c - 显示字符串中每个字符的二进制

Question

我已经弄清楚如何获取用户的输入(字符串)并从中获取每个字母并给出每个字母的二进制等价物。

问题是我希望每个字母在屏幕上显示时每行一个字母来给出一个二进制数。我希望得到帮助。

例如:

C 0 1 0 0 0 0 1 1
h 0 1 1 0 1 0 0 0
a 0 1 1 0 0 0 0 1
n 0 1 1 0 1 1 1 0
g 0 1 1 0 0 1 1 1

这是我使用的代码:

#include <stdio.h>

int main()
{

    //get input
    printf( "Enter a string: " );
    char s[255];
    scanf( "%[^\n]s" , s );

    //print output 
    printf( "'%s' converted to binary is: " , s );

    //for each character, print it's binary aoutput
    int i,c,power;

    for( i=0 ; s[i]!='\0' ; i++ )
    {
        //c holds the character being converted
        c = s[i];

        //for each binary value below 256 (because ascii values < 256)
        for( power=7 ; power+1 ; power-- )
        //if c is greater than or equal to it, it is a 1
        if( c >= (1<<power) )
        {
            c -= (1<<power); //subtract that binary value
            printf("1");
        }
        //otherwise, it is a zero
        else
            printf("0");
    } 

    return 0;
}

Answer 1

您只需在处理每个字符后添加一个 printf("\n") 语句即可。

for( i=0 ; s[i]!='\0' ; i++ )
    {
        //c holds the character being converted
        c = s[i];

        //for each binary value below 256 (because ascii values < 256)
        for( power=7 ; power+1 ; power-- )
        //if c is greater than or equal to it, it is a 1
        if( c >= (1<<power) )
        {
            c -= (1<<power); //subtract that binary value
            printf("1");
        }
        //otherwise, it is a zero
        else
            printf("0");

        /* Add the following statement, for this to work as you expected */
        printf("\n");
    }