char const * const * const varName

Question

我有一個方法具有以下簽名：

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

我怎么解釋這個： char const *const *const strings ？。

謝謝，帕萬。

Answer 1

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

所以基本上它意味着所有指針和指針指向的字符串是常量，這意味着函數不能以任何方式修改傳遞的字符串（除非它被轉換）。

例如

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

這將腐爛成char **

當傳遞給

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

Answer 2

在char const *const *const strings ， strings是指向char指針的指針。 如果沒有const限定符，它將如下所示：

char **strings;

const限定符禁止在解除引用的特定級別修改解除引用的值：

**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

換句話說，第三個const表示指針本身是不可變的，第二個const表示指向指針是不可變的，第一個表示指向的字符是不可變的。

Answer 3

關鍵字const使得此關鍵字之后的聲明變為常量。 代碼解釋比單詞更好：

/////// 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