对位的水平读取字符 [英] Reading characters on a bit level

问题描述

我希望能够从键盘输入一个字符，并显示二进制code代表说，例如格式00000001关键。

I would like to be able to enter a character from the keyboard and display the binary code for said key in the format 00000001 for example.

此外，我也想在某种程度上，让我输出，如果他们是真的还是假的读取位。

Furthermore i would also like to read the bits in a way that allows me to output if they are true or false.

例如

01010101 = false,true,false,true,false,true,false,true

我会后如何我试图做我自己的想法，但我完全不知道，我现在还在用C试验，这是我编程的第一口味在如此低的水平刻度。

I would post an idea of how i have tried to do it myself but I have absolutely no idea, i'm still experimenting with C and this is my first taste of programming at such a low level scale.

三江源

推荐答案

这code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;
}

您不一定需要三环路 - 你可以将它们结合起来，如果你想要的，如果你只是做了两种输出之一，那么你就不需要数组，你可以只使用每个位值，你掩盖它。但我认为这使事情变得独立，并希望更容易理解。

You don't necessarily need three loops - you could combine them together if you wanted, and if you're only doing one of the two kinds of output, then you wouldn't need the array, you could just use each bit value as you mask it off. But I think this keeps things separate and hopefully easier to understand.

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