我可以用printf打印十六进制或八进制数字。是否有格式标签打印为二进制或任意基数?
我正在运行gcc。
printf("%d %x %o\n", 10, 10, 10); //prints "10 A 12\n"
printf("%b\n", 10); // prints "%b\n"
我可以用printf打印十六进制或八进制数字。是否有格式标签打印为二进制或任意基数?
我正在运行gcc。
printf("%d %x %o\n", 10, 10, 10); //prints "10 A 12\n"
printf("%b\n", 10); // prints "%b\n"
当前回答
之前发布的答案都不是我想要的,所以我写了一个。在printf中使用%B是超级简单的!
/*
* File: main.c
* Author: Techplex.Engineer
*
* Created on February 14, 2012, 9:16 PM
*/
#include <stdio.h>
#include <stdlib.h>
#include <printf.h>
#include <math.h>
#include <string.h>
static int printf_arginfo_M(const struct printf_info *info, size_t n, int *argtypes)
{
/* "%M" always takes one argument, a pointer to uint8_t[6]. */
if (n > 0) {
argtypes[0] = PA_POINTER;
}
return 1;
}
static int printf_output_M(FILE *stream, const struct printf_info *info, const void *const *args)
{
int value = 0;
int len;
value = *(int **) (args[0]);
// Beginning of my code ------------------------------------------------------------
char buffer [50] = ""; // Is this bad?
char buffer2 [50] = ""; // Is this bad?
int bits = info->width;
if (bits <= 0)
bits = 8; // Default to 8 bits
int mask = pow(2, bits - 1);
while (mask > 0) {
sprintf(buffer, "%s", ((value & mask) > 0 ? "1" : "0"));
strcat(buffer2, buffer);
mask >>= 1;
}
strcat(buffer2, "\n");
// End of my code --------------------------------------------------------------
len = fprintf(stream, "%s", buffer2);
return len;
}
int main(int argc, char** argv)
{
register_printf_specifier('B', printf_output_M, printf_arginfo_M);
printf("%4B\n", 65);
return EXIT_SUCCESS;
}
其他回答
快速简单的解决方法:
void printbits(my_integer_type x)
{
for(int i=sizeof(x)<<3; i; i--)
putchar('0'+((x>>(i-1))&1));
}
适用于任何大小类型以及有符号整型和无符号整型。'&1'需要处理有符号整型,因为移位可能会进行符号扩展。
有很多方法可以做到这一点。这里有一个超级简单的方法,用于从有符号或无符号32位类型中打印32位或n位(如果有符号,则不输入负号,只打印实际的位),并且不返回回车符。注意,i在移位前递减:
#define printbits_n(x,n) for (int i=n;i;i--,putchar('0'|(x>>i)&1))
#define printbits_32(x) printbits_n(x,32)
如果返回一个包含稍后存储或打印的比特的字符串呢?你可以分配内存并返回它,用户必须释放它,或者你返回一个静态字符串,但如果它再次被调用,或者被另一个线程调用,它会被破坏。两种方法显示:
char *int_to_bitstring_alloc(int x, int count)
{
count = count<1 ? sizeof(x)*8 : count;
char *pstr = malloc(count+1);
for(int i = 0; i<count; i++)
pstr[i] = '0' | ((x>>(count-1-i))&1);
pstr[count]=0;
return pstr;
}
#define BITSIZEOF(x) (sizeof(x)*8)
char *int_to_bitstring_static(int x, int count)
{
static char bitbuf[BITSIZEOF(x)+1];
count = (count<1 || count>BITSIZEOF(x)) ? BITSIZEOF(x) : count;
for(int i = 0; i<count; i++)
bitbuf[i] = '0' | ((x>>(count-1-i))&1);
bitbuf[count]=0;
return bitbuf;
}
电话:
// memory allocated string returned which needs to be freed
char *pstr = int_to_bitstring_alloc(0x97e50ae6, 17);
printf("bits = 0b%s\n", pstr);
free(pstr);
// no free needed but you need to copy the string to save it somewhere else
char *pstr2 = int_to_bitstring_static(0x97e50ae6, 17);
printf("bits = 0b%s\n", pstr2);
我的解决方案:
long unsigned int i;
for(i = 0u; i < sizeof(integer) * CHAR_BIT; i++) {
if(integer & LONG_MIN)
printf("1");
else
printf("0");
integer <<= 1;
}
printf("\n");
void print_ulong_bin(const unsigned long * const var, int bits) {
int i;
#if defined(__LP64__) || defined(_LP64)
if( (bits > 64) || (bits <= 0) )
#else
if( (bits > 32) || (bits <= 0) )
#endif
return;
for(i = 0; i < bits; i++) {
printf("%lu", (*var >> (bits - 1 - i)) & 0x01);
}
}
应该工作-未经测试。
之前发布的答案都不是我想要的,所以我写了一个。在printf中使用%B是超级简单的!
/*
* File: main.c
* Author: Techplex.Engineer
*
* Created on February 14, 2012, 9:16 PM
*/
#include <stdio.h>
#include <stdlib.h>
#include <printf.h>
#include <math.h>
#include <string.h>
static int printf_arginfo_M(const struct printf_info *info, size_t n, int *argtypes)
{
/* "%M" always takes one argument, a pointer to uint8_t[6]. */
if (n > 0) {
argtypes[0] = PA_POINTER;
}
return 1;
}
static int printf_output_M(FILE *stream, const struct printf_info *info, const void *const *args)
{
int value = 0;
int len;
value = *(int **) (args[0]);
// Beginning of my code ------------------------------------------------------------
char buffer [50] = ""; // Is this bad?
char buffer2 [50] = ""; // Is this bad?
int bits = info->width;
if (bits <= 0)
bits = 8; // Default to 8 bits
int mask = pow(2, bits - 1);
while (mask > 0) {
sprintf(buffer, "%s", ((value & mask) > 0 ? "1" : "0"));
strcat(buffer2, buffer);
mask >>= 1;
}
strcat(buffer2, "\n");
// End of my code --------------------------------------------------------------
len = fprintf(stream, "%s", buffer2);
return len;
}
int main(int argc, char** argv)
{
register_printf_specifier('B', printf_output_M, printf_arginfo_M);
printf("%4B\n", 65);
return EXIT_SUCCESS;
}
void PrintBinary( int Value, int Places, char* TargetString)
{
int Mask;
Mask = 1 << Places;
while( Places--) {
Mask >>= 1; /* Preshift, because we did one too many above */
*TargetString++ = (Value & Mask)?'1':'0';
}
*TargetString = 0; /* Null terminator for C string */
}
调用函数“拥有”字符串…:
char BinaryString[17];
...
PrintBinary( Value, 16, BinaryString);
printf( "yadda yadda %s yadda...\n", BinaryString);
取决于您的CPU, PrintBinary中的大多数操作呈现给一个或很少几个机器指令。