另一个经验法则是检查const的位置:

* =>之前存储的值为常量 * =>指针本身是常量

// Some more complex constant variable/pointer declaration.
// Observing cases when we get error and warning would help
// understanding it better.

int main(void)
{
  char ca1[10]= "aaaa"; // char array 1
  char ca2[10]= "bbbb"; // char array 2

  char *pca1= ca1;
  char *pca2= ca2;

  char const *ccs= pca1;
  char * const csc= pca2;
  ccs[1]='m';  // Bad - error: assignment of read-only location ‘*(ccs + 1u)’
  ccs= csc;    // Good

  csc[1]='n';  // Good
  csc= ccs;    // Bad - error: assignment of read-only variable ‘csc’

  char const **ccss= &ccs;     // Good
  char const **ccss1= &csc;    // Bad - warning: initialization from incompatible pointer type

  char * const *cscs= &csc;    // Good
  char * const *cscs1= &ccs;   // Bad - warning: initialization from incompatible pointer type

  char ** const cssc=   &pca1; // Good
  char ** const cssc1=  &ccs;  // Bad - warning: initialization from incompatible pointer type
  char ** const cssc2=  &csc;  // Bad - warning: initialization discards ‘const’
                               //                qualifier from pointer target type

  *ccss[1]= 'x'; // Bad - error: assignment of read-only location ‘**(ccss + 8u)’
  *ccss= ccs;    // Good
  *ccss= csc;    // Good
  ccss= ccss1;   // Good
  ccss= cscs;    // Bad - warning: assignment from incompatible pointer type

  *cscs[1]= 'y'; // Good
  *cscs= ccs;    // Bad - error: assignment of read-only location ‘*cscs’
  *cscs= csc;    // Bad - error: assignment of read-only location ‘*cscs’
  cscs= cscs1;   // Good
  cscs= cssc;    // Good

  *cssc[1]= 'z'; // Good
  *cssc= ccs;    // Bad - warning: assignment discards ‘const’
                 //                qualifier from pointer target type
  *cssc= csc;    // Good
  *cssc= pca2;   // Good
  cssc= ccss;    // Bad - error: assignment of read-only variable ‘cssc’
  cssc= cscs;    // Bad - error: assignment of read-only variable ‘cssc’
  cssc= cssc1;   // Bad - error: assignment of read-only variable ‘cssc’
}

这里是一个详细的解释与代码

/*const char * p;
char * const p; 
const char * const p;*/ // these are the three conditions,

// const char *p;const char * const p; pointer value cannot be changed

// char * const p; pointer address cannot be changed

// const char * const p; both cannot be changed.

#include<stdio.h>

/*int main()
{
    const char * p; // value cannot be changed
    char z;
    //*p = 'c'; // this will not work
    p = &z;
    printf(" %c\n",*p);
    return 0;
}*/

/*int main()
{
    char * const p; // address cannot be changed
    char z;
    *p = 'c'; 
    //p = &z;   // this will not work
    printf(" %c\n",*p);
    return 0;
}*/



/*int main()
{
    const char * const p; // both address and value cannot be changed
    char z;
    *p = 'c'; // this will not work
    p = &z; // this will not work
    printf(" %c\n",*p);
    return 0;
}*/

const * char是无效的C代码，没有意义。也许你想问const char *和char const *之间的区别，或者可能是const char *和char * const之间的区别?

参见:

什么是const指针(相对于指向const对象的指针)? C中的Const c++中const声明的区别 c++的const问题 为什么我可以改变一个const char*变量的值?

许多答案提供了具体的技术，经验法则等，以理解变量声明的特殊实例。但是有一个通用的技巧来理解任何声明:

顺时针/螺旋规则

A)

const char *a;

根据顺时针/螺旋规则，a是指向常量字符的指针。这意味着字符是不变的，但指针可以改变。即a = "其他字符串";没问题，但[2]= 'c';将无法编译

B)

char * const a;

根据规则，a是指向字符的const指针。例如，你可以做一个[2]= 'c';但是你不能用a = "other string";

第一个是语法错误。也许你指的是两者的区别

const char * mychar

and

char * const mychar

在这种情况下，第一个指针是指向不能更改的数据的指针，第二个指针将始终指向相同的地址。