如何使用struct const指针作为函数参数来控制struct成员行为？

Question

I have a C code, somewhat similar to this:

struct st
{
    int *var;
}

void fun(st *const ptr)
{
    // considering memory for struct is already initialized properly.
    ptr->var = NULL; // NO_ERROR
    ptr = NULL; // ERROR, since its a const pointer.
}

void main()
{ 
    //considering memory for struct is initialized properly
    fun(ptr);
}

I dont want to declare int *var as const in the structure definition, so as not to mess with the huge code base. Not looking to make any change in the structure definition Is there any way in C, to get an error for the NO_ERROR line ptr->var = NULL; // NO_ERROR ptr->var = NULL; // NO_ERROR ?

Answer 1

只需使用const声明参数，因此ptr是指向const对象的指针：

void fun(const struct st* ptr)

Answer 2

Your declaration makes ptr to be constant, but not the object to which it points. Also don't miss struct keyword

void fun(struct st *const ptr);

Instead you should use

void fun(const struct st *ptr);

Such declaration allows to change pointer but not the object to which it points.

Answer 3

Keep this in mind:

• With type* const ptr , you cannot change the pointer but you can change the pointed data
• With const type* ptr , you can change the pointer but you cannot change the pointed data

So all you need is to replace struct st* const ptr with const struct st* ptr .

Answer 4

Superb! Thank you so much guys! I did this - void fun(const struct st *const ptr); and I was able to get an error while changing both ptr and ptr->var .. Just what I needed..