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I was learning boost's variant and found that it makes implicit conversion from char* to std::string in this snippet:

boost::variant<int, char*, std::string> v; // implicitly converted from char* to std::string, so we can't store char* and std::string both? v = "123"; std::cout << "boost::variant value: " << boost::get<std::string>(v) << std::endl; v = 1; std::cout << "boost::variant value: " << boost::get<int>(v) << std::endl; v = "123"; std::cout << "boost::variant value: " << boost::get<char*>(v) << std::endl;

So the exception will be raised by the boost::get<char*>(v) call. I wondered a bit and tried to make it more fair: change char* to const char* in the typename list:

boost::variant<int, const char*, std::string> v; v = "123"; std::cout << "boost::variant value: " << boost::get<std::string>(v) << std::endl;

Now exception is correct. But my question is: why compiler/library chooses std::string instead of char* as I specified? I understand, that v = "123" is assignment of const char* but why it converts to std::string and not char*?

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    Why do people insist on adding the c tag to questions that have nothing to do with C? Commented Feb 25, 2015 at 10:54
  • Well, sorry, I was pointing at "c++" first but it added "c". Next I forgot to remove it. Commented Feb 25, 2015 at 11:00
  • If that is a reproducible problem (if you can describe the exact steps you took to add the c++ tag, and where the c tag got added by the system instead), I encourage you do report it as a bug. It's never happened to me, and I can't manage to even make c and c++ both show up at the same time in the tag selection, but I won't rule out the possibility of you doing something I haven't thought of trying. Commented Feb 25, 2015 at 11:03
  • Thanks for the suggestion. I don't remember already what I did exactly but I remember I was pointing at "c++", but maybe I also I did something else in the moment (click, key press, etc).. Okay, if I will reproduce it I'll do what you suggested to me. I also did not mean to put "c" tag here but now I think about - why not? The question is about c-type conversion between char* and const char*, is not it? Commented Feb 25, 2015 at 11:07
  • No, C and C++ are different languages, and even for the types that exist in both languages, the conversions between those types are different in C++ than they are in C. If C did have an implicit conversion from const char * to char * (it doesn't, luckily), that would have no impact on C++. If C makes "123" a char[4] rather than const char[4] (it does, unfortunately), that too has no impact on C++ and is not relevant to your question. Commented Feb 25, 2015 at 11:09

2 Answers 2

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const char * doesn't implicitly convert to char * because that would be unsafe. The whole point of const is that you don't inadvertently attempt to modify the object. Suppose that const char * did implicitly convert to char *. Then what's preventing programs such as these:

#include <cstdio> void f(char *s) { *s = 'a'; } int main() { const char *str = "Hello!"; f(str); std::puts(str); }

which are obviously a massive error, from compiling without even so much as a warning?

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Because there's an explicit conversion from const char * to std::string from std::string and dropping const is never done implicitly for obvious reasons.

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