错误:'unary*'的无效类型参数(有'int')

Question

我有一个C程序:

#include <stdio.h>
int main(){
  int b = 10;             //assign the integer 10 to variable 'b'

  int *a;                 //declare a pointer to an integer 'a'

  a=(int *)&b;            //Get the memory location of variable 'b' cast it
                          //to an int pointer and assign it to pointer 'a'

  int *c;                 //declare a pointer to an integer 'c'

  c=(int *)&a;            //Get the memory location of variable 'a' which is
                          //a pointer to 'b'.  Cast that to an int pointer 
                          //and assign it to pointer 'c'.

  printf("%d",(**c));     //ERROR HAPPENS HERE.  

  return 0;
}    

编译器产生错误:

error: invalid type argument of ‘unary *’ (have ‘int’)

有人可以解释这个错误的含义吗？

Answer 1

由于c是保存整数指针的地址,因此其类型应为int**:

int **c;
c = &a;

整个计划变为:

#include <stdio.h>                                                              
int main(){
    int b=10;
    int *a;
    a=&b;
    int **c;
    c=&a;
    printf("%d",(**c));   //successfully prints 10
    return 0;
}

Answer 2

Barebones C程序产生上述错误:

#include <iostream>
using namespace std;
int main(){
    char *p;
    *p = 'c';

    cout << *p[0];  
    //error: invalid type argument of `unary *'
    //peeking too deeply into p, that's a paddlin.

    cout << **p;    
    //error: invalid type argument of `unary *'
    //peeking too deeply into p, you better believe that's a paddlin.
}

ELI5:

大师将一块闪亮的圆形石头放在一个小盒子里,然后送给学生.大师说:"打开盒子,取下石头".学生这样做了.

然后主人说:"现在打开石头,取下石头".学生说:"我不能开石头".

然后学生开悟了.

Answer 3

我重新格式化了你的代码.

错误位于此行:

printf("%d", (**c));

要修复它,请更改为:

printf("%d", (*c));

*从地址中检索值.**从地址中检索另一个值的值(在这种情况下为地址).

另外,()是可选的.

#include <stdio.h>

int main(void)
{
    int b = 10; 
    int *a = NULL;
    int *c = NULL;

    a = &b;
    c = &a;

    printf("%d", *c);

    return 0;
} 

编辑:

这条线:

c = &a;

必须替换为:

c = a;

这意味着指针'c'的值等于指针'a'的值.因此,'c'和'a'指向相同的地址('b').输出是:

10

编辑2:

如果你想使用双*:

#include <stdio.h>

int main(void)
{
    int b = 10; 
    int *a = NULL;
    int **c = NULL;

    a = &b;
    c = &a;

    printf("%d", **c);

    return 0;
} 

输出:

10