C：声明一个常量指针的常量字符数组 [英] C: declare a constant pointer to an array of constant characters

问题描述

我想了解数组声明，常量性，以及它们产生的变量类型。

以下是允许的（我的编译器）：

  S01的char [] =ABC; // typeof运算（S01）=的char *
为const char S02 [] =ABC; // typeof运算（S02）=为const char *（==字符常量*）
字符常量S03 [] =ABC; // typeof运算（S03）=字符常量*（==为const char *）
 

另外，我们可以手动声明数组的大小：

 的char S04 [4] =ABC; // typeof运算（S04）=的char *
为const char S05 [4] =ABC; // typeof运算（S05）=为const char *（==字符常量*）
字符常量S06 [4] =ABC; // typeof运算（S06）=字符常量*（==为const char *）
 

我如何得到一个结果类型的变量为const char * const的？以下是不允许的（我的编译器）：

 为const char S07常量[] =ABC;
字符常量S08常量[] =ABC;
为const char S09 [] =常量ABC;
字符常量S10 [] =常量ABC;
为const char S11常量[4] =ABC;
字符常量S12 const的[4] =ABC;
为const char S13 [4]常量=ABC;
字符常量S14 [4]常量=ABC;
 

感谢

解决方案

 为const char * const的S15 =ABC;
 

I am trying to understand array declarations, constness, and their resulting variable types.

The following is allowed (by my compiler):

      char s01[] = "abc" ;  // typeof(s01) = char*
const char s02[] = "abc" ;  // typeof(s02) = const char* (== char const*)
char const s03[] = "abc" ;  // typeof(s03) = char const* (== const char*)

Alternatively, we can declare the array size manually:

      char s04[4] = "abc" ;  // typeof(s04) = char*
const char s05[4] = "abc" ;  // typeof(s05) = const char* (== char const*)
char const s06[4] = "abc" ;  // typeof(s06) = char const* (== const char*)

How do I get a resulting variable of type const char* const? The following are not allowed (by my compiler):

const char s07 const[] = "abc" ;
char const s08 const[] = "abc" ;
const char s09[] const = "abc" ;
char const s10[] const = "abc" ;
const char s11 const[4] = "abc" ;
char const s12 const[4] = "abc" ;
const char s13[4] const = "abc" ;
char const s14[4] const = "abc" ;

Thanks

解决方案

const char *const s15 = "abc";

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