Question

I keep getting the error “Conversion from string literal to char* is deprecated” in my code. The purpose of the code is to use a pointer-to-pointer to assign string1 and string2 a word, then print it out. How can I fix this?

Here is my code:

#include <iostream>
using namespace std;

struct WORDBLOCK
{
    char* string1;
    char* string2;
};

void f3()
{
    WORDBLOCK word;

    word.string1 = "Test1";
    word.string2 = "Test2";


    char *test1 = word.string1;
    char *test2 = word.string2;

    char** teststrings;

    teststrings = &test1;
    *teststrings = test2;

    cout << "The first string is: "
         << teststrings
         << " and your second string is: "
         << *teststrings
         << endl;  
}

Answer 1

C++ string literals are arrays of const char, which means you can’t legally modify them.

If you want to safely assign a string literal to a pointer (which involves an implicit array-to-pointer conversion), you need to declare the target pointer as const char*, not just as char*.

Here’s a version of your code that compiles without warnings:

#include <iostream>

using namespace std;

struct WORDBLOCK
{
    const char* string1;
    const char* string2;
};

void f3()
{
    WORDBLOCK word;

    word.string1 = "Test1";
    word.string2 = "Test2";

    const char *test1 = word.string1;
    const char *test2 = word.string2;

    const char** teststrings;

    teststrings = &test1;
    *teststrings = test2;

    cout << "The first string is: "
         << teststrings
         << " and your second string is: "
         << *teststrings
         << endl;
}

Consider what could happen if the language didn’t impose this restriction:

#include <iostream>
int main() {
    char *ptr = "some literal";  // This is invalid
    *ptr = 'S';
    std::cout << ptr << "\n";
}

A (non-const) char* lets you modify the data that the pointer points to. If you could assign a string literal (implicitly converted to a pointer to the first character of the string) to a plain char*, you’d be able to use that pointer to modify the string literal with no warnings from the compiler. The invalid code above, if it worked, would print

Some literal

— and it might actually do so on some systems. On my system, though, it dies with a segmentation fault because it attempts to write to read-only memory (not physical ROM, but memory that’s been marked as read-only by the operating system).

(An aside: C’s rules for string literals are different from C++’s rules. In C, a string literal is an array of char, not an array of const char — but attempting to modify it has undefined behavior. This means that in C you can legally write char *s = "hello"; s[0] = 'H';, and the compiler won’t necessarily complain — but the program is likely to die with a segmentation fault when you run it. This was done to maintain backward compatibility with C code written before the const keyword was introduced. C++ had const from the very beginning, so this particular compromise wasn’t necessary.)