c - 间接更新变量

Question

我正在学习 C，我被分配了一项任务。

“通过添加一个存储 x 地址的新变量来修改程序。然后使用您的变量更新(间接)i 的值，然后打印出新值以证明您的修改已生效。”这是我必须修改的代码:

#include <stdio.h>
int main()
{
    int i, j;
    int * p, * q;
    int ** x;

    i = 100;
    j = 200;
    p = &i;
    q = &j;
    x = &p;

    *p = *p + *q;
    *q = **x / 2;
    **x = *p + j;

    printf(" i = %d\n",   i);
    printf("&i = %p\n",  &i);
    printf(" j = %d\n",   j);
    printf("&j = %p\n",  &j);

    printf(" p = %p\n",   p);
    printf("&p = %p\n",  &p);
    printf("*p = %d\n",  *p);
    printf(" q = %p\n",   q);
    printf("&q = %p\n",  &q);
    printf("*q = %d\n",  *q);

    printf(" x = %p\n",   x);
    printf("&x = %p\n",  &x);
    printf("*x = %p\n",  *x);
    printf("**x= %d\n", **x);
    return 0;
}

这是我尝试过的。我声明了存储地址的变量。然后我给它分配了 x of x (newVariable) 的地址，然后尝试创建地址 i 的增量更新。编译程序时出现以下错误:

ptr3.c:14:18: warning: assignment makes integer from pointer without a cast [-Wint-conversion]
     *newVariable = &x; /* Assign new variable address of x */
                  ^
ptr3.c:19:19: error: lvalue required as unary ‘&’ operand
     newVariable = &i++; /* Autoincrement address by 1 */
                   ^
ptr3.c:21:12: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘int *’ [-Wformat=]
     printf("&newVariable = %d\n", newVariable); /* New printf statement */
        ^

代码:

#include <stdio.h>
int main()
{
    int i, j;
    int * p, * q;
    int ** x; 
    int * newVariable; /* New variable of type address*/

    i = 100;
    j = 200;
    p = &i;
    q = &j;
    x = &p;
    newVariable = &x; /* Assign new variable address of x */

    *p = *p + *q;
    *q = **x / 2;
    **x = *p + j;
    newVariable = &i++; /* Autoincrement address by 1 */

    printf("&newVariable = %d\n", newVariable); /* New printf statement */

    printf(" i = %d\n",   i);
    printf("&i = %p\n",  &i);
    printf(" j = %d\n",   j);
    printf("&j = %p\n",  &j);

    printf(" p = %p\n",   p);
    printf("&p = %p\n",  &p);
    printf("*p = %d\n",  *p);
    printf(" q = %p\n",   q);
    printf("&q = %p\n",  &q);
    printf("*q = %d\n",  *q);

    printf(" x = %p\n",   x);
    printf("&x = %p\n",  &x);
    printf("*x = %p\n",  *x);
    printf("**x= %d\n", **x);

    return 0;
}

版本 3:

#include <stdio.h>
int main()
{
    int i, j;
    int * p, * q;
    int ** x; 
    int * newVariable; /* New variable of type address*/

    i = 100;
    j = 200;
    p = &i;
    q = &j;
    x = &p;
    newVariable = &x; /* Assign new variable address of x */

    *p = *p + *q;
    *q = **x / 2;
    **x = *p + j;
    newVariable = &i++; /* Autoincrement address by 1 */

    printf("&newVariable = %d\n", newVariable); /* New printf statement */

    printf(" i = %d\n",   i);
    printf("&i = %p\n",  &i);
    printf(" j = %d\n",   j);
    printf("&j = %p\n",  &j);

    printf(" p = %p\n",   p);
    printf("&p = %p\n",  &p);
    printf("*p = %d\n",  *p);
    printf(" q = %p\n",   q);
    printf("&q = %p\n",  &q);
    printf("*q = %d\n",  *q);

    printf(" x = %p\n",   x);
    printf("&x = %p\n",  &x);
    printf("*x = %p\n",  *x);
    printf("**x= %d\n", **x);

    return 0;
}

谁能告诉我我做错了什么？我看不出程序有什么问题。

Answer 1

如果新变量将保存 x 的地址，它需要是一个指向 x 类型的指针，它是 int **。所以:

int ***newVariable;

您正在正确分配 x 的地址，其中:

newVariable = &x;

要修改i，让我们稍微想一想。原来的x本身指向指向指向 i 的其他内容(在本例中为 p)。所以newVariable 添加了一个额外的级别，这意味着您只需要取消对它的三个级别的引用以访问 i。例如，如果你想要通过递增来修改 i，然后可以使用:

***newVariable += 3; // increment i (indirectly) by 3

要打印 i 的值，你可以直接使用它本身，或者 p which指向它，或者 x 指向 p，或者 newVariable指向 x。所以这些语句将产生相同的结果:

printf("i = %d\n", i);
printf("i = %d\n", *p);
printf("i = %d\n", **x);
printf("i = %d\n", ***newVariable);