'char' array issue, custom print function and strcpy_s problem - c++

Question

I just started C++ and now I'm making a simple program. But don't know how to fix this problem.
I'm not a native english speaker so some sentences may not be understandable.
int main()
{
char test[5][4] = { "abc", "def", "ghi", "jkl", "mno" };
for (int i = 0; i < 5; ++i)
std::cout << test[i] << "\t";
return 0;
}
with this simple code, I made a print function
void printTest(char* pArr)
{
for (int i = 0; i < 5; ++i)
std::cout << pArr[i] << "\t";
}
Then in my main function, I typed printTest(*test);
but the result was 'a b c d'
while my expectation was 'abc def ghi jkl mno'
So I fixed printTest function like below
(changed const char* test[5][4] = { ... }
in main function)
void printTest(const char** pArr)
{
for (int i = 0; i < 5; ++i)
std::cout << pArr[i] << "\t";
}
which worked well.
The problem is, I want to use strcpy_s fucntion also.
strcpy_s(test[0], "xyx"); like this.
As strcpy_s get char* for the first parameter (not const char*),
I think I need to use char* for my array, and print function.
While that thing cause a wrong printing issue.
Did i wrote something wrong in my printTest function?
Or Is there any way that I can use strcpy_s function with const char* parameter?
PS. I used std::string and made it as I expected.
But I want to have a understanding and control of Char array.

Answer 1

I don't think void printTest(const char** pArr) will work with
char test[5][4].
The c++ compiler should refuse something like
void printTest(const char** pArr);
char test[5][4];
printTest(test);
Because test is a pointer to char [4],
while printTest() expects a pointer to char *.
You may have interesting in this function:
void printTest2(const char (*pArr)[4] )
{
for (int i = 0; i < 5; ++i)
std::cout << pArr[i] << "\t";
std::cout << "\n";
}
And the const keyword tells compiler (and what more important, the reader of your code) that "you won't modify the contents of 'pArr'". So compiler will not allow you to strcpy to pArr[i]. And this will compile and run.
void printTest3(char (*pArr)[4] )
{
for (int i = 0; i < 5; ++i)
{
strcpy(pArr[i], "123");
std::cout << pArr[i] << "\t";
}
std::cout << "\n";
}
...
printTest3(test);

Answer 2

I think you are a but confused about he use of const. Don't worry, it's happened to pretty much every one.
What's important to understand is that a non-const variable can be implicitly cast to a const variable at any time, but the reverse is not possible.
For your function PrintTest()
// Here, the const char** declaration is correct, your function only prints pArr,
// and the caller should not expect the function to modify his precious data.
// Another thing to remember: pArr is valid (and thus constant) only within
// the scope of PrintTest(). It simply does not exist anywhere else.
void printTest(const char** pArr)
{
for (int i = 0; i < 5; ++i)
std::cout << pArr[i] << "\t";
}
// this naturally allows printing of const and non-const data, as one would expects.
const char const_strings[2][4] = { "abc", "def" };
printTest(const_strings); // works!
// Note that if printTest() required a non_const pointer, the line above
// would not compile.
// const_strings being const, there is no way to modify it using strcpy_s();
// The code below also works fine,
char modifiable_strings[2][4] = { "tuv", "xyz" };
printTest(modifiable_strings); // works!
strcpy_s(modifiable_strings[0], "abc"); // is OK, because modifiable is not const.
printTest(modifiable_strings); // works, with different output!