How to properly cast char ** x to const char ** - c++

Question

I have the following variable.
char **arr
Then I want to perform some modification on array which means it can't be declared as a constant.
Now I have a function that accepts the argument of type const char ** arr. But I have no control over the signature of this function.
Now when I cast arr to const char ** arr g++ generates a warning which is [-Werror=cast-qual].
For more clarification consider following MCVE:
#include<cstdio>
void print(const char** x){
printf("%s", x[0]);
}
int main(int argc, char **argv){
if(argc>1){
print((const char **)argv);
}
return 0;
}
//Then compile it as follow:
$ g++ -Wcast-qual test.cpp
//gives the following output:
MCVE.cpp: In function ‘int main(int, char**)’:
MCVE.cpp:5:36: warning: cast from type ‘char**’ to type ‘const char**’ casts away qualifiers [-Wcast-qual]
const char ** q = (const char**) argv;
So my question is why this generates a warning? Is there any risk in doing this?
And how to achieve a behavior I want to achieve?

Answer 1

Allowing a cast from char** to const char** provides a loophole to modify a const char or a const char*.
Sample code:
const char c = 'A';
void foo(const char** ptr)
{
*ptr = &c; // Perfectly legal.
}
int main()
{
char* ptr = nullptr;
foo(&ptr); // When the function returns, ptr points to c, which is a const object.
*ptr = 'B'; // We have now modified c, which was meant to be a const object.
}
Hence, casting a char** to const char** is not a safe cast.
You can use
if(argc>1)
{
const char* ptr = argv[0];
print(&ptr);
}
for your code to compile without the cast-qual warning.
If you need to pass more than just the first argument, you'll need to construct an array of const char* and use it.
if(argc>1)
{
int N = <COMPUTE N FIRST>;
const char** ptr = new const char*[N];
for (int i = 0; i < N; ++i )
{
ptr[i] = argv[i];
}
print(ptr);
delete [] ptr; // Make sure to deallocate dynamically allocated memory.
}