通过使用指针修改两个函数中的结构

试试这个



      以下为原文

    try this
 

void function_1(Type_structure* My_struct);
void function_2(Type-structure* My_struct2);
 
void main(void)
{
    // declarations
    Type_structure My_structure= {0x01 , 0x00, 0x00, 1, 0, 0, 0, 0, 0}; //typedef

    // Initialize the device
    SYSTEM_Initialize();

    while(1)
    {
            function_1(&My_structure); 
    }
}


void function_1(Type_structure* My_struct)
{
    function_2(My_struct);
}
 
void function_2(Type-structure* My_struct2)
{
   // do something with My_struct2
}
 
 
 

对不起，拼写错误，无法编辑



      以下为原文

    sorry, typos and couldn't edit
 
void function_2(Type_structure* My_struct2);
 
void function_2(Type_structure* My_struct2)
{
   // do something with My_struct2
}
 

谢谢你的回复，但没用。我想我得用双指针了？你觉得双指针怎么样？



      以下为原文

    Thank you for ur reply but it doesn't work. I think I have to use a double pointer ? what do you think about double pointers ?   

不管用吗？你想让它做什么？还是不编译？发布你的实际代码并解释你想要达到的目标。不确定你的双指针是什么意思。



      以下为原文

    doesn't work?  what do you want it to do?
 
or doesn't compile?
 
post your actual code and explain what you want to achieve.
 
not sure what you mean by double pointers.
 

也许吧



      以下为原文

    Maybe static Type_structure My_structure= ...

它不编译。现在我返回这个结构，但如果我成功地完成了我想做的事情，那就更好了。如果我有时间的话，我会回来的。非常感谢你！祝你度过美好的一天



      以下为原文

    It doesn't compile. For the moment I return the struct but it would have been better If i had succeed with the thing that I wanted to do .  I will come back here if I have some time ... Thank you very much ! And have a nice day :D 

编译器给出了什么错误消息？



      以下为原文

   


What error message did the compiler give?

函数2（Type结构*）位于哪里？它甚至知道你的Type结构吗？



      以下为原文

    Where is function_2(Type_Structure*) located?
Does it even know about your Type_Structure?

函数原型应该是全局的，或者是包含马丁的。第2页：“你怎么看待双指针？”他们还好吧，你觉得麦当劳怎么样？



      以下为原文

    Function prototypes should be global at the top or from an include.h like martin explained. post #2.
 
"What do you think about double pointers ?"
They are ok, what do you think of McDonalds?
 
 

呵呵，我想他指的是指向指针的指针。



      以下为原文

   
Hehe, I think he meant pointers to pointers

没有TyBufff，你每次都需要“StReXXX”，其中“XXX”是TaGyType是眼睛糖果，我更喜欢它。标签是可选的。问候，戴夫



      以下为原文

    Without typedef, you need the "struct xxx" every time, where "xxx" is the tag

struct _stag {
    int x1;
    int x2;
    int x3;
};

void function_1(struct _stag* My_struct);
void function_2(struct _stag* My_struct2);
 
void system_init(void);

void main(void)
{
    // declarations
    struct _stag My_structure = {42, 1, 0};
    int i;
    system_init();

    for (i = 0; i < 10; i++) {
        function_1(&My_structure);
        printf("%dn", My_structure.x3);
    }
 
    while (1) {}
 
}

void function_1(struct _stag* s)
{
    s->x1 = s->x2;
    s->x2 = s->x3;
    function_2(s);
}
 
void function_2(struct _stag* s)
{
    s->x3 = s->x1 + s->x2;
}

 
Typedef is eye candy; I like it much better.  The tag is optional.

typedef struct  {
    int x1;
    int x2;
    int x3;
} stype;

void function_1(stype* My_struct);
void function_2(stype* My_struct2);
 
void system_init(void);

void main(void)
{
    // declarations
    stype My_structure = {42, 1, 0};
    int i;
    system_init();

    for (i = 0; i < 10; i++) {
        function_1(&My_structure);
        printf("%dn", My_structure.x3);
    }
 
    while (1) {}
}


void function_1(stype* s)
{
    s->x1 = s->x2;
    s->x2 = s->x3;
    function_2(s);
}
 
void function_2(stype* s)
{
    s->x3 = s->x1 + s->x2;
}


 
Regards,
 
Dave

指向指针，他可以毁灭世界。



      以下为原文

    Pointer to pointers.
const char *say={"Klatuu","Barada","Nikto"};
 
activateGort(say);
 
activateGort(const char **Gort){
puts(*Gort++);
puts(*Gort++);
puts(*Gort++);
}
He could destroy the World.

指向指针，他可以毁灭世界。



      以下为原文

    Pointer to pointers.
const char *say={"Klatuu","Barada","Nikto"};
 
activateGort(say);
 
activateGort(const char **Gort){
puts(*Gort++);
puts(*Gort++);
puts(*Gort++);
}
He could destroy the World.

你缺少数组的括号：或者更好：



      以下为原文

    You're missing brackets for the array:const char *say[]={"Klatuu","Barada","Nikto"};or better:const char * const say[]={"Klatuu","Barada","Nikto"};

我是故意这样做的，看看你能不能认出它。咧嘴笑：



      以下为原文

    I did that on purpose to see if you would spot it.  grin:

：）



      以下为原文

   


:)

一定是想到了双层奶酪汉堡。



      以下为原文

    Must have been thinking of double cheese burgers.

你可以试试这样的东西



      以下为原文

    You can try something like this 
struct my_struct{
 int i;
 int j;
 int k;
};

void f2 (my_struct *structure)
{
 structure->j = 20;
}
void f1 (my_struct *structure)
{
 structure->i = 10;
 f2(structure);
}

int main (void)
{
 my_struct example = {1, 2, 3};
 f1(&example);
 printf("i = %d , j = %d, k = %d",example.i,example.j,example.k);
 return 0;
}