为什么C ++不为字符数组接受有符号或无符号字符

Question

I am using C++ in native mode with Visual Studio 2017. That compiler compiles the statement below without complaint:

const char * AnArrayOfStrings[]  = {"z1y2x3w4", "Aname"};

However, if I change the above statement to specify that char is signed or unsigned, the compiler emits a C2440 error. For instance, the statements below, do not compile:

const signed   char * AnArrayOfStrings2[] = {"z1y2x3w4", "Aname"};

const unsigned char * AnArrayOfStrings2[] = {"z1y2x3w4", "Aname"};

I fail to see the reason for the compiler refusing to compile the statement when the sign of char is made explicit.

My question is: is there a good reason that I have failed to see for the compiler refusing to compile those statements ?

Thank you for your help (I did research in StackOverflow, the C++ documentation, I used Google and, consulted about a dozen C/C++ books in an effort to find the answer myself but, a reason still eludes me.)

Answer 1

"z1y2x3w4" is const char[9] and there is no implicit conversion from const char* to const signed char* .

You could use reinterpret_cast

const signed char * AnArrayOfStrings[]  = {reinterpret_cast<const signed char *>("z1y2x3w4"),
                                           reinterpret_cast<const signed char *>("Aname")};

Answer 2

If you compile the above code

const signed   char * AnArrayOfStrings2[] = {"z1y2x3w4", "Aname"};  

in C with gcc using options -Wall then it will give the following warning

test.c:5:49: warning: pointer targets in initialization differ in signedness [-Wpointer-sign]
  const unsigned   char * AnArrayOfStrings2[] = {"z1y2x3w4", "Aname"};
                                                 ^
test.c:5:49: note: (near initialization for 'AnArrayOfStrings2[0]')
test.c:5:61: warning: pointer targets in initialization differ in signedness [-Wpointer-sign]
  const unsigned   char * AnArrayOfStrings2[] = {"z1y2x3w4", "Aname"};  

The type of elements of AnArrayOfStrings2 and "z1y2x3w4" are different. AnArrayOfStrings2[0] is of type const signed char * while "z1y2x3w4" is of type const char[9] .
The same code will raise error in C++. You will need explicit cast to make it work in C++.

---

To explain why

const char * AnArrayOfStrings[]  = {"z1y2x3w4", "Aname"}; 

works I will take s simple example

const char c[] = "asc";
const char *p1 = c;           // OK
signed const char *p2 = c;    // Error
unsigned const char *p3 = c;  // Error

In the second line of the above snippet, c will convert to const char * thus making p1 and c compatible types.
In third line the type of p2 and c are incompatible and compiler will raise an error in C++ (a warning in C). Same will happen with line 4.

If we take another example for int type

const int i[] = {1,2,3};
const int *ii = i            // OK
signed const int *si = i;    // OK
unsigned const int *usi = i; // Error  

First two pointer initializations work as int without any specifier is equivalent to signed int (but this is not true with char ) and therefore types are compatible. Intialization fails in last case as const int * or signed const int * is incompatible with unsigned const int * .