C Programming – What Does 2 Const Mean

code:

const char * const key;

There are 2 const in above pointer, I saw things like this the first time.

I know the first const makes the value pointed by the pointer immutable,
but did the second const make the pointer itself immutable?

Anyone can help to explain this?

@Update:

And I wrote a program that proved the answer is correct.

#include <stdio.h>

void testNoConstPoiner() {
    int i = 10;

    int *pi = &i;
    (*pi)++;
    printf("%d\n", i);
}

void testPreConstPoinerChangePointedValue() {
    int i = 10;

    const int *pi = &i;

    // this line will compile error
    // (*pi)++;
    printf("%d\n", *pi);
}


void testPreConstPoinerChangePointer() {
    int i = 10;
    int j = 20;

    const int *pi = &i;
    pi = &j;
    printf("%d\n", *pi);
}

void testAfterConstPoinerChangePointedValue() {
    int i = 10;

    int * const pi = &i;
    (*pi)++;
    printf("%d\n", *pi);
}

void testAfterConstPoinerChangePointer() {
    int i = 10;
    int j = 20;

    int * const pi = &i;
    // this line will compile error
    // pi = &j
    printf("%d\n", *pi);
}

void testDoublePoiner() {
    int i = 10;
    int j = 20;

    const int * const pi = &i;
    // both of following 2 lines will compile error
    // (*pi)++;
    // pi = &j
    printf("%d\n", *pi);
}

int main(int argc, char * argv[]) {
    testNoConstPoiner();

    testPreConstPoinerChangePointedValue();
    testPreConstPoinerChangePointer();

    testAfterConstPoinerChangePointedValue();
    testAfterConstPoinerChangePointer();

    testDoublePoiner();
}

Uncomment lines in 3 of the functions, will get compile error with tips.