标签：nim expr 浮点数 author ast ddb statement including enc
只能用指定字符，实现函数功能
/*
* CS:APP Data Lab
*
* <Please put your name and userid here>
*
* bits.c - Source file with your solutions to the Lab.
* This is the file you will hand in to your instructor.
*
* WARNING: Do not include the <stdio.h> header; it confuses the dlc
* compiler. You can still use printf for debugging without including
* <stdio.h>, although you might get a compiler warning. In general,
* it‘s not good practice to ignore compiler warnings, but in this
* case it‘s OK.
*/
#if 0
/*
* Instructions to Students:
*
* STEP 1: Read the following instructions carefully.
*/
You will provide your solution to the Data Lab by
editing the collection of functions in this source file.
INTEGER CODING RULES:
Replace the "return" statement in each function with one
or more lines of C code that implements the function. Your code
must conform to the following style:
int Funct(arg1, arg2, ...) {
/* brief description of how your implementation works */
int var1=Expr1;
...
int varM=ExprM;
varJ=ExprJ;
...
varN=ExprN;
return ExprR;
}
Each "Expr" is an expression using ONLY the following:
1. Integer constants 0 through 255 (0xFF), inclusive. You are
not allowed to use big constants such as 0xffffffff.
2. Function arguments and local variables (no global variables).
3. Unary integer operations ! ~
4. Binary integer operations & ^ | + << >>
Some of the problems restrict the set of allowed operators even further.
Each "Expr" may consist of multiple operators. You are not restricted to
one operator per line.
You are expressly forbidden to:
1. Use any control constructs such as if, do, while, for, switch, etc.
2. Define or use any macros.
3. Define any additional functions in this file.
4. Call any functions.
5. Use any other operations, such as &&, ||, -, or ?:
6. Use any form of casting.
7. Use any data type other than int. This implies that you
cannot use arrays, structs, or unions.
You may assume that your machine:
1. Uses 2s complement, 32-bit representations of integers.
2. Performs right shifts arithmetically.
3. Has unpredictable behavior when shifting if the shift amount
is less than 0 or greater than 31.
EXAMPLES OF ACCEPTABLE CODING STYLE:
/*
* pow2plus1 - returns 2^x + 1, where 0 <= x <= 31
*/
int pow2plus1(int x) {
/* exploit ability of shifts to compute powers of 2 */
return (1 << x) + 1;
}
/*
* pow2plus4 - returns 2^x + 4, where 0 <= x <= 31
*/
int pow2plus4(int x) {
/* exploit ability of shifts to compute powers of 2 */
int result=(1 << x);
result += 4;
return result;
}
FLOATING POINT CODING RULES
For the problems that require you to implement floating-point operations,
the coding rules are less strict. You are allowed to use looping and
conditional control. You are allowed to use both ints and unsigneds.
You can use arbitrary integer and unsigned constants. You can use any arithmetic,
logical, or comparison operations on int or unsigned data.
You are expressly forbidden to:
1. Define or use any macros.
2. Define any additional functions in this file.
3. Call any functions.
4. Use any form of casting.
5. Use any data type other than int or unsigned. This means that you
cannot use arrays, structs, or unions.
6. Use any floating point data types, operations, or constants.
NOTES:
1. Use the dlc (data lab checker) compiler (described in the handout) to
check the legality of your solutions.
2. Each function has a maximum number of operations (integer, logical,
or comparison) that you are allowed to use for your implementation
of the function. The max operator count is checked by dlc.
Note that assignment (‘=‘) is not counted; you may use as many of
these as you want without penalty.
3. Use the btest test harness to check your functions for correctness.
4. Use the BDD checker to formally verify your functions
5. The maximum number of ops for each function is given in the
header comment for each function. If there are any inconsistencies
between the maximum ops in the writeup and in this file, consider
this file the authoritative source.
/*
* STEP 2: Modify the following functions according the coding rules.
*
* IMPORTANT. TO AVOID GRADING SURPRISES:
* 1. Use the dlc compiler to check that your solutions conform
* to the coding rules.
* 2. Use the BDD checker to formally verify that your solutions produce
* the correct answers.
*/
#endif
//1
/*
* bitXor - x^y using only ~ and &
* Example: bitXor(4, 5)=1
* Legal ops: ~ &
* Max ops: 14
* Rating: 1
*/
int bitXor(int x, int y) {
//求异或
return ~((~(x&(~y)))&(~((~x)&y)));
}
/*
* tmin - return minimum two‘s complement integer
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 4
* Rating: 1
*/
int tmin(void) {
//求最小的返回补码的数
return 1<<31;
}
//2
/*
* isTmax - returns 1 if x is the maximum, two‘s complement number,
* and 0 otherwise
* Legal ops: ! ~ & ^ | +
* Max ops: 10
* Rating: 1
*/
int isTmax(int x) {
//如果x是2的补码中的最大整数（0x7fffffff），则返回1,否则返回0
return !(x^(0x80000000-1));
}
/*
* allOddBits - return 1 if all odd-numbered bits in word set to 1
* where bits are numbered from 0 (least significant) to 31 (most significant)
* Examples allOddBits(0xFFFFFFFD)=0, allOddBits(0xAAAAAAAA)=1
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 12
* Rating: 2
*/
int allOddBits(int x) {
//x中所有奇数位为1则返回1，否则返回0
return !((x&0xAAAAAAAA)^0xAAAAAAAA);
}
/*
* negate - return -x
* Example: negate(1)=-1.
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 5
* Rating: 2
*/
int negate(int x) {
//求负数
return ~x+1;
}
//3
/*
* isAsciiDigit - return 1 if 0x30 <= x <= 0x39 (ASCII codes for characters ‘0‘ to ‘9‘)
* Example: isAsciiDigit(0x35)=1.
* isAsciiDigit(0x3a)=0.
* isAsciiDigit(0x05)=0.
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 15
* Rating: 3
*/
int isAsciiDigit(int x) {
//如果0x30 <= x <= 0x39返回1
//>>31取符号位
return !((x+(~0x30+1))>>31|(0x39+(~x+1)));
}
/*
* conditional - same as x ? y : z
* Example: conditional(2,4,5)=4
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 16
* Rating: 3
*/
int conditional(int x, int y, int z) {
//x?y:z;
//!!x可以把非0x变为1,0不变
return ((!x+(~1+1))&y)|((!!x+(~1+1))&z);
}
/*
* isLessOrEqual - if x <= y then return 1, else return 0
* Example: isLessOrEqual(4,5)=1.
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 24
* Rating: 3
*/
int isLessOrEqual(int x, int y) {
//x<=y返回1，否则0
//直接减法会溢出
int a=x^y;//x==y
int b=!!(((x>>31)^(y>>31))&(x>>31));//x>=0,y<0
int c=!!((!((x>>31)^(y>>31)))&((x+(~y+1))>>31));//同号
return !a|b|c;
}
//4
/*
* logicalNeg - implement the ! operator, using all of
* the legal operators except !
* Examples: logicalNeg(3)=0, logicalNeg(0)=1
* Legal ops: ~ & ^ | + << >>
* Max ops: 12
* Rating: 4
*/
int logicalNeg(int x) {
//实现!的功能
//-2147483648取反会溢出，正负的符号位相同
return (~((((~x+1)>>31)|(x>>31))&1))&1;
}
/* howManyBits - return the minimum number of bits required to represent x in
* two‘s complement
* Examples: howManyBits(12)=5
* howManyBits(298)=10
* howManyBits(-5)=4
* howManyBits(0) =1
* howManyBits(-1)=1
* howManyBits(0x80000000)=32
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 90
* Rating: 4
*/
int howManyBits(int x) {
//求二进制下多少位
int b16,b8,b4,b2,b1,sign=x>>31;
x^=sign;
b16=(!!(x>>16))<<4;
x>>=b16;
b8=(!!(x>>8))<<3;
x>>=b8;
b4=(!!(x>>4))<<2;
x>>=b4;
b2=(!!(x>>2))<<1;
x>>=b2;
b1=(!!(x>>1))<<0;
x>>=b1;
return b16+b8+b4+b2+b1+x+1;
}
//float
/*
* floatScale2 - Return bit-level equivalent of expression 2*f for
* floating point argument f.
* Both the argument and result are passed as unsigned int‘s, but
* they are to be interpreted as the bit-level representation of
* single-precision floating point values.
* When argument is NaN, return argument
* Legal ops: Any integer/unsigned operations incl. ||, &&. also if, while
* Max ops: 30
* Rating: 4
*/
unsigned floatScale2(unsigned uf) {
//求2*uf
int sign=uf&(1<<31);
//符号位
int exp=(uf&0x7F800000)>>23;
//阶码
if(exp==0)
return uf<<1|sign;
//若为非规格数，直接给uf乘以2后加上符号位即可
if(exp==255)
return uf;
// 若为inf或者NaN，直接返回自身
exp++;
// 若uf乘以2为255，则返回inf
if(exp==255)
return (0x7F800000|sign);
return sign|(exp<<23)|(uf&0x807FFFFF);
// 返回阶码加1后的原符号数
return 2;
}
/*
* floatFloat2Int - Return bit-level equivalent of expression (int) f
* for floating point argument f.
* Argument is passed as unsigned int, but
* it is to be interpreted as the bit-level representation of a
* single-precision floating point value.
* Anything out of range (including NaN and infinity) should return
* 0x80000000u.
* Legal ops: Any integer/unsigned operations incl. ||, &&. also if, while
* Max ops: 30
* Rating: 4
*/
int floatFloat2Int(unsigned uf) {
int exp=((uf&0x7F800000)>>23)-127;
int sign=uf>>31;
int frac=((uf&0x007FFFFF)|0x00800000);
if(!(uf&0x7FFFFFFF))
return 0;
//若原浮点数为0，则返回0
if(exp>31)
return 0x80000000;
//若原浮点数指数大于31，返回inf
if(exp<0)
return 0;
//若浮点数小于0，则返回0；
if(exp>23)
frac=frac<<(exp-23);
//将小数转化为整数
else
frac=frac>>(23-exp);
if(!((frac>>31)^sign))
return frac;
//判断是否溢出，若符号位没有变化，则没有溢出，返回正确的值
else if(frac>>31)
return 0x80000000;
//原数为正值，现在为负值，返回溢出值
else
return ~frac+1;
// 原数为负值，现在为正值，返回相反数
}
/*
* floatPower2 - Return bit-level equivalent of the expression 2.0^x
* (2.0 raised to the power x) for any 32-bit integer x.
*
* The unsigned value that is returned should have the identical bit
* representation as the single-precision floating-point number 2.0^x.
* If the result is too small to be represented as a denorm, return
* 0. If too large, return +INF.
*
* Legal ops: Any integer/unsigned operations incl. ||, &&. Also if, while
* Max ops: 30
* Rating: 4
*/
unsigned floatPower2(int x) {
//判断是否溢出
int exp=x+127;
// 计算阶码
if(exp<=0)
return 0;
// 阶码小于等于0，则返回0
if(exp>=255)
return 0xFF<<23;//inf
// 阶码大于等于255，则返回INF
return exp<<23;
}
标签：nim expr 浮点数 author ast ddb statement including enc
原文地址：https://www.cnblogs.com/diakla/p/14660644.html
