Errors C2664 and E0167, stumped

Question

I'm taking a c++ beginner's distance class and can't solve this compile error. It's written exactly as in the example book, and when my brother cuts and pastes it into his VS2015 it works fine, but in my VS2017 it doesn't. I have uninstalled and re-installed VS2017 community to no avail.

I have only been coding for 2 weeks so I'm very, very new to this.

The error codes I get are:

Error E0167 argument of type "const char " is incompatible with parameter of type "char
Error C2664 'void Hello(char )': cannot convert argument 1 from 'const char [8]' to 'char '

The code:

// FUNCTION: Hello, prints out a welcome message on the screen.
void Hello(char* name)
{
 cout << "Hello " << name << "!";
}

// FUNCTION: Main, program start.
int main()
{
 Hello("Krister");
 cin.get();
 return 0;
}

I really hope someone can help me with this.

All the best, Jepp

What, exactly, is unclear? You are passing `const char*` to a function which accepts `char*`. Those types are incompatible. — Algirdas Preidžius, Feb 26 '18 at 11:58
change the method signature to `void Hello(const char* name)` and it should work fine. VS2017 is by default more rigid to const correctness. — Alex, Feb 26 '18 at 11:59
All literal strings are *constant* literal strings. They are in reality *constant* arrays of characters, and pointers to them must be pointers to `const` characters. — Some programmer dude, Feb 26 '18 at 12:00
Big thanks for the help guys, I really appreciate it. It seems my text book is having some big problems. — Jeppen, Feb 26 '18 at 12:23
@LightnessRacesinOrbit Our personal barista called in sick today so we are not fully awake. — Ron, Feb 26 '18 at 12:26

score 8 · Answer 1 · answered Feb 26 '18 at 11:59

8

A string literal may be referred to by a const char*.

It may not be referred to by a char*; this was possible in old versions of C, and some older C++ compilers permitted it with a warning. In modern times it is completely prohibited.

By passing "Krister" to a function taking char*, you are asking the compiler to try to convert one to the other; it is failing, due to the above rule, as evidenced by the error message.

Chuck a const in there for great success.

If that code came from a textbook, lose it. Here is a list of good C++ books.

answered Feb 26 '18 at 11:59

Lightness Races in Orbit

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I'm totally blown away by the poor quality of this text book, there are so many errors, but this really took the case. Big thanks for the explanation! – Jeppen Feb 26 '18 at 12:21
1

@Jeppen Please read the [What should I do when someone answers my question?](https://stackoverflow.com/help/someone-answers) topic. – Ron Feb 26 '18 at 12:22
@Jeppen: No problem. Unfortunately there is no shortage of poor teaching material out there (about 50% of it is on YouTube) – Lightness Races in Orbit Feb 26 '18 at 12:26

score 6 · Answer 2 · answered Feb 26 '18 at 12:19

Just to complement the answer by Lightness Races in Orbit, which describes what you definitely should do. The reason for the difference between VS2015 and VS2017 is that the later sets the /permissive flag off by default, unlike the former.

It means that VS2017 may very well reject code that "your brother's" VS2015 accepts, unless the project options are tinkered with.

I recommend you keep the flag in its off state. Strict conformance is good, it makes you pick up better habits and write more portable C++.

Errors C2664 and E0167, stumped

2 Answers2

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