Big endian: _1234AB;

11. Q3 Endian
Try to write an C function to determine whether the machine running this code is big endian or small endian. The function should have no more than 2 lines. (It’s okey to do this step by step, ignore the restriction when you start) Int BIG_small(){ _________(1)___________; _________(2)___________; } // return 0 if small, 1 for big

12. Q3 Endian
Int i = 1;
Return !(*((char *)&i));

13. Q3 Endian Fault 1: Byte_pointer y = &0x01000000; return y[0]; Fault 2:
Char *p = &0x1100;
return p;

14. Q3 Endian Fault 3: Unsigned char* x = 0x100; Return (int)x[0];
Int a = 0x0100;
Return ((char *)&a)[0];

15. Q3 Endian Fault 5: Int x = 0xFF; Return !(&x)[0]; Fault 6:
Unsigned short x = 0x ;
Return (unsigned char)x;

16. Q3 Endian Fault 7: Unsigned int x =1; Return (int)((char*)&x);
Int x = 0x01;
Return !(x ^ show_byte(x,3));

17. Q3 Endian Fault 9: Unsigned char c = 0x1200; Retur c[0]; Fault 10:
Short int a = 0x1122;
Return (char)a == 0x11;

18. Q3 Endian Fault 11: Int x = 0x12345678; Return (&x)[0]==0x12;
Return ((char*) &x)[0];

19. Q4 Encodings Expression 2’s-complement Signed 8-bit Decimal 3+2
Suppose all the numbers are 8‐bit long. Give the result of the following expressions of C language in full 8‐bit 2’s complement form and signed decimal form.
Expression
2’s-complement
Signed 8-bit Decimal
3+2 5
99/2
(-23)/2
round to zero
-127-1
125+3
125>>3

20. Q5 Sign vs. unsign
So easy
(unsigned)-7 -8 -7
(unsigned)-8

21. Q6 casting
The following C code pieces are executed on a typical 32‐bit big endian machine with 2’s complement encoding. Please give the output.
int main() {
int x = 257;
char y = -10;
int z = 128;
char a = (char)x;
short b =(short)y;
unsigned short ud = (unsigned short)b;
char c = (char)z;
unsigned int w = (c > 0) ? 0 : 1;
printf("a=%d,b=%d,ud=%x,c=%d,w=%d\n",a,b,ud,c,w); }

22. Q6 casting Correct: A=1, b=-10, ud=0xfff6,c =-128,w=1 Incorrect:
Ud = 0xfffffff6
What about little endian?

23. Q7 integer Arithmetic
easy

24. Q8 Float num Normalized: V = (-1)sign * (1.fraction) * 2e-bias
Where bias = 2K-1-1
Denormalized:
V = (-1)sign * (0.fraction) * 2E
where E = 1 – (2K-1-1)

25. Q8 Float num Convert decimal fraction into binary form
* 2 =
* 2 =
0.375 * 2 =
0.75 * 2 =
0.5 * 2 =
= ( )2

26. Q8 Float num
(12.625)10 = ( )2
= * 23
(0x4c18)16 = ( )2
= * 27

27. Q8 float num
Sum = * * 23
= * 27
= * 27 = 0x4C50
Mul = * * 210
= * 210
= * 210 = 0x5350 or
= * 134 = = 0x5350