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体中。 get_binary_representation()
工作正常,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;
}
答案 0 :(得分:3)
小心:
0
到 length-1
(两者都包含)可用于具有 length
元素的数组。 result[i]
在 i = length
时超出范围。reuslt[0]
未初始化。malloc()
族的铸造结果是 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;
}