Why is the default char* in a parameter must be const?

Question

class MyString {
private:
    char *m_pchString;
    int m_nLength;
public:
    MyString(char* pchString="0") {      //problem on this parameter
        m_nLength = strlen(pchString)+1;
        m_pchString = new char[m_nLength];
        strncpy(m_pchString, pchString, m_nLength);
        m_pchString[m_nLength-1] = 0;
    }
    ~MyString() {
        delete[] m_pchString;
        m_pchString = 0;
    }
    char* GetString() {return m_pchString;}
    int GetLength() {return m_nLength;}
};

If I complied this, the compiler will send me a warning:

warning: deprecated conversion from string constant to 'char*'

unless I modify the parameter from char *pchString = "0" to const char *pchString = "0"

Why is the default char* in a parameter must be const?

This has nothing to do with default parameters. String literals in C++ are `const` no matter where you use them. — nobody, Apr 09 '15 at 01:58
@Juen Khaw, you may want to change your "About me:" to `const char *chStat = "Begineer";` :) — vsoftco, Apr 09 '15 at 02:06
[Are string literals const?](http://stackoverflow.com/questions/4493139/) — Remy Lebeau, Apr 09 '15 at 02:23

score 6 · Accepted Answer · edited May 23 '17 at 12:06

Because string literals like "some string" are immutable, and trying to modify one (you can attempt to modify them if passing via non-const reference) leads to undefined behaviour. That's why the standard C++ deprecated this conversion.

Try this for fun (but pls not in production code), live here:

#include <iostream>

int main() 
{
    char* str = "test";
    str[0] = 'w';
    std::cout << str; // oops
}

Related: Why are string literals const?

score 0 · Answer 2 · answered Apr 09 '15 at 02:00

When you use a string literal, such as "0", it is put in the read only parts of the program. Making any changes to such data leads to undefined behavior. By trying to initialize a variable of type char* to a sting literal, you are allowing the possibility of the read only data to be modified. That's what the compiler is complaining about.

It's similar to:

const int num;
int* ptr = &num;  // Wrong

score -3 · Answer 3 · answered Apr 09 '15 at 01:57

-3

The reason being that char* is considered a outdated practice compared to using the string class.

answered Apr 09 '15 at 01:57

Nick Krause

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1
8

This has nothing to do with the `string` class at all. – nobody Apr 09 '15 at 02:00

Why is the default char* in a parameter must be const?

3 Answers3