如何将void **数组（函数的返回值）赋给变量？

Needed: A function which returns two different values (int,str)

需要：返回两个不同值的函数（int，str）

So foo calculates smth. and stores the addresses of the two values in the return array.

所以foo计算smth。并将两个值的地址存储在返回数组中。

Now I want to read the values back into variables of these types.

现在我想将值读回到这些类型的变量中。

void** foo(){
    void** bar = malloc(2 * sizeof(void*));
    ...
    return bar;
}


int main(int argc, char* argv[]){
    void** result = foo();
    int* val1 = (int*)result[0];
    char* val2 = (char*)result[1];
}

This snippet is not compiling.

此代码段未编译。

Excuse me:

the problem was not the code but that I declared it in a case branch without any compution instructions.

问题不是代码，而是我在没有任何计算指令的案例分支中宣布它。

int main(int argc, char* argv[]){
    switch(xyz){
      case a:
        void** result = foo();
        int* val1 = (int*)result[0];
        char* val2 = (char*)result[1];
      break;
    }
}

Which seems not to make sense in a case.

在一个案例中似乎没有意义。

Thanks though.

谢谢。

2 个解决方案

#1

If you come up with the need to use a void**, it is a strong indication of poor program design. Just forget about this idea. void pointers in general is something you should avoid.

如果你想出使用void **的话，那就是程序设计不佳的强烈迹象。忘了这个想法。一般来说，void指针是你应该避免的。

What you actually want to do seems to be something like this:

你真正想做的事情似乎是这样的：

typedef struct
{
  int i;
  char c;
} something_t;

something_t* foo (void)
{
  something_t* result = malloc (sizeof(*result));
  ...
  return result;
}

int main(int argc, char* argv[]){
  something_t* result = foo();
  int* val1 = &result->i;
  char* val2 = &result->c;
}

If the struct members need to be pointers, simply change the struct.

如果struct成员需要是指针，只需更改结构。

#2

You should use structure to do your job

你应该用结构来完成你的工作

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

#define STRING_WIDTH 32

struct test
{
    int val1;
    char *val2;
};

struct test *foo(void)
{
    struct test *bar = malloc(sizeof(struct test));

    if (bar != NULL)
    {
        bar->val1 = 0;
        bar->val2 = calloc (STRING_WIDTH, 1);

        if (bar->val2 == NULL)
            return NULL;
    }

    return bar;
}


int main( void )
{
    struct test *result = foo();

    if (result != NULL)
    {
        // USE YOUR STRUCT
    }

    return 0;
}

Take note that mallocated memory must be freed

请注意，必须释放mallocated内存

#1