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
这段代码可以处理64位的需求。 我创建了两个函数:pBin和pBinFill。两者都做同样的事情,但是pBinFill用最后一个参数提供的填充字符填充前导空格。 测试函数生成一些测试数据,然后使用pBinFill函数将其打印出来。
#define kDisplayWidth 64
char* pBin(long int x,char *so)
{
char s[kDisplayWidth+1];
int i = kDisplayWidth;
s[i--] = 0x00; // terminate string
do { // fill in array from right to left
s[i--] = (x & 1) ? '1' : '0'; // determine bit
x >>= 1; // shift right 1 bit
} while (x > 0);
i++; // point to last valid character
sprintf(so, "%s", s+i); // stick it in the temp string string
return so;
}
char* pBinFill(long int x, char *so, char fillChar)
{
// fill in array from right to left
char s[kDisplayWidth+1];
int i = kDisplayWidth;
s[i--] = 0x00; // terminate string
do { // fill in array from right to left
s[i--] = (x & 1) ? '1' : '0';
x >>= 1; // shift right 1 bit
} while (x > 0);
while (i >= 0) s[i--] = fillChar; // fill with fillChar
sprintf(so, "%s", s);
return so;
}
void test()
{
char so[kDisplayWidth+1]; // working buffer for pBin
long int val = 1;
do {
printf("%ld =\t\t%#lx =\t\t0b%s\n", val, val, pBinFill(val, so, '0'));
val *= 11; // generate test data
} while (val < 100000000);
}
输出:
00000001 = 0x000001 = 0b00000000000000000000000000000001
00000011 = 0x00000b = 0b00000000000000000000000000001011
00000121 = 0x000079 = 0b00000000000000000000000001111001
00001331 = 0x000533 = 0b00000000000000000000010100110011
00014641 = 0x003931 = 0b00000000000000000011100100110001
00161051 = 0x02751b = 0b00000000000000100111010100011011
01771561 = 0x1b0829 = 0b00000000000110110000100000101001
19487171 = 0x12959c3 = 0b00000001001010010101100111000011
其他回答
这里有一个快速的技巧来演示如何做你想做的事情。
#include <stdio.h> /* printf */
#include <string.h> /* strcat */
#include <stdlib.h> /* strtol */
const char *byte_to_binary
(
int x
)
{
static char b[9];
b[0] = '\0';
int z;
for (z = 128; z > 0; z >>= 1)
{
strcat(b, ((x & z) == z) ? "1" : "0");
}
return b;
}
int main
(
void
)
{
{
/* binary string to int */
char *tmp;
char *b = "0101";
printf("%d\n", strtol(b, &tmp, 2));
}
{
/* byte to binary string */
printf("%s\n", byte_to_binary(5));
}
return 0;
}
C标准库中没有这样输出二进制的格式化函数。printf家族支持的所有格式操作都是针对人类可读的文本。
你可以使用一个小表格来提高速度。类似的技术在嵌入式世界中也很有用,例如,反转一个字节:
const char *bit_rep[16] = {
[ 0] = "0000", [ 1] = "0001", [ 2] = "0010", [ 3] = "0011",
[ 4] = "0100", [ 5] = "0101", [ 6] = "0110", [ 7] = "0111",
[ 8] = "1000", [ 9] = "1001", [10] = "1010", [11] = "1011",
[12] = "1100", [13] = "1101", [14] = "1110", [15] = "1111",
};
void print_byte(uint8_t byte)
{
printf("%s%s", bit_rep[byte >> 4], bit_rep[byte & 0x0F]);
}
1我主要指的是嵌入式应用程序,其中优化器不是那么激进,速度差异是可见的。
/* 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" */
A small utility function in C to do this while solving a bit manipulation problem. This goes over the string checking each set bit using a mask (1< void printStringAsBinary(char * input) { char * temp = input; int i = 7, j =0;; int inputLen = strlen(input); /* Go over the string, check first bit..bit by bit and print 1 or 0 **/ for (j = 0; j < inputLen; j++) { printf("\n"); while (i>=0) { if (*temp & (1 << i)) { printf("1"); } else { printf("0"); } i--; } temp = temp+1; i = 7; printf("\n"); } }