问题
Given an example
const int limit = 500;
const int * const cpci = &limit;
const int * const * pcpci = &cpci;
I am having difficulty understanding what the last line means.
Basically in array terms the value pcpci it's just an array of (const int * const)'s. But i can't seem to make multiple copies inside pcpci since it is not supposed to be a constant pointer.
For Example
const int limit = 500;
const int * const cpci = &limit;
const int * const * pcpci = &cpci;
const int limit2 = 600;
const int * const cpci2 = &limit2;
*(pcpci+1) = &cpci2;
In the last line of the above code i got "error lvalue must be modifiable". But i was wondering why is this happening since pcpci is not a constant pointer and only it's elements should be constant and non modifiable.
回答1:
First of all this statement
*(pcpci+1) = &cpci;
has undefined behaviour because you may not dereference a pointer that does not points to an object. You could use this construction if pcpci would point to an element of an array and pcpci + 1 also point to the next element of the same array.
So it would be more coorectly to write
*(pcpci) = &cpci;
However the object pointed to by pcpci is a constant object with type T const where T is const int * so it may not be reassigned.
That it would be more clear you can rewrite definition
const int * const * pcpci = &cpci;
the following way
typedef const int * const ConstPointerToConstObject;
ConstPointerToConstObject * pcpci = &cpci;
So if to derefernce pcpci you will get an object of type ConstPointerToConstObject that may not be changed because it is a constant pointer to constant object.
回答2:
But i was wondering why is this happening since pcpci is not a constant pointer
No, but *(pcpci+1) is. It has type const int* const. Obviously you can't assign anything to that.
来源：https://stackoverflow.com/questions/27080915/understanding-pointer-to-constant-pointer-to-integer-constant-const-int-const