在 C 中创建使用指针转换 Pascal 字符串的方法

Question

我一直在研究一种应该将 Pascal 字符串转换为 C 字符串的方法。我还被告知返回的char * 应该指向一个新分配的char 数组，其中包含一个以空字符结尾的C 字符串。被调用者负责在这个数组上调用free()。

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

char *pascal_convert(void *x)
{
    int *y;
    x = y;
    char *z;
    *z = *((int*)x);
    char *arr = malloc(sizeof(*z));
    for (int i = 0; i < *y; i++)
    {
        arr[i] = z[i];
    }
    char* fin = arr;

    return fin;
}

Answer 1

需要很多调整

char *pascal_convert(void *x)
{
    // int *y;
    // x = y;   This assignment is backwards
    unsigned char *y = x;  // Need unsigned char (unless your pascal uses wider type here)

    // y = z;
    // char *z;
    // *z = *((int*)x);
    size_t size = *y++;  // Size is just the first element

    // char *arr = malloc(sizeof(*z));
    char *arr = malloc(size + 1);  // Allocate + 1 for the null chacter

    if (arr) {  // test need as `malloc()` may fail
      // for (int i = 0; i < *y; i++) { arr[i] = z[i]; }
      memcpy(arr, y, size); 
      arr[size] = '\0';  // append null character
    } 

    // char* fin = arr;  // No need for new variable
    // return fin;
    return arr;
}