在位级别上读取字符

此代码是C89:

/* we need this to use exit */
#include <stdlib.h>
/* we need this to use CHAR_BIT */
#include <limits.h>
/* we need this to use fgetc and printf */
#include <stdio.h>

int main() {
    /* Declare everything we need */
    int input, index;
    unsigned int mask;
    char inputchar;

    /* an array to store integers telling us the values of the individual bits.
       There are (almost) always 8 bits in a char, but it doesn't hurt to get into
       good habits early, and in C, the sizes of the basic types are different
       on different platforms. CHAR_BIT tells us the number of bits in a byte. 
     */
    int bits[CHAR_BIT];

    /* the simplest way to read a single character is fgetc, but note that
       the user will probably have to press "return", since input is generally 
       buffered */
    input = fgetc(stdin);
    printf("%d\n", input);

    /* Check for errors. In C, we must always check for errors */
    if (input == EOF) {
        printf("No character read\n");
        exit(1);
    }

    /* convert the value read from type int to type char. Not strictly needed,
       we can examine the bits of an int or a char, but here's how it's done. 
     */
    inputchar = input;

    /* the most common way to examine individual bits in a value is to use a 
       "mask" - in this case we have just 1 bit set, the most significant bit
       of a char. */
    mask = 1 << (CHAR_BIT - 1);

    /* this is a loop, index takes each value from 0 to CHAR_BIT-1 in turn,
       and we will read the bits from most significant to least significant. */
    for (index = 0; index < CHAR_BIT; ++index) {
        /* the bitwise-and operator & is how we use the mask.
           "inputchar & mask" will be 0 if the bit corresponding to the mask
           is 0, and non-zero if the bit is 1. ?: is the ternary conditional
           operator, and in C when you use an integer value in a boolean context,
           non-zero values are true. So we're converting any non-zero value to 1. 
         */
        bits[index] = (inputchar & mask) ? 1 : 0;

        /* output what we've done */
        printf("index %d, value %u\n", index, inputchar & mask);

        /* we need a new mask for the next bit */
        mask = mask >> 1;
    }

    /* output each bit as 0 or 1 */
    for (index = 0; index < CHAR_BIT; ++index) {
        printf("%d", bits[index]);
    }
    printf("\n");

    /* output each bit as "true" or "false" */
    for (index = 0; index < CHAR_BIT; ++index) {
        printf(bits[index] ? "true" : "false");
        /* fiddly part - we want a comma between each bit, but not at the end */
        if (index != CHAR_BIT - 1) printf(",");
    }
    printf("\n");
    return 0;
}

您不一定需要三个循环—如果需要,可以将它们组合在一起,如果只进行这两种输出中的一种,那么就不需要数组,只需在屏蔽时使用每个位值。但我认为这会使事情分开,希望更容易理解。

对于位调整,使用无符号类型通常更安全,因为有符号负值的移动具有依赖于实现的效果。平原 char 可以是签名的,也可以是未签名的(传统上,它在Macintosh平台上是未签名的,但在PC上是签名的)。因此,首先将你的角色投射到 unsigned char 类型。

那么,你的朋友就是位布尔运算符( & , | , ^ 和 ~ )轮班操作员( << 和 >> )例如,如果您的角色在变量中 x ,然后要得到第5位,只需使用: ((x >> 5) & 1) . 移位运算符将值向右移动,将五个较低的位丢弃,并将您感兴趣的位移动到“最低位置”(也称为“最右边”)。按位与1简单地将所有其他位设置为0,因此结果值为0或1,即您的位。请注意,我从左有效(最右边)到最有效(最左边)对位进行编号,并且以零开始,而不是一。

如果您假设您的字符是8位,则可以将代码编写为:

unsigned char x = (unsigned char)your_character;
int i;

for (i = 7; i >= 0; i --) {
    if (i != 7)
        printf(",");
    printf("%s", ((x >> i) & 1) ? "true" : "false");
}

您可能会注意到,由于我从右到左对位进行编号,但您希望从左到右输出,所以循环索引必须减少。

注意,根据C标准, 无符号字符 有 至少 8位,但可能有更多(现在,只有少数嵌入式DSP的字符不是8位)。为了更加安全,在代码的开头添加这个(作为顶级声明):

#include <limits.h>
#if CHAR_BIT != 8
#error I need 8-bit bytes!
#endif

如果目标系统恰好是那些特殊的嵌入式DSP之一,这将阻止成功编译。作为注释上的注释,C标准中的“字节”一词表示“与 无符号字符 “,这样,在C语言中,一个字节可能有八个以上的位(一个字节并不总是八位字节)。这是一个传统的困惑源。

这可能不是最安全的方法-没有健全/大小/类型检查-但它应该仍然有效。

unsigned char myBools[8];
char myChar;  

// get your character - this is not safe and you should
// use a better method to obtain input...
// cin >> myChar; <- C++
scanf("%c", &myChar);

// binary AND against each bit in the char and then
// cast the result. anything > 0 should resolve to 'true'
// and == 0 to 'false', but you could add a '> 1' check to be sure.
for(int i = 0; i < 8; ++i)
{
   myBools[i] = ( (myChar & (1 << i) > 0) ? 1 : 0 );
}

这将为字符提供一个无符号字符数组(0或1(true或false))。