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为x位分配n个字节

bit-manipulation c

Yanick Rochon · 技术社区 · 15 年前

首先,这可以是任何语言的通用算法,但是我正在学习C,如果有一些C特定的特性,我想知道!

我正在编写一个函数,它将为给定的位数分配足够的内存;到 long long * 变量。位数不能是 < 1 . 我测试了算法:

int bits;  // the function argument, checked for value > 0
size_t dataSize;  // the value passed to the malloc function

for (bits = 1; bits<100; bits++) {
   if (bits < sizeof(long long)) {
      dataSize = 1;
   } else {
      dataSize = (bits + (sizeof(long long) - (bits % sizeof(long long)))) / sizeof(long long);
   }

   printf("%d = %d\n", bits, (int) dataSize);
}

看起来不错…但丑:) 有什么方法可以更优雅地实现这一点? 谢谢您!

8 回复 | 直到 15 年前

Christoph 15 年前

calloc() malloc()

#include <limits.h>

const size_t BITS_PER_BLOCK = sizeof (long long) * CHAR_BIT;
size_t count = bits / BITS_PER_BLOCK + !!(bits % BITS_PER_BLOCK);
unsigned long long *blocks = calloc(count, sizeof *blocks);

!! 1 BITS_PER_BLOCK

size_t count = (bits + BITS_PER_BLOCK - 1) / BITS_PER_BLOCK;

user470379 15 年前

int bits;
size_t dataSize;

dataSize = bits / (sizeof(long long) * 8);
if( bits % (sizeof(long long) * 8) ) { //Don't add 1 if it was evenly divisible
    dataSize++;
}
dataSize *= sizeof(long long)

long long

Å imon Tóth 15 年前

long long

int bits = n;
int items = (((bits - 1) / CHAR_BIT) / sizeof(long long)) + 1;

MahlerFive 15 年前

nategoose 15 年前

size_t len_needed(int bits) {
   size_t x=  bits/(sizeof(long long) * 8);
   x = (bits%(sizeof(long long) * 8) ? 1 : 0;

   return x;
}

? :

pmg 15 年前

int bits;  // the function argument, checked for value > 0
size_t dataSize;  // the value passed to the malloc function

for (bits = 1; bits<100; bits++) {
   if (bits < sizeof(long long)) {
      dataSize = 1;
   } else {
      dataSize = (bits + (sizeof(long long) - (bits % sizeof(long long)))) / sizeof(long long);
   }

   printf("%d = %d (%d)\n", bits, (int) dataSize, 1 + bits/sizeof (long long));
   /*             ^^^^^                         ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ */
}

MSN 15 年前

Number_of_long_longs_for_x= (x + sizeof(long long) - 1)/sizeof(long long)

long long log2(ULLONG_MAX+1) sizeof(long long)

alvin 15 年前

#define SIZEOF_LL  sizeof(long long)
nbytes  =  (xbits  + 8         - 1) / 8;
nllongs =  (nbytes + SIZEOF_LL - 1) / SIZEOF_LL;

nbytes =  (xbits  + 7) / 8;
nllongs = (nbytes + 7) / 8;