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Question

我试图在C中将字符串转换为二进制。此函数必须返回一个字符串（char *），如“010010101”等。此外，我想打印返回的内容。我无法确定这段代码

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char* stringToBinary(char* s)
{
    if(s == NULL) return 0; /* no input string */
    char *binary = malloc(sizeof(s)*8);
    strcpy(binary,"");
    char *ptr = s;
    int i;

    for(; *ptr != 0; ++ptr)
    {

        /* perform bitwise AND for every bit of the character */
        for(i = 7; i >= 0; --i){
            (*ptr & 1 << i) ? strcat(binary,"1") : strcat(binary,"0");
        }


    }

    return binary;
}

Answer 1

对输入没有任何假设，只需打印字节中的位：

-m(no)fused-madd

Answer 2

你的代码似乎很好。你只是错误的金额。这里有更正：

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

char* stringToBinary(char* s) {
    if(s == NULL) return 0; /* no input string */
    size_t len = strlen(s);
    char *binary = malloc(len*8 + 1); // each char is one byte (8 bits) and + 1 at the end for null terminator
    binary[0] = '\0';
    for(size_t i = 0; i < len; ++i) {
        char ch = s[i];
        for(int j = 7; j >= 0; --j){
            if(ch & (1 << j)) {
                strcat(binary,"1");
            } else {
                strcat(binary,"0");
            }
        }
    }
    return binary;
}

样品运行：

"asdf"           => 01100001011100110110010001100110
"tester"         => 011101000110010101110011011101000110010101110010
"Happy New Year" => 0100100001100001011100000111000001111001001000000100111001100101011101110010000001011001011001010110000101110010

Answer 3

我不编译，但我希望这会激励你所以：一个字符是8位。 XXXXXXXX 使用掩码比较每个位更容易，更快 XXXXXXXX ＆安培; 00000001 只有当第1位为1时才是真的。接下来会的 XXXXXXXX ＆安培; 00000010

char * stringToBinary( char * s )
{   
    // Variables.
    char * aux, *binary, mask;
    int size, i, y;

    if(s == NULL){
        // If arg is null, nothing to do.
        return NULL;
    }

    // Calculate the size of the str.
    size= strlen(s);
    // alloc the str that contain the answer.
    binary= malloc( (size*8)+1 );
    // If no memory, nothing to do.
    if( binary == NULL ){
        // No memory
        return NULL;
    }

    // save a copy of the arg.
    aux= s;
    // for each char in the str arg.
    for( i=0; i<size; i++ ){
        // In each char to examinate, reset the mask.
        mask= 0x0001;
        // For each bit of a char.
        for( y=0; y<8; y++ ){
            // Compare each bit with the mask.
            if( (*aux) & mask ){
                // add 0 to the answer.
                strcat(bynary,"1");
            }else{
                // add 1 to the answer.
                strcat(bynary,"0");
            }
            // shit the mask 1 pos to left.
            mask= mask<<1;
         }
         aux++
    }
    return binary;
}

在C中将字符串转换为二进制

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3 个答案: