C结构为void *指针

本文介绍了C结构为void *指针的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有一个结构定义为：

typedef struct { int type; void* info; } Data;

然后我有我想要分配到的void *使用下面的函数其他几个结构：

and then i have several other structs that i want to assign to the void* using the following function:

Data* insert_data(int t, void* s) { Data * d = (Data*)malloc(sizeof(Data)); d->type = t; d->info = s; return d; } struct { ... } Struct#;

然后我就打电话

insert_data(1, variable_of_type_Struct#);

当我编译这个它会发出警告

When i compile this it gives a warning

warning: assignment from incompatible pointer type

我试着投变量在插入到（void *的），但没有奏效。

i tried to cast the variable in the insert to (void*) but didn't work

insert_data(1, (void *) variable_of_type_Struct#);

我

如何才能摆脱这种警告？

How can i get rid of this warning?

感谢

推荐答案

在传递的结构，而不是它的一个副本（即不按值传递）地址：

Pass in the address of the struct, not a copy of it (i.e. not passed by value):

insert_data(1, &variable_of_type_Struct);