char const *const *const varName

Question

我有一个具有以下签名的方法：

size_t advanceToNextRuleEntryRelatedIndex( size_t index, size_t nStrings, char const *const *const strings)

我该如何解释这个：char const *const *const strings？

谢谢，帕万。

Answer 1

char const *const *const strings
 ^    v    ^  v   ^  v
 |    |    |  |   |  |
 +-----    +--+   +--+

所以基本上它意味着所有的指针和指针所指向的字符串都是常量，这意味着函数不能以任何方式修改传递的字符串(除非它被强制转换)。

例如：

char* p{"string1","string2"};

它会腐烂成焦炭**

当传递给

int n = 0; 
advanceToNextRuleEntryRelatedIndex( n, 2, p);

Answer 2

在char const *const *const strings中，strings是指向char指针的指针。如果没有const限定符，它将如下所示：

char **strings;

const限定符禁止在特定的取消引用级别修改取消引用的值：

**strings = (char) something1; // not allowed because of the first const
*strings = (char *) something2; // not allowed because of the second const
strings = (char **) something3; // not allowed because of the third const

换句话说，第三个常量表示指针本身是不可变的，第二个常量表示指向的指针是不可变的，第一个常量表示指向的字符是不可变的。

Answer 3

关键字const将此关键字之后的声明转换为一个常量。代码比文字解释得更好：

/////// Test-code. Place anywhere in global space in C/C++ code, step with debugger
char a1[] = "test1";
char a2[] = "test2";

char *data[2] = {a1,a2};

// Nothing const, change letters in words, replace words, re-point to other block of words
char **string = &data[0]; 

// Can't change letters in words, but replace words, re-point to other block of words
const char **string1 = (const char **) &data[0];
// Can neither change letters in words, not replace words, but re-point to other block of words
const char * const* string2 = (const char * const*) &data[0];
// Can change nothing, however I don't understand the meaning of the 2nd const
const char const* const* const string3 = (const char const* const* const ) &data[0];


int foo()
{

    // data in debugger is:               {"test1","test2"}
    **string = 'T';         //data is now {"Test1","test2"}
    //1 **string1 = 'T';    //Compiler error: you cannot assign to a variable that is const (VS2008)
    *string1=a2;            //data is now {"test2","test2"}
    //2 **string2='T';      //Compiler error: you cannot assign to a variable that is const (VS2008)
    //3 *string2=a2;        //Compiler error: you cannot assign to a variable that is const (VS2008)
    string2=string1;

    //4 **string3='T';      //Compiler error: you cannot assign to a variable that is const (VS2008)
    //5 *string3=a2;        //Compiler error: you cannot assign to a variable that is const (VS2008)
    //6 string3=string1;    //Compiler error: you cannot assign to a variable that is const (VS2008)
    return 0;
}

static int dummy = foo();

/////// END OF Test-code