建议的duplicate和当前给出的答案没有解决为什么首先给出的例子没有问题.主要是为什么没有推理：

“const int **是指向const int *的指针,它与int *不同”

也适用于：

“const int *是一个指向const int的指针,它与int只是一个不同的东西”

我从不同的角度接近它,希望得到另一种解释.

带有示例的代码.

#include <stdio.h>
 
void f_a (int const a){
 
    /*
     *  Can't do:
     *      a = 3;  //error: assignment of read-only parameter ‘a’
     *
     * Explanation: I can't change the value of a in the scope of the function due to the const
    */
    printf("%d\n",a);
}
 
void f_ptr_a_type1 (int const * ptr_a){
    /*
     * Can do this:
     *     ptr_a’ = 0x3;
     * which make dereferencig to a impossible.
     *     printf("%d\n",* ptr_a’);  -> segfault
     * But const won't forbid it.
     *
     *  Can't do:
     *      *ptr_a’ = 3;  //error: assignment of read-only parameter ‘* ptr_a’
     *
     * Explanation: I can't change the value of a by pointer dereferencing and addignment due to the int const
    */
}
 
void f_ptr_a_type2 (int * const ptr_a){
    /*
     * Can do this:
     *     *a = 3;
     *
     *  Can't do:
     *      ptr_a = 3;  //error: assignment of read-only parameter ‘ptr_a’
     *
     * Explanation: I can't change the value because the const is protecting the value of the pointer in the funcion scope
    */
}
 
void f_ptr_ptr_a (int const ** ptr_ptr_a){
    /*
     * Can do this:
     *     ptr_ptr_a = 3;
     *     * ptr_ptr_a = 0x3;
     *
     *  Can't do:
     *      ** ptr_ptr_a = 0x3;  //error: assignment of read-only parameter ‘**ptr_a’
     *
     * Explanation: Makes sense. Just follows the pattern from prevIoUs functions.
    */
}
 
int main()
{
    int a = 7;
    f_a(a);
 
    int * ptr_a = &a;
    f_ptr_a_type1(&a);
    f_ptr_a_type2(&a);
 
    int ** ptr_ptr_a = &ptr_a;
    f_ptr_ptr_a(ptr_ptr_a);  //warning: passing argument 1 of ‘f_ptr_ptr_a’ from incompatible pointer type [-Wincompatible-pointer-types]
}

接受的广泛接受的答案是这样的：

int ** isn’t the same as const int** and you can’t safely cast it

我的问题是为什么功能突然关注？

这里没有抱怨int不是int const：

int a = 7;
f_a(a);

它没有在这里抱怨因为int *既不是int const *也不是int * const：

int * ptr_a = &a;
f_ptr_a_type1(&a);
f_ptr_a_type2(&a);

但突然间,它开始在双指针案件中抱怨.

>使用这个术语和示例寻找解释？
>为什么函数突然开始担心写入
超出范围的东西的权限？