运用位运算实现以下函数，不得使用if,while,for等，不得使用乘除减等运算，不得调用任何函数。不得使用负数及大于255(0xff)的数，不得使用小数等。在每个函数前有对此函数的要求限制及说明。如下：Legal ops为在该函数中允许使用的操作符，Max ops表示这些操作符最多可使用的次数（各操作符使用的次数之和），当然你可以使用任意多个“=”号，你也可以定义任意多个变量，Rating为实现的难度（给定难度值仅供参考），下面有更详细的英文说明。
仅可能少地使用操作符！！我已经实现了所有的函数，希望网友能提供更好的方法，欢迎交流学习（QQ:121853868）
如下：
/*
* bang - Compute !x without using !
* Examples: bang(3) = 0, bang(0) = 1
* Legal ops: ~ & ^ | + <<>>
* Max ops: 12
* Rating: 4
*/
int bang(int x) {
int a,b,c;
a = (~x+1);
b = a>>31;
c = x>>31;
return 1&~((b&1)|(c&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.
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.
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 an integer by more
than the word size.
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 < }
/*
* 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 < result += 4;
return result;
}
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 operators (! ~ & ^ | + <<>>)
that you are allowed to use for your implementation of the function.
The max operator count is checked by dlc. Note that '=' 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. 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.
/*
* 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 btest test harness to check that your solutions produce
* the correct answers. Watch out for corner cases around Tmin and Tmax.
*/
以上为说明。
以下为需要实现的函数：
/*
* bang - Compute !x without using !
* Examples: bang(3) = 0, bang(0) = 1
* Legal ops: ~ & ^ | + <<>>
* Max ops: 12
* Rating: 4
*/
int bang(int x) {
int a,b,c;
a = (~x+1);
b = a>>31;
c = x>>31;
return 1&~((b&1)|(c&1));
}
/*
* bitCount - returns count of number of 1's in word
* Examples: bitCount(5) = 2, bitCount(7) = 3
* Legal ops: ! ~ & ^ | + <<>>
* Max ops: 40
* Rating: 4
*/
int bitCount(int x) {
int a = 255+(255<<8);
int b = a^(a<<8);
int c = b^(b<<4);
int d = c^(c<<2);
int e = d^(d<<1);
x = (x&e)+((x>>1)&e);
x = (x&d)+((x>>2)&d);
x = (x&c)+((x>>4)&c);
x = (x&b)+((x>>8)&b);
x = (x&a)+((x>>16)&a);
return x;
}
/*
* copyLSB - set all bits of result to least significant bit of x
* Example: copyLSB(5) = 0xFFFFFFFF, copyLSB(6) = 0x00000000
* Legal ops: ! ~ & ^ | + <<>>
* Max ops: 5
* Rating: 2
*/
int copyLSB(int x) {
return ~(x&1)+1;
}
/*
* divpwr2 - Compute x/(2^n), for 0 <= n <= 30
* Round toward zero
* Examples: divpwr2(15,1) = 7, divpwr2(-33,4) = -2
* Legal ops: ! ~ & ^ | + <<>>
* Max ops: 15
* Rating: 2
*/
int divpwr2(int x, int n) {
int a=(x>>31)&1;
return (x+(a<>n;
}
/*
* evenBits - return word with all even-numbered bits set to 1
* Legal ops: ! ~ & ^ | + <<>>
* Max ops: 8
* Rating: 2
*/
int evenBits(void) {
return (85<<24)+(85<<16)+(85<<8)+85;
/*or (85<<12)|(85<<8)|(85<<4)|85); */
}
/*
* fitsBits - return 1 if x can be represented as an
* n-bit, two's complement integer.
* 1 <= n <= 32
* Examples: fitsBits(5,3) = 0, fitsBits(-4,3) = 1
* Legal ops: ! ~ & ^ | + <<>>
* Max ops: 15
* Rating: 2
*/
int fitsBits(int x, int n) {
int y = x>>(n+~1+1);
return !y^!(~y);
}
/*
* getByte - Extract byte n from word x
* Bytes numbered from 0 (LSB) to 3 (MSB)
* Examples: getByte(0x12345678,1) = 0x56
* Legal ops: ! ~ & ^ | + <<>>
* Max ops: 6
* Rating: 2
*/
int getByte(int x, int n) {
int y = x>>(n<<3);
return y&255;
}
/*
* isGreater - if x > y then return 1, else return 0
* Example: isGreater(4,5) = 0, isGreater(5,4) = 1
* Legal ops: ! ~ & ^ | + <<>>
* Max ops: 24
* Rating: 3
*/
int isGreater(int x, int y) {
int z = x+~y+1;
int x1 = (x>>31)&1;
int y1 = (y>>31)&1;
int z1 = (z>>31)&1;
int k = (x1^y1)&(z1^x1);
return (!k&(!z1+~(!z)+1))|(z1&k);
}
/*
* isNonNegative - return 1 if x >= 0, return 0 otherwise
* Example: isNonNegative(-1) = 0. isNonNegative(0) = 1.
* Legal ops: ! ~ & ^ | + <<>>
* Max ops: 6
* Rating: 3
*/
int isNonNegative(int x){
int a = x>>31;
return !(a&1);
}
/*
* isNotEqual - return 0 if x == y, and 1 otherwise
* Examples: isNotEqual(5,5) = 0, isNotEqual(4,5) = 1
* Legal ops: ! ~ & ^ | + <<>>
* Max ops: 6
* Rating: 2
*/
int isNotEqual(int x, int y) {
int z = x+~y+1;
return !!z;
}
/*
* isPower2 - returns 1 if x is a power of 2, and 0 otherwise
* Examples: isPower2(5) = 0, isPower2(8) = 1, isPower2(0) = 0
* Note that no negative number is a power of 2.
* Legal ops: ! ~ & ^ | + <<>>
* Max ops: 60
* Rating: 4
*/
int isPower2(int x) {
int y =x>>31;
return (!!x)&!(1&y)&!(x&(x+(~1)+1));
}
/*
* log2 - return floor(log base 2 of x), where x > 0
* Example: log2(16) = 4
* Legal ops: ! ~ & ^ | + <<>>
* Max ops: 90
* Rating: 4
*/
int log2(int x) {
int y = x;
int j = !!y;
int h = (j<<31)>>31;
int n = 1;
int k = !(x>>16);
int s = (k<<31)>>31;
x = ((~s)&x)+(s&(x<<16));
n = n+(s&16);
k = !(x>>24);
s = (k<<31)>>31;
x = (~s&x)+(s&(x<<8));
n = n+(s&8);
k = !(x>>28);
s = (k<<31)>>31;
x = (~s&x)+(s&(x<<4));
n = n+(s&4);
k = !(x>>30);
s = (k<<31)>>31;
x = (~s&x)+(s&(x<<2));
n = n+(s&2);
return h&(31+!!(x>>31)+(~n+1));
}
/*
* logicalShift - shift x to the right by n, using a logical shift
* Can assume that 1 <= n <= 31
* Examples: logicalShift(0x87654321,4) = 0x08765432
* Legal ops: ~ & ^ | + <<>>
* Max ops: 16
* Rating: 3
*/
int logicalShift(int x, int n) {
int y = 1<<(31+~n+1);
int z = y|(y+~1+1);
int a = x>>n;
return a&z;
}
/*
* satAdd - adds two numbers but when positive overflow occurs, returns
* maximum possible value, and when negative overflow occurs,
* it returns minimum positive value.
* Examples: satAdd(0x40000000,0x40000000) = 0x7fffffff
* satAdd(0x80000000,0xffffffff) = 0x80000000
* Legal ops: ! ~ & ^ | + <<>>
* Max ops: 30
* Rating: 4
*/
int satAdd(int x, int y){
int z = x+y;
int x1 = (x>>31)&1;
int y1 = (y>>31)&1;
int z1 = (z>>31)&1;
int xy = !(x1^y1);
int k = xy&(z1^x1);
int a = k<<31;
int b =a|(a+~k+1);
int maxP = a+~1+1;
return (~b&z)|(b&(maxP+!z1));
}
/*
* tc2sm - Convert from two's complement to sign-magnitude
* where the MSB is the sign bit
* You can assume that x > TMin
* Example: tc2sm(-5) = 0x80000005.
* Legal ops: ! ~ & ^ | + <<>>
* Max ops: 15
* Rating: 4
*/
int tc2sm(int x) {
int y = x >> 31 ;
int sign = y&1;
int mag = (x+y)^y;
return (sign <<31) | mag;
}
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