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
接下来将向您展示内存布局:
#include <limits>
#include <iostream>
#include <string>
using namespace std;
template<class T> string binary_text(T dec, string byte_separator = " ") {
char* pch = (char*)&dec;
string res;
for (int i = 0; i < sizeof(T); i++) {
for (int j = 1; j < 8; j++) {
res.append(pch[i] & 1 ? "1" : "0");
pch[i] /= 2;
}
res.append(byte_separator);
}
return res;
}
int main() {
cout << binary_text(5) << endl;
cout << binary_text(.1) << endl;
return 0;
}
其他回答
void
print_binary(unsigned int n)
{
unsigned int mask = 0;
/* this grotesque hack creates a bit pattern 1000... */
/* regardless of the size of an unsigned int */
mask = ~mask ^ (~mask >> 1);
for(; mask != 0; mask >>= 1) {
putchar((n & mask) ? '1' : '0');
}
}
c
// Based on https://stackoverflow.com/a/112956/1438550
#include <stdio.h>
#include <stdint.h>
const char *int_to_binary_str(int x, int N_bits){
static char b[512];
char *p = b;
b[0] = '\0';
for(int i=(N_bits-1); i>=0; i--){
*p++ = (x & (1<<i)) ? '1' : '0';
if(!(i%4)) *p++ = ' ';
}
return b;
}
int main() {
for(int i=31; i>=0; i--){
printf("0x%08X %s \n", (1<<i), int_to_binary_str((1<<i), 32));
}
return 0;
}
期望的行为:
Run:
gcc -pthread -Wformat=0 -lm -o main main.c; ./main
Output:
0x80000000 1000 0000 0000 0000 0000 0000 0000 0000
0x40000000 0100 0000 0000 0000 0000 0000 0000 0000
0x20000000 0010 0000 0000 0000 0000 0000 0000 0000
0x10000000 0001 0000 0000 0000 0000 0000 0000 0000
0x08000000 0000 1000 0000 0000 0000 0000 0000 0000
0x04000000 0000 0100 0000 0000 0000 0000 0000 0000
0x02000000 0000 0010 0000 0000 0000 0000 0000 0000
0x01000000 0000 0001 0000 0000 0000 0000 0000 0000
0x00800000 0000 0000 1000 0000 0000 0000 0000 0000
0x00400000 0000 0000 0100 0000 0000 0000 0000 0000
0x00200000 0000 0000 0010 0000 0000 0000 0000 0000
0x00100000 0000 0000 0001 0000 0000 0000 0000 0000
0x00080000 0000 0000 0000 1000 0000 0000 0000 0000
0x00040000 0000 0000 0000 0100 0000 0000 0000 0000
0x00020000 0000 0000 0000 0010 0000 0000 0000 0000
0x00010000 0000 0000 0000 0001 0000 0000 0000 0000
0x00008000 0000 0000 0000 0000 1000 0000 0000 0000
0x00004000 0000 0000 0000 0000 0100 0000 0000 0000
0x00002000 0000 0000 0000 0000 0010 0000 0000 0000
0x00001000 0000 0000 0000 0000 0001 0000 0000 0000
0x00000800 0000 0000 0000 0000 0000 1000 0000 0000
0x00000400 0000 0000 0000 0000 0000 0100 0000 0000
0x00000200 0000 0000 0000 0000 0000 0010 0000 0000
0x00000100 0000 0000 0000 0000 0000 0001 0000 0000
0x00000080 0000 0000 0000 0000 0000 0000 1000 0000
0x00000040 0000 0000 0000 0000 0000 0000 0100 0000
0x00000020 0000 0000 0000 0000 0000 0000 0010 0000
0x00000010 0000 0000 0000 0000 0000 0000 0001 0000
0x00000008 0000 0000 0000 0000 0000 0000 0000 1000
0x00000004 0000 0000 0000 0000 0000 0000 0000 0100
0x00000002 0000 0000 0000 0000 0000 0000 0000 0010
0x00000001 0000 0000 0000 0000 0000 0000 0000 0001
接下来将向您展示内存布局:
#include <limits>
#include <iostream>
#include <string>
using namespace std;
template<class T> string binary_text(T dec, string byte_separator = " ") {
char* pch = (char*)&dec;
string res;
for (int i = 0; i < sizeof(T); i++) {
for (int j = 1; j < 8; j++) {
res.append(pch[i] & 1 ? "1" : "0");
pch[i] /= 2;
}
res.append(byte_separator);
}
return res;
}
int main() {
cout << binary_text(5) << endl;
cout << binary_text(.1) << endl;
return 0;
}
const char* byte_to_binary(int x)
{
static char b[sizeof(int)*8+1] = {0};
int y;
long long z;
for (z = 1LL<<sizeof(int)*8-1, y = 0; z > 0; z >>= 1, y++) {
b[y] = (((x & z) == z) ? '1' : '0');
}
b[y] = 0;
return b;
}
我喜欢的代码由paniq，静态缓冲区是一个好主意。但是，如果你想在一个printf()中有多个二进制格式，它就失败了，因为它总是返回相同的指针并覆盖数组。
下面是一个C风格的下拉列表，它可以在分割缓冲区上旋转指针。
char *
format_binary(unsigned int x)
{
#define MAXLEN 8 // width of output format
#define MAXCNT 4 // count per printf statement
static char fmtbuf[(MAXLEN+1)*MAXCNT];
static int count = 0;
char *b;
count = count % MAXCNT + 1;
b = &fmtbuf[(MAXLEN+1)*count];
b[MAXLEN] = '\0';
for (int z = 0; z < MAXLEN; z++) { b[MAXLEN-1-z] = ((x>>z) & 0x1) ? '1' : '0'; }
return b;
}
