HackMD
2018q1 第 2 週測驗題
測驗 一
Image Not Showing
Possible Reasons
- The image file may be corrupted
- The server hosting the image is unavailable
- The image path is incorrect
- The image format is not supported
請完成下方程式碼，依循 IEEE 754 單精度規範，輸出
- 當
- 當
注意：這裡假設 u2f 函式返回的浮點數值與其無號數輸入有著相同的位數，也就是至少 32-bit
#include <math.h>
static inline float u2f(unsigned int x) { return *(float *) &x; }
float exp2_fp(int x) {
unsigned int exp /* exponent */, frac /* fraction */;
/* too small */
if (x < 2 - pow(2, Y0) - 23) {
exp = 0;
frac = 0;
/* denormalize */
} else if (x < Y1 - pow(2, Y2)) {
exp = 0;
frac = 1 << (unsigned) (x - (2 - pow(2, Y3) - Y4));
/* normalized */
} else if (x < pow(2, Y5)) {
exp = pow(2, Y6) - 1 + x;
frac = 0;
/* too large */
} else {
exp = Y7;
frac = 0;
}
/* pack exp and frac into 32 bits */
return u2f((unsigned) exp << 23 | frac);
}
作答區
Y0 = ?
(a)0(b)1(c)2(d)3(e)4(f)5(g)6(h)7
Y1 = ?
(a)0(b)1(c)2(d)3(e)4(f)5(g)6(h)7
Y2 = ?
(a)0(b)1(c)2(d)3(e)4(f)5(g)6(h)7
Y3 = ?
(a)0(b)1(c)2(d)3(e)4(f)5(g)6(h)7
Y4 = ?
(a)20(b)21(c)22(d)23(e)24(f)25(g)26(h)27
Y5 = ?
(a)0(b)1(c)2(d)3(e)4(f)5(g)6(h)7
Y6 = ?
(a)0(b)1(c)2(d)3(e)4(f)5(g)6(h)7
Y7 = ?
(a)0x00(b)0xF0(c)0xFA(d)0xFB(e)0xFC(f)0xFD(g)0xFE(h)0xFF
Reference:
延伸題: 給定一個單精度的浮點數值，輸出較大且最接近的
測驗 二
考慮以下程式碼對一個浮點數進行 * 0.5 的操作，補完程式:
typedef unsigned int float_bits;
float_bits float_x0_5(float_bits f) {
unsigned sig = f >> 31;
unsigned rest = f & 0x7FFFFFFF;
unsigned exp = f >> 23 & 0xFF;
unsigned frac = f & 0x7FFFFF;
/* NaN or INF */
if (exp == A0) return f;
/* round to even. Last 2 bits of frac are considered:
* 00 => 0 just >> 1
* 01 => 0 (round to even) just >> 1
* 10 => 1 just >> 1
* 11 => 1 + 1 (round to even) just >> 1 and plus 1
*/
int addition = (frac & 0x3) == 0x3;
/* Denormalized */
if (exp == 0) {
frac >>= A1;
frac += addition;
/* Normalized to denormalized */
} else if (exp == 1) {
rest >>= A2;
rest += addition;
exp = rest >> A3 & A4;
frac = rest & 0x7FFFFF;
/* Normalized */
} else {
exp -= A5;
}
return sig << A6 | exp << 23 | frac;
}
作答區
A0 = ?
(a)0x00(b)0xF0(c)0xFA(d)0xFB(e)0xFC(f)0xFD(g)0xFE(h)0xFF
A1 = ?
(a)0(b)1(c)2(d)3(e)4(f)5(g)6(h)7
A2 = ?
(a)0(b)1(c)2(d)3(e)4(f)5(g)6(h)7
A3 = ?
(a)20(b)21(c)22(d)23(e)24(f)25(g)26(h)27
A4 = ?
(a)0x00(b)0xF0(c)0xFA(d)0xFB(e)0xFC(f)0xFD(g)0xFE(h)0xFF
A5 = ?
(a)0(b)1(c)2(d)3(e)4(f)5(g)6(h)7
A6 = ?
(a)31(b)30(c)29(d)28(e)27(f)26(g)25(h)24
Reference:
測驗 三
考慮到某些實數的二進位表示形式如同 01)，而 0011)，考慮到以下 y 值，求出對應的十進位分數值。
- y =
010011=> - y =
101=> - y =
0110=>
X1 = ?
(a)11(b)23(c)63(d)97(e)57(f)31(g)5(h)13
X2 = ?
(a)17(b)19(c)13(d)11(e)3(f)7(g)23(h)97
X3 = ?
(a)23(b)19(c)17(d)13(e)11(f)7(g)5(h)3
Reference:
- Binary Fractions
[83]
測驗 四
在 你所不知道的C語言: 遞迴呼叫篇 提到以下程式碼:
#include <stdio.h>
int p(int i, int N) {
return (i < N && printf("%d\\n", i)
&&!p(i + 1, N)) || printf("%d\\n", i);
}
int main() { return p(1, 10) && printf("\\n"); }
預期會得到以下輸出:
1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 10 -> 9 -> 8 -> 7 -> 6 -> 5 -> 4 -> 3 -> 2 -> 1 ->
注意 10 只會出現一次。請填補下方程式碼，使得輸出為 1 -> 2 -> 3 -> 4 -> 5 ->，並且包含換行符號 (\n)
#include <stdio.h>
int p(int i, int N) {
return (i < N && printf("%d -\> ", i)
&& Q1(i + Q2, N + Q3));
}
int main() { return p(1, 10) && printf("\\n"); }
提示: Q1 是函式名稱，可能包含 ~ (bitwise NOT) 操作，Q2 和 Q3 是有號整數
作答區
Q1 = ?
(a)p(b)~p
Q2 = ?
(a)-5(b)-4(c)-3(d)-2(e)-1(f)0(g)1(h)2(i)3(j)4
Q3 = ?
(a)-5(b)-4(c)-3(d)-2(e)-1(f)0(g)1(h)2(i)3(j)4
