CSAPP(深入理解计算机系统)第二版家庭作业答案-第二章

答案均由本人完成，并且实验或者调试，欢迎参考！

2.55-2.57略2.58int is_little_endian(){int i=1;return *((char*)&i);}2.59int test_2_59(int x,int y){return (x&0xff)|(y&(~0xff));}2.60unsigned replace_byte(unsigned x,unsigned char b,int i){    unsigned int mask=0xff;    unsigned int val=b;    unsigned int shift_val=i<<3;    mask=mask<<shift_val;    val=val<<shift_val;    return (x&~mask)|val;}2.61A.!~x;B.!x;C.!~(x>>24)D.!(x&0xff)2.62bool int_shifts_are_logical(){int x=-1;x=x>>(sizeof(int)<<3-1);return !~x; }2.63int sra(int x,int k){    int xsrl=(unsigned)x>>k;    int w=sizeof(int)<<3;    //if(x<0)    //    return xsrl|~((1<<(w-k-1))-1);    //return xsrl;//不用逻辑运算符int z=1<<(w-k-1);return xsrl|~((z&xsrl)-1);}unsigned srl(unsigned x,int k){    unsigned xsra=(int)x>>k;    int w=sizeof(int)<<3;    return xsra&((1<<(w-k))-1);}2.64//求任意偶数位是否为1，从0开始计算int any_even_one(unsigned x){while(x){if(x&1)return 1;x=x>>2;}return 0;}2.65//求是否包含偶数个1int even_ones(unsigned x){int w=32;while(w>1){w=w>>1;x=(x>>w)^x;}return !(x&1);}2.66//求最高有效位的1int leftmost_one(unsigned x){    x |= (x >> 1);    x |= (x >> 2);    x |= (x >> 4);    x |= (x >> 8);    x |= (x >> 16);  //这里一定会得到从最高位的1到最低位的全部为1    return x^(x>>1);}2.67A.在int为32位的机器上不应该左移32位B.int bad_int_size_is_32(){int set_msb=1<<31;return set_msb && set_msb<0;}C.int bad_int_size_is_32(){int z=1<<15;z=z<<15;int set_msb=z<<1;int beyond_msb=z<<2;return set_msb && !beyond_msb;}2.68int lower_bits(int x,int n){return x&((2<<(n-1))-1);}2.69unsigned rotate_right(unsigned x,int n){int w=sizeof(unsigned)<<3;return x>>n | x<<(w-n-1)<<1;}2.70int fits_bits(int x,int n){x=x>>(n-1); //x全为1或者全为0能表示 return !(x && ~x);}2.71A.Word是unsigned类型所以会进行无符号扩展，比如原字节是0xff(-1)，这个函数将返回0xff(255)，而非题目要求B.int xbyte(packed_t word,int bytenum){return ((int)(word<<((3-bytenum)<<3)))>>24;}2.72A.size_t是unsigned类型，相减结果会提升至unsigned，而unsigned类型总是>=0B.void copy_int(int val,void *buf,unsigned maxbytes){if(maxbytes>=sizeof(val))memcpy(buf,(void*)&val,sizeof(val));}2.73int saturating_add(int x,int y){int sum=x+y;if( x>=0 && y>=0 && sum<0)return INT_MAX;if(x<0 && y<0 && sum>=0)return INT_MIN;return sum;}2.74int tsub_ovf(int x,int y){int sub=x-y;if(x >=0 && y<0 && sub<0)return 1;if( x<0 && y>=0 && sub>=0)return 1;return 0;}2.75unsigned unsigned_high_PRod(unsigned x,unsigned y)  {      int w = sizeof(unsigned) << 3;  int sx=(int)x;int sy=(int)y;int t= signed_high_prod(sx,sy);return t+ ((sx>>(w-1))&1)*sy+((sy>>(w-1))&1)*sx;  }2.76A.(x<<2)+xB.(x<<3)+xC.(x<<5)-(x<<1)D.(x<<3)-(x<<6)2.77int divide_power2(int x,int k){int bias= (x>>31) & ((1<<k)-1);return (x+bias)>>k;}2.78int mul5div8(int x){int x=(x<<2)+x;int bias= (x>>31) & 7;return (x+bias)>>3;}2.79int fiveeighths(int x){int bias= (x>>31) & 7;x=(x+bias)>>3;return (x<<2)+x;}2.80A.~((1<<n)-1)B.((1<<n)-1)<<m2.81A.x=0,y=TminB.总为1，位级表示相同C.~x=-x-1~y=-y-1~(x+y)=-x-y-1-x-1-y-1!=-x-y-1总为0D.总为1，无符号和有符号减法位级表示相同，而且最后都转成int类型，结果一致E.总为1，右移再左移必有1位丢失或者不丢失，丢失的这1位权值为正，值会变小2.82A.等比数列求和a1*(1-q^n)/(1-q)a1=Y/(2^k) q=1/(2^k)假设n非常大，1-q^n可看作1a1/(1-q)所以通项公式是Y/(2^k-1)B.y=001 1/7y=1001 3/5y=000111 1/92.83return sx==sy?(ux==uy)||((ux>uy)^sx):((ux<<1)==0 && (uy<<1)==0)||sx<sy;或者return ((ux<<1)==0 && (uy<<1)==0) || (!sx && sy) || (!sx && !sy && ux >= uy) ||(sx && sy && ux <= uy);2.84A.E M fV2 1.01 0.01 5e=1+2^(k-1)f=01000..00B.E M fVn 1.111.. 0.11.. 2^(n+1)-1e=n+2^(k-1)-1f=11111..11C.E M fV2^(k-1)-2 1 0 2^(2^(k-1)-2)e=2^k-3f=02.85最小的正非规格化数：2^(-2^14+2)*1/2^63最小的正规格化数：2^(-2^14+2)最大的规格化数：2^(2^14-1)*(2-1/2^63)2.86描述 Hex M E V-0 0x8000 0 -62 --最小的值>1 0x3f01 257/256 0 257*2^(-8)256 0x4700 1 8 --最大的非规格化数 0x00ff255/256 -62255*2^(-70)负无穷 0xff00 -- -- --0x3AA0 -- 13/8 -5 13*2^(-8)2.87格式A 格式B位 值 位值1 01110 001 -9/161 0110 0010 -9/160 10110 101 208 0 1110 10102081 00111 110 -7/10241 0000 0111     -7/10240 00000 101 3/(2^17)0 0000 0001 1/(2^10)1 11011 000 -2^12 1 1111 0000负无穷0 11000 100 3*2^80 1111 0000 正无穷       2.88A.0x转float有精度丢失，x=2^24+1=16777217B.0x=Tmax,y=Tmax,x+y溢出得到负数，而dx+dy不溢出C.1double可精确表示2^53内的所有整数，而相加不可能超出这个范围D.032位乘法可能导致结果超出2^53，会导致舍入E.0dx或dy为02.89float fpwr2(int x){unsigned exp,frac;unsigned u;if(x<-149){exp=0;frac=0;}else if(x<-126){exp=0;frac= 1<<(149+x);}else if(x<128){exp=x+127;frac=0;}else{exp=255;frac=0;}u=exp<<23|frac;return u2f(u);}2.900x40490fdb0100 0000 0100 1001 0000 1111 1101 1011A.11.0010010000111111011011B.11.001001001001001001001001..C.从低9位，即2^(-9)2.91float_bits float_absval(float_bits f){unsigned exp=f>>23&0xff;unsigned frac=f&0x7fffff;if(exp==0xff && frac!=0)return f;return f&0x7fffffff;}2.92float_bits float_negate(float_bits f){unsigned exp=f>>23&0xff;unsigned frac=f&0x7fffff;if(exp==0xff && frac!=0)return f;return f^0x80000000;}2.93float_bits float_half(float_bits f){unsigned sign=f>>31;unsigned exp=f>>23&0xff;unsigned frac=f&0x7fffff;if(exp==0xff && frac!=0)return f;if(exp==0xff && frac==0)return f;       //正无穷或者负无穷if(exp ==0 || exp==1){if(exp==1)frac=frac|0x800000;if((frac & 3)==3)frac=(frac+1)>>1;elsefrac=frac>>1;}if(exp>0) {exp-=1;}return (sign<<31) | (exp<<23) | frac;}2.94float_bits float_twice(float_bits f){unsigned sign=f>>31;unsigned exp=f>>23&0xff;unsigned frac=f&0x7fffff;if(exp==0xff && frac!=0)return f;if(exp==0xff && frac==0)return f;       //正无穷或者负无穷if(exp ==0){ frac=frac<<1;}else{exp+=1;if(exp==0xff)frac=0;}return (sign<<31) | (exp<<23) | frac;}2.95float_bits float_i2f(int i){if(i==0)return 0;unsigned sign=0;unsigned frac=i;if(i&0x80000000){sign=1;frac=~frac+1;}unsigned c=0;while (!(frac&0x80000000)){frac=frac<<1;++c;}unsigned exp= (31-c)+127;frac=frac<<1;frac=(frac>>9)+((frac&0x100)&&((frac&0x200)||(frac&0xff)));//向偶数舍入，进位if(frac&0x800000) //如果进位++exp;return (sign<<31) | (exp<<23) | (frac&0x7fffff);}2.96int float_f2i(float_bits f){unsigned sign=f>>31;unsigned exp=f>>23&0xff;unsigned frac=f&0x7fffff;if(exp<0x7f)return 0;else if(exp>158 ||(exp==158&&frac>0))return 0x80000000;int c=exp-0x7f;if(c>23)frac= ((1<<23)|frac)<<(c-23);elsefrac= (frac>>(23-c))|(1<<c);return sign?(~frac+1):frac;}
原创地址：http://blog.csdn.net/maidou0921/article/details/53907832