显示在C号的二进制重新presentation？ [英] Display the binary representation of a number in C?

问题描述

可能重复：结果
  <一href=\"http://stackoverflow.com/questions/111928/is-there-a-printf-converter-to-print-in-binary-format\">Is还有一个printf转换器的二进制格式打印？

不过学习C，我想知道：

Still learning C and I was wondering:

，是有可能做类似下面的？

Given a number, is it possible to do something like the following?

char a = 5;
printf("binary representation of a = %b",a);
> 101

或将我写我自己的方法做转化为二进制？

Or would i have to write my own method to do the transformation to binary?

推荐答案

是（写你自己的），像下面这样功能齐全。

Yes (write your own), something like the following complete function.

#include <stdio.h> /* only needed for the printf() in main(). */
#include <string.h>

/* Create a string of binary digits based on the input value.
   Input:
       val:  value to convert.
       buff: buffer to write to must be >= sz+1 chars.
       sz:   size of buffer.
   Returns address of string or NULL if not enough space provided.
*/
static char *binrep (unsigned int val, char *buff, int sz) {
    char *pbuff = buff;

    /* Must be able to store one character at least. */
    if (sz < 1) return NULL;

    /* Special case for zero to ensure some output. */
    if (val == 0) {
        *pbuff++ = '0';
        *pbuff = '\0';
        return buff;
    }

    /* Work from the end of the buffer back. */
    pbuff += sz;
    *pbuff-- = '\0';

    /* For each bit (going backwards) store character. */
    while (val != 0) {
        if (sz-- == 0) return NULL;
        *pbuff-- = ((val & 1) == 1) ? '1' : '0';

        /* Get next bit. */
        val >>= 1;
    }
    return pbuff+1;
}

这主要添加到它的结束，看看它在运行：

Add this main to the end of it to see it in operation:

#define SZ 32
int main(int argc, char *argv[]) {
    int i;
    int n;
    char buff[SZ+1];

    /* Process all arguments, outputting their binary. */
    for (i = 1; i < argc; i++) {
        n = atoi (argv[i]);
        printf("[%3d] %9d -> %s (from '%s')\n", i, n,
            binrep(n,buff,SZ), argv[i]);
    }

    return 0;
}

与运行它的progname 0 7 12 52 123来获得：

[  1]         0 -> 0 (from '0')
[  2]         7 -> 111 (from '7')
[  3]        12 -> 1100 (from '12')
[  4]        52 -> 110100 (from '52')
[  5]       123 -> 1111011 (from '123')

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