c - 将指针及其地址转换为整数指针

Question

以下两个作业有什么区别？

int main()
{
    int a=10;
    int* p=  &a;

    int* q = (int*)p; <-------------------------
    int* r = (int*)&p; <-------------------------
}

我对这两个声明的行为感到非常困惑。
我什么时候应该使用一个而不是另一个？

Answer 1

int* q = (int*)p;

是正确的，尽管太冗长了。int* q = p足够了。q和都是指针p。int

int* r = (int*)&p;

不正确（从逻辑上讲，虽然它可能会编译），因为&pis anint**但是r是int*. 我想不出你想要这个的情况。

Answer 2

#include <stdio.h>
int main()
{
    int a = 10;   /* a has been initialized with value 10*/

    int * p = &a; /* a address has been given to variable p which is a integer type pointer
                   * which means, p will be pointing to the value on address of a*/

    int * q = p ; /*q is a pointer to an integer, q which is having the value contained by p,                     * q--> p --> &a; these will be *(pointer) to value of a which is 10;

    int * r = (int*) &p;/* this is correct because r keeping address of p, 
                         * which means p value will be pointer by r but if u want
                         * to reference a, its not so correct.
                         * int ** r = &p; 
                         * r-->(&p)--->*(&p)-->**(&p)                             
                         */
       return 0;
}

Answer 3

int main()
{
    int a=10;
    int* p=  &a;

    int* q  = p; /* q and p both point to a */
    int* r  = (int*)&p; /* this is not correct: */
    int **r = &p; /* this is correct, r points to p, p points to a */

    *r = 0; /* now r still points to p, but p points to NULL, a is still 10 */
}

Answer 4

类型很重要。

表达式p具有类型int *（指向 的指针int），因此表达式&p具有类型int **（指向指向的指针的指针int）。这些是不同的、不兼容的类型；如果没有显式强制转换，您不能将类型的值分配给类型int **的变量int *。

正确的做法是写

int  *q = p;
int **r = &p;

除非您知道为什么需要将值转换为不同的类型，否则您永远不应该在赋值中使用显式强制转换。