我正在编写一个函数来打印c中的位，我只允许使用write函数。我的功能对其他数字不起作用。
void print_bits(unsigned char octet)
{
int oct;
int div;
div = 128;
oct = octet;
while (!(div <= 1))
{
if (div <= oct)
{
write(1, "1", 1);
oct = oct % div;
}
else
{
write(1, "0", 1);
div = div / 2;
}
}
}
答案 0 :(得分：4)
我已经重新编写了Is there a printf converter to print in binary format?
中的代码void print_bits(unsigned char octet)
{
int z = 128, oct = octet;
while (z > 0)
{
if (oct & z)
write(1, "1", 1);
else
write(1, "0", 1);
z >>= 1;
}
}
答案 1 :(得分：1)
包含limits.h CHAR_BIT，允许您概括函数以允许传递任何长度的值，但将输出限制为所需的字节数。传递文件描述符将允许写入任何打开的描述符（或者只是通过STDOUT_FILENO写入stdout。
void writebits (const unsigned long v, int fd)
{
if (!v) { putchar ('0'); return; };
size_t sz = sizeof v * CHAR_BIT;
unsigned long rem = 0;
while (sz--)
if ((rem = v >> sz))
write (fd, (rem & 1) ? "1" : "0", 1);
}
例如：
#include <stdio.h>
#include <unistd.h>
/* CHAR_BIT */
#ifndef CHAR_BIT
# define CHAR_BIT 8
#endif
void writebits (const unsigned long v, int fd)
{
if (!v) { putchar ('0'); return; };
size_t sz = sizeof v * CHAR_BIT;
unsigned long rem = 0;
while (sz--)
if ((rem = v >> sz))
write (fd, (rem & 1) ? "1" : "0", 1);
}
int main (void) {
unsigned v = 0xcafebabe;
writebits (v, STDOUT_FILENO);
putchar ('\n');
writebits ((unsigned char)(v >> 24), STDOUT_FILENO);
putchar ('\n');
return 0;
}
示例输出
$ ./bin/writebits
11001010111111101011101010111110
11001010
答案 2 :(得分：0)
如果你想打印位然后使用按位运算符，你可以这样做......
这样的例子：
for(i=31 ; i>=0 ; i--)
{
if(num & 1<< i) /* num & 1 << position
printf("1 ");
else
printf("0 ");
}
printf("\n");
答案 3 :(得分：0)
在互联网上找到某个地方：
void printBits( void const * const ptr, size_t const size )
{ printf( "hexVal = %#08x; decVal = %i \n", * ( uint* )ptr, * ( uint* )ptr );
unsigned char *b = ( unsigned char* ) ptr;
unsigned char byte;
int i, j;
for (i=size-1;i>=0;i--)
for (j=7;j>=0;j--)
{ byte = (b[i] >> j) & 1;
printf( " %u ", byte );
}
puts("");
for( int ind = 31; ind > 9; ind-- )
printf( "%d ", ind );
for( int ind = 9; ind > -1; ind-- )
printf( " %d ", ind );
puts(""); puts("");
};
void compareValsBits( const void * ptrA, size_t sizeA, const void * ptrB, size_t sizeB )
{ if ( sizeA != sizeB ) return;
printf( "comparing: \n[%#08x](%i)\n[%#08x](%i)\n====\n", * ( uint* )ptrA, * ( uint* )ptrA, * ( uint* )ptrB, * ( uint* )ptrB );
unsigned char *a = ( unsigned char* ) ptrA;
unsigned char *b = ( unsigned char* ) ptrB;
unsigned char byteA, byteB;
int i, j;
for (i=sizeA-1;i>=0;i--)
for (j=7;j>=0;j--)
{ byteA = (a[i] >> j) & 1;
printf( " %u ", byteA );
}
puts("");
for (i=sizeA-1;i>=0;i--)
for (j=7;j>=0;j--)
{ byteB = (b[i] >> j) & 1;
printf( " %u ", byteB );
}
puts("");
for( int ind = 31; ind > 9; ind-- )
printf( "%d ", ind );
for( int ind = 9; ind > -1; ind-- )
printf( " %d ", ind );
puts("");
for (i=sizeA-1;i>=0;i--)
for (j=7;j>=0;j--)
{ byteA = (a[i] >> j) & 1;
byteB = (b[i] >> j) & 1;
if ( byteA == byteB )
printf( " " );
else
printf( " x " );
}
printf( "\n====\n" );
};
一些简单的宏展开设置和重置位： https://stackoverflow.com/questions/1872220/is-it-possible-to-iterate-over-arguments-in-variadic-macros
#include <stdio.h>
typedef __uint32_t uint;
//==============================================================================================
#define STRINGIZE(arg) #arg
#define CONCATENATE(arg1, arg2) arg1##arg2
#define FOR_EACH_1(macroName, param, x, ...) macroName(param, x)
#define FOR_EACH_2(macroName, param, x, ...) macroName(param, x); FOR_EACH_1(macroName, param, __VA_ARGS__);
#define FOR_EACH_3(macroName, param, x, ...) macroName(param, x); FOR_EACH_2(macroName, param, __VA_ARGS__);
#define FOR_EACH_4(macroName, param, x, ...) macroName(param, x); FOR_EACH_3(macroName, param, __VA_ARGS__);
#define FOR_EACH_5(macroName, param, x, ...) macroName(param, x); FOR_EACH_4(macroName, param, __VA_ARGS__);
#define FOR_EACH_6(macroName, param, x, ...) macroName(param, x); FOR_EACH_5(macroName, param, __VA_ARGS__);
#define FOR_EACH_7(macroName, param, x, ...) macroName(param, x); FOR_EACH_6(macroName, param, __VA_ARGS__);
#define FOR_EACH_8(macroName, param, x, ...) macroName(param, x); FOR_EACH_7(macroName, param, __VA_ARGS__);
#define FOR_EACH_NARG(...) FOR_EACH_NARG_(__VA_ARGS__, FOR_EACH_RSEQ_N())
#define FOR_EACH_NARG_(...) FOR_EACH_ARG_N(__VA_ARGS__)
#define FOR_EACH_ARG_N(_1, _2, _3, _4, _5, _6, _7, _8, N, ...) N
#define FOR_EACH_RSEQ_N() 8, 7, 6, 5, 4, 3, 2, 1, 0
#define FOR_EACH_(N, macroName, param, x, ...) CONCATENATE(FOR_EACH_, N)(macroName, param, x, __VA_ARGS__)
#define FOR_EACH(macroName, param, x, ...) FOR_EACH_(FOR_EACH_NARG(x, __VA_ARGS__), macroName, param, x, __VA_ARGS__)
//============================================================================================
//--------------------------------------------------------------------------------------------
//------------------------ MACRO KERNELS ----------------------------------------------------
//--------------------------------------------------------------------------------------------
#define SET_BIT( VARIABLE, bitNo ) { VARIABLE = ( ( ( VARIABLE >> bitNo ) & 0x1u ) != 0x1u ) \
? VARIABLE |= 0x1u << bitNo \
: VARIABLE = VARIABLE; \
}
#define RESET_BIT( VARIABLE, bitNo ) { VARIABLE = ( ( ( VARIABLE >> bitNo ) & 0x1u ) != 0x0u ) \
? VARIABLE &= ~( 0x1u << bitNo ) \
: VARIABLE = VARIABLE; \
}
#define BIT_VAL( VARIABLE, bitNo ) ( VARIABLE >> bitNo ) & 0x1u
//--------------------------------------------------------------------------------------------
//%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
//%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
//%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
uint a = 0x0u;
int main( void )
{ FOR_EACH( SET_BIT, a, 0, 1, 2 );
printf( "0x%.8X\n", a );
FOR_EACH( RESET_BIT, a, 1, 2 );
printf( "0x%.8X\n", a );
return 0;
};//end of main()
//%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
//%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
//%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
答案 4 :(得分：0)
这个解决方案设置了一个自动调整参数类型的填充。 long long 确保大多数数字都可以轻松处理。它有一个静态缓冲区，因此调用者没有要释放的内存。应该是相当通用的，因为它将数字格式化为 char * 以用于 printf 或根据需要放置。只是一个基本的降序 do while 循环，编译器展开应该很简单，因为 printf 可以根据需要格式化其余部分。 Is there a printf converter to print in binary format?
的优化版本inline char* binstr(long long number) {
# define BINSTR_BODY_DEFINITION \
static const char BYTE_BOUNDRY = 8; /* reuse fixed byte boundary to determine padding `*/ \
static const char ZERO = '0'; /* reuse fixed byte 0 */ \
static const char ONE = '1'; /* reuse fixed byte 1 */ \
static const char digits_length = sizeof number * BYTE_BOUNDRY; \
/* reuse predetermined byte digits length */ \
static char src[digits_length + 1]; /* instantiate reusable array */ \
int i = digits_length; /* instantiate temp byte counter */ \
src[i--] = 0x00; /* terminate string */ \
\
do { \
/* fill in array from right to left */ \
src[i--] = number & 1 ? ONE : ZERO; \
number >>= 1; /* shift right 1 bit */ \
} while (number > 0); \
while (i >= 0) src[i--] = ZERO; /* fill with ZERO */ \
return src;
BINSTR_BODY_DEFINITION
}
inline char* binstr(long int number) {
BINSTR_BODY_DEFINITION
}
inline char* binstr(int number) {
BINSTR_BODY_DEFINITION
}
inline char* binstr(short number) {
BINSTR_BODY_DEFINITION
# undef BINSTR_BODY_DEFINITION
}
int main()
{
long int val = 1;
do {
// experimenting with different casts will produce different output lengths
printf("%ld=\t\t%#lx=\t 0b%s\n", val, val, binstr(long long( val )));
val *= 11; // generate test data
} while (val < 100000000);
return 0;
}
打印出以下内容：
1 = 0x1 = 0b0000000000000000000000000000000000000000000000000000000000000001
11 = 0xb = 0b0000000000000000000000000000000000000000000000000000000000001011
121 = 0x79 = 0b0000000000000000000000000000000000000000000000000000000001111001
1331 = 0x533 = 0b0000000000000000000000000000000000000000000000000000010100110011
等等...
答案 5 :(得分：-2)
#include<stdio.h>
int countOne =0;
int countZero =0;
void print_bits(int n,unsigned int num)
{
if(n == 31) return;
print_bits(n+1,num);
(num & (1<<n)) ? printf("1"):printf("0");
}
void countBits(int n,unsigned int num)
{
if(n == 31) return;
countBits(n+1,num);
if(num & (1<<n))
{
countOne++;
}
else
{
if(countOne)
countZero++;
}
}
void printbits(int num)
{
printf("%d : ",num);
print_bits(-1,num);
printf("\n");
countBits(-1,num);
printf("\n");
}
int main()
{
printbits(189);
printf("NumberOf1's[%d] Numberof0's[%d]\n",countOne,countZero);
}