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
至于我,我为此编写了一些通用代码
#include<stdio.h>
void int2bin(int n, int* bin, int* bin_size, const int bits);
int main()
{
char ch;
ch = 'A';
int binary[32];
int binary_size = 0;
int2bin(1324, binary, &binary_size, 32);
for (int i = 0; i < 32; i++)
{
printf("%d ", binary[i]);
}
return 0;
}
void int2bin(int n, int* bin,int *bin_size,const int bits)
{
int i = 0;
int temp[64];
for (int j = 0; j < 64; j++)
{
temp[j] = 0;
}
for (int l = 0; l < bits; l++)
{
bin[l] = 0;
}
while (n > 0)
{
temp[i] = n % 2;
n = n / 2;
i++;
}
*bin_size = i;
//reverse modulus values
for (int k = 0; k < *bin_size; k++)
{
bin[bits-*bin_size+k] = temp[*bin_size - 1 - k];
}
}
其他回答
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。
/* Convert an int to it's binary representation */
char *int2bin(int num, int pad)
{
char *str = malloc(sizeof(char) * (pad+1));
if (str) {
str[pad]='\0';
while (--pad>=0) {
str[pad] = num & 1 ? '1' : '0';
num >>= 1;
}
} else {
return "";
}
return str;
}
/* example usage */
printf("The number 5 in binary is %s", int2bin(5, 4));
/* "The number 5 in binary is 0101" */
使用以下功能:
void conbin(int num){
if(num != 0)
{
conbin(num >> 1);
if (num & 1){
printf("1");
}
else{
printf("0");
}
}
}
我只是想把我的解贴出来。它用于获取一个字节的0和1,但多次调用此函数可以用于更大的数据块。我将它用于128位或更大的结构。还可以修改它,使用size_t作为输入参数和指向想要打印的数据的指针,因此它可以与大小无关。但这对我很管用。
void print_binary(unsigned char c)
{
unsigned char i1 = (1 << (sizeof(c)*8-1));
for(; i1; i1 >>= 1)
printf("%d",(c&i1)!=0);
}
void get_binary(unsigned char c, unsigned char bin[])
{
unsigned char i1 = (1 << (sizeof(c)*8-1)), i2=0;
for(; i1; i1>>=1, i2++)
bin[i2] = ((c&i1)!=0);
}
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;
}