使用 void* 类型的参数调用 void* 类型的函数

Question

我有一个将从main.c 调用的函数。 函数原型是

void *function (void *args)
{ 
    Struct strData *data = args; 
    char* str;
    int a;
    str = data->name; 
    a = data->value;
    if (data->value == 1)
    { 
        data->value = 10;
        return data;
    }
}

而结构定义为

Struct strData
{
    int data;
    char* name;
};

如何从 main 调用上述函数？以及如何将整个结构从函数返回到 void*？以及如何将值发送到参数？我是否需要对函数内部参数的值进行类型转换？

Answer 1

这里试图回答你的问题：

#include <stdio.h>

struct strData
{
//changed member name from data to value due to conflicting call in funtion
    int value;
    char* name;
};

//function takes a void pointer as parameter and returns a void pointer
//I don't understand why you have written such code below.
void *function (void *args)
{ 
    struct strData *data = args; 
    char* str;
    int a;
    str = data->name; 
    a = data->value;
    if( data->value == 1)
    {   
        data->value = 10; 
    }   

    //return statement required at the end. Why?
    return data;
}

int main(){
    //created a variable of strData
    struct strData data;
    //initialised random values
    data.value = 1;
    data.name = "stackoverflow.com";
    //pass address of the variable. Why? typecast to struct. Why? Finally get value using *. Why?
    data = *(struct strData*)function(&data);
    printf("%s %d", data.name, data.value);
    return 0;
}

回答所有质疑自己的原因。