无法打印 typedef 结构的 char* 类型字段

Question

以下代码将十进制转换为二进制并将其二进制表示形式存储到结构中。 get_binary_representation() 可以正常工作，您可以通过取消注释打印val 的每一位的行来亲自查看。但是，当我尝试打印存储在 binary 中的二进制表示时，没有打印任何内容。

#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
#include <inttypes.h>
#include <stdbool.h>


typedef struct {
    char* val;
    int   length;
} binary;


char* get_binary_representation(uint16_t val, int length) {
    
    char* result; 

    if (length < 1)
        return NULL;

    if (val == 0) {
        result = (char*) malloc(sizeof(char) * 2);
        result[0] = '0';
        result[1] = '\0';
        return result;
    }

    result = (char*) malloc(sizeof(char) * length);
    result[length - 1] = '\0';

    for (int i = length; i > 0; i--) {
        result[i] = ((val & 1) == 1) ? '1' : '0';
        // printf("Bit %d = %c\n", i, result[i]);
        val >>= 1;
    }

    return result;
}


binary i16_to_binary(uint16_t* val) {
    binary result; // tried replacing it with pointer and using malloc, didn't help
    result.val = get_binary_representation(*val, 16);    
    /* There is a function that finds length of number in binary 
       but I excluded it from MRE for simplicity sake. 
       16 for length because uint16_t is 16 bit integer*/
    return result;
}


int main(int argc, char** argv) {
    uint16_t a = 16; uint16_t b = 254; uint16_t c = 37;
    
    binary a_bin = i16_to_binary(&a);
    binary b_bin = i16_to_binary(&b);
    binary c_bin = i16_to_binary(&c);
    
    /* should print 0000 0000 0001 0000, 0000 0000 0111 1110, 0000 0000 0010 0101
       without spaces and commas */
    printf("%s %s %s\n", a_bin.val, b_bin.val, c_bin.val);

    return 0;
}

Answer 1

小心：

• 不要导致缓冲区溢出。只有索引0 到length-1（包括两者）可用于具有length 元素的数组。当i = length 时，result[i] 超出范围。
• 初始化所有要读取的元素。 reuslt[0] 未初始化。
• malloc() 家庭的铸造结果是considered as a bad practice。

部分：

    if (val == 0) {
        result = (char*) malloc(sizeof(char) * 2);
        result[0] = '0';
        result[1] = '\0';
        return result;
    }

    result = (char*) malloc(sizeof(char) * length);
    result[length - 1] = '\0';

    for (int i = length; i > 0; i--) {
        result[i] = ((val & 1) == 1) ? '1' : '0';
        // printf("Bit %d = %c\n", i, result[i]);
        val >>= 1;
    }

应该是：

    if (val == 0) {
        result = malloc(sizeof(char) * 2); /* no casting the result */
        result[0] = '0';
        result[1] = '\0';
        return result;
    }

    result = malloc(sizeof(char) * (length + 1)); /* no casting the result and allocate enough elements */
    result[length] = '\0'; /* put NUL to proper place */

    for (int i = length - 1; i >= 0; i--) { /* adjust the indice */
        result[i] = ((val & 1) == 1) ? '1' : '0';
        // printf("Bit %d = %c\n", i, result[i]);
        val >>= 1;
    }