Implicit conversion from char** to const char**

Why my compiler(GCC) doesnt implicitly cast from char** to const char**?

Thie following code:

#include <iostream>

void print(const char** thing) {
    std::cout << thing[0] << std::endl;
}

int main(int argc, char** argv) {
    print(argv);
}

Gives the following error:

oi.cpp: In function ‘int main(int, char**)’:
oi.cpp:8:12: error: invalid conversion from ‘char**’ to ‘const char**’ [-fpermissive]
oi.cpp开发者_运维百科:3:6: error:   initializing argument 1 of ‘void print(const char**)’ [-fpermissive]

Such a conversion would allow you to put a const char* into your array of char*, which would be unsafe. In print you could do:

thing[0] = "abc";

Now argv[0] would point to a string literal that cannot be modified, while main expects it to be non-const (char*). So for type safety this conversion is not allowed.

@Fred Overflow's link to the FAQ is a complete answer. But (sorry Marshall) it's not the most clear explanation. I don't know if mine is more clear, but I hope so.

The thing is, if p is a char* pointer, then it can be used to modify whatever it's pointing at.

And if you could obtain a pointer pp that points to p, but with pp of type char const**, then you could use pp to assign to p the address of a const char.

And with that, you could then use p to modify the const char. Or, you would think you could. But that const char could even be in read-only memory…

In code:

char const        c = 'a';
char*             p = 0;
char const**      pp = &p;               // Not allowed. :-)

*pp = &c;        // p now points to c.
*p = 'b';        // Uh oh.

#include <iostream>

void print(const char** thing) {
    std::cout << thing[0] << std::endl;
}

int main(int argc, char** argv) {
    print(argv);    // Dang, doesn't compile!
}

just do …

#include <iostream>

void print( char const* const* thing )
{
    std::cout << thing[0] << std::endl;
}

int main( int argc, char** argv )
{
    print( argv );    // OK. :-)
}

Cheers & hth.,

Note, that although

void dosmth(const char** thing);

int main(int argc, char** argv) {
  dosmth(argv);

is forbidden, you can and should do

void dosmth(const char* const* thing);

int main(int argc, char** argv) {
  dosmth(argv);

Which is probably what you wanted anyway. The point here is that thing now refers to a const char* array which is itself immutable and which referenced values char are themselves immutable. So, for a "look at it, but do not change it" scenario, const char* const* is the type to use.

Note: I used the more common (but in my opinion inferior) standard of trying to write the const modifier as left as possible. Personally, I recommend writing char const* const* instead of const char* const* as it is hugely more concise that way.

Because it might allow us to modify a constant value. Read here to understand why: http://c-faq.com/ansi/constmismatch.html