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40 Cards in this Set
- Front
- Back
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C for loop |
int factorial1(int n)
{ int res = 1; int i = 1; for(i=1;i<=n;i++) res=res*i; return res; } |
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C while loop |
int factorial2(int n)
{ int res = 1; int i = 1; while(i<=n) { res = res*i; i++; } return res; } |
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C do loop |
int factorial3(int n)
{ int res = 1; int i = 1; do { res = res*i; i++; } while(i<=n); return res; } |
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C Recursion |
int factorial4(int n)
{ if (n >= 1) return n*factorial4(n-1); else return 1; } |
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CHAR_BIT |
8 |
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SCHAR_MIN |
-128 |
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SCHAR_MAX |
+127 |
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UCHAR_MAX |
255 |
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CHAR_MIN |
-128 |
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CHAR_MAX |
+127 |
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MB_LEN_MAX |
16 |
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SHRT_MIN |
-32768 |
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SHRT_MAX |
+32767 |
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USHRT_MAX |
65535 |
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INT_MIN |
-2147483648 |
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INT_MAX |
+2147483647 |
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UINT_MAX |
4294967295 |
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LONG_MIN |
-9223372036854775808 |
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LONG_MAX |
+9223372036854775807 |
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ULONG_MAX |
18446744073709551615 |
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How to print the value of limits.h |
int main() { printf("The number of bits in a byte %d\n", CHAR_BIT); } |
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2.59 Write a C expression that will yield a word consisting of the least significant byte of x, and the remaining bytes of y. For operands x = 0x89ABCDEF and y = 0x76543210, this would give 0x765432EF |
int prob259(int x, int y)
{ return (x & 0xFF) | (y & ~ 0xFF); } |
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2.60 Suppose we number the bytes in a w-bit word from 0 (least significant) to w/8 − 1(most significant). Write code for the following C function, which will return anunsigned value in which byte i of argument x has been replaced by byte b: unsigned replace_byte (unsigned x, int i, unsigned char b); Here are some examples showing how the function should work:replace_byte(0x12345678, 2, 0xAB) --> 0x12AB5678replace_byte(0x12345678, 0, 0xAB) --> 0x123456AB |
unsigned replace_byte(unsigned x, int i, unsigned char b)
{ return ( x & ~(0xFF << (i << 3))) | (b << (i << 3)); } |
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2.61
Write C expressions that evaluate to 1 when the following conditions are true, andto 0 when they are false. Assume x is of type int. A. Any bit of x equals 1. B. Any bit of x equals 0. C. Any bit in the least significant byte of x equals 1. D. Any bit in the most significant byte of x equals 0. Your code should follow the bit-level integer coding rules (page 120), with theadditional restriction that you may not use equality (==) or inequality (!=) tests. |
int prob261A(int x)
{ return !!x ; }
int prob261B(int x) { return !x; }
int prob261C(int x) { return !!(x & 0xFF); } |
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2.68
Write code for a function with the following prototype: /** Mask with least signficant n bits set to 1 * Examples: n = 6 --> 0x2F, n = 17 --> 0x1FFFF * Assume 1 <= n <= w*/ int lower_one_mask(int n); Your function should follow the bit-level integer coding rules (page 120). Becareful of the case n = w. |
int lower_one_mask(int n){
return (2<<(n-1))-1; } |
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* bitAnd - x&y using only ~ and |
* Example: bitAnd(6, 5) = 4 * Legal ops: ~ | * Max ops: 8 * Rating: 1 |
int bitAnd(int x, int y) {
return ~(~x | ~y); } |
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bitNor - ~(x|y) using only ~ and & * Example: bitNor(0x6, 0x5) = 0xFFFFFFF8 * Legal ops: ~ & * Max ops: 8 * Rating: 1 |
int bitNor ( int x, int y) { return (~x & ~y); } |
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* upperBits - pads n upper bits with 1's
* You may assume 0 <= n <= 32 * Example: upperBits(4) = 0xF0000000 * Legal ops: ! ~ & ^ | + << >> * Max ops: 10 * Rating: 1 |
int upperBits ( int n) { int mask = (!!n) << 31 >> 31; return mask & (( 1 << 31) >> (n +~0)); |
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* allEvenBits - return 1 if all even-numbered bits in word set to 1
* Examples allEvenBits(0xFFFFFFFE) = 0, allEvenBits(0x55555555) = 1 * Legal ops: ! ~ & ^ | + << >> * Max ops: 12 * Rating: 2 |
int allEvenBits (int x) { int mask = 0x55; mask |= mask <<8; mask |= mask <<16; int y = x | mask; return !(x^y); |
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* byteSwap - swaps the nth byte and the mth byte * Examples: byteSwap(0x12345678, 1, 3) = 0x56341278 * byteSwap(0xDEADBEEF, 0, 2) = 0xDEEFBEAD * You may assume that 0 <= n <= 3, 0 <= m <= 3 * Legal ops: ! ~ & ^ | + << >> * Max ops: 25 * Rating: 2 |
byteSwap(int x, int n, int m) { int mask, xx, nn,mm, nx, mx; nn = n << 3; mm = m << 3; nx = x >> nn; mx = x >> mm; xx = ((mx&0xFF) << nn)|((nx&0xFF)< <mm); mask = (0xFF << nn) | (0xFF << mm); return (x & ~mask) | xx; } |
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logicalShift - shift x to the right by n, using a logical shift
* Can assume that 0 <= n <= 31 * Examples: logicalShift(0x87654321,4) = 0x08765432 * Legal ops: ! ~ & ^ | + << >> * Max ops: 20 * Rating: 3 |
int logicalShift ( int x, int n) { int mask ((1<<31)>> n) <<1; return (x >> n ) & ~ mask; |
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* bitParity - returns 1 if x contains an odd number of 0's * Examples: bitParity(5) = 0, bitParity(7) = 1 * Legal ops: ! ~ & ^ | + << >> * Max ops: 20 * Rating: 4 |
int bitParity(int x) { x ^= x >> 16; x ^= x >> 8; x ^= x >> 4; x ^= x >> 2; x ^= x >> 1; return x & 1; } |
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* tmin - return minimum two's complement integer
* Legal ops: ! ~ & ^ | + << >> * Max ops: 4 * Rating: 1 |
int tmin (void) { return 1 << 31; } |
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* isNegative - return 1 if x < 0, return 0 otherwise * Example: isNegative(-1) = 1.
* Legal ops: ! ~ & ^ | + << >> * Max ops: 6 * Rating: 2 |
int isNegative (int x) { return (x >> 31) & 0x1; } |
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* sign - return 1 if positive, 0 if zero, and -1 if negative
* Examples: sign(130) = 1 * sign(-23) = -1 * Legal ops: ! ~ & ^ | + << >> * Max ops: 10 * Rating: 2 |
int sign (int x) { return (x>> 31) | !!x; } |
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* isEqual - return 1 if x == y, and 0 otherwise * Examples: isEqual(5,5) = 1, isEqual(4,5) = 0 * Legal ops: ! ~ & ^ | + << >> * Max ops: 5 * Rating: 2 |
int isEqual (int x, int y) { return !(x^y); } |
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* 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 divpwer2 (int x, int n) { int mask = (1 << n) + ~0; int bias = (x >> 31) & mask; return (x+bias) >> n; } |
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* 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) { int sgnx = (x>>31)&0x1; int sgny = (y>>31)&0x1; int d = x + (~y+0x1); int sgnd = (d>>31)&0x1; return !(x^y) | (sgnx&!sgny) | (!(sgnx^sgny)&sgnd); } |
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* 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) { int mask = !x + ~0; return (y & mask) | (z & ~mask); |
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* addOK - Determine if can compute x+y without overflow
* Example: addOK(0x80000000,0x80000000) = 0, * addOK(0x80000000,0x70000000) = 1, * Legal ops: ! ~ & ^ | + << >> * Max ops: 20 * Rating: 3 |
int addOk ( int x, int y) { int xneg = x >>31; int yneg = y >> 31; int xpyneg = (y + x )>>31; return ! ((~xneg & ~yneg & xpyneg) | (xneg & yneg & ~xpyneg)); |
