如何将字符指针数组作为 void * 传递，然后转换回字符指针数组？

Question

我将一个字符指针数组传递给 sqlite3_exec，它接受 1 个参数并将其显示为 void *，但随后我想将其作为字符指针数组访问回调函数。

char *output_params[] = {"one", "two"};
result = sqlite3_exec(db, sql_statement, callback, output_params, &zErrMsg);

....

static int callback(void *param, int argc, char **argv, char **azColName) {
    // How do I access my character array?
    char *output_params[2] = (char **)param;
}

通过后如何访问？

Answer 1

这对我有用：

int callback(void *param, int argc, char **argv, char **azColName)
{
    const char** p = (const char **)param;
    printf("%s\n", p[0]);
    printf("%s\n", p[1]);
}

这是演示该概念的简单程序。

#include <stdio.h>

void foo(void* in)
{
    char **p = (char**)in;
    printf("%s\n", p[0]);
    printf("%s\n", p[1]);
}

void main(int argc, char** argv)
{
   char *output_params[] = {"one", "two"};
   foo(output_params);
}

如何将字符指针数组作为 void * 传递，然后转换回字符指针数组？

How do I pass an array of character pointers as void *, then cast back to array of character pointers?

c

arrays

pointers

void

void-pointers