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

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;
}
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。

void DisplayBinary(unsigned int n)
{
    int l = sizeof(n) * 8;
    for (int i = l - 1 ; i >= 0; i--) {
        printf("%x", (n & (1 << i)) >> i);
    }
}

这可能不是很有效,但很简单。试试这个:

tmp1 = 1;
while(inint/tmp1 > 1) {
    tmp1 <<= 1;
}
do {
    printf("%d", tmp2=inint/tmp1);
    inint -= tmp1*tmp2;
} while((tmp1 >>= 1) > 0);
printf(" ");

使用以下功能:

void conbin(int num){  
        if(num != 0)
        {
            conbin(num >> 1);     
            if (num & 1){
            printf("1");
            }
            else{
            printf("0");
            }
        }
    }