Which kind of explicit conversion is done here?

Question

This is the code that compile and works in my MVS2015:

#define STRING_TEST "Test String"

using namespace std;

void test(char * str) {
    cout << str;
}

int main ()
{
    test(STRING_TEST);
}

I guess that STRING_TEST was converted into a char[11]. Instead, since I pass it without &, it creates a char * pointer? Or what am I missing?

`STRING_TEST` is a macro. It doesn't get "converted", it gets *substituted* with its replacement text. — Kerrek SB, Sep 23 '16 at 12:37
why the macro? It would be exactly the same to write `test("Test String");` — 463035818_is_not_an_ai, Sep 23 '16 at 12:37
You need to enable more compiler warnings, this is not legal C++[11 and later] and deprecated in C++98/03. — Baum mit Augen, Sep 23 '16 at 12:38
Note, a string literal is of type `const char[]`, conversion to `char*` is dangerous and deprecated. — Hatted Rooster, Sep 23 '16 at 12:39
I think that there's a good explanation on c++ preprocessor [here](http://stackoverflow.com/questions/6264249/how-does-the-compilation-linking-process-work). — Gabriel Vasconcelos, Sep 23 '16 at 12:41
[Does not compile](http://melpon.org/wandbox/permlink/9t908JG7EL2o31g3) on a real C++ compiler. — Kerrek SB, Sep 23 '16 at 12:41
@GillBates: It may be a real compiler (see below), but it's not a real C++ compiler. It compiles something that was called C++ in 1989 when it was first loaded onto the DOS emulator it requires to run, but when we say C++ now we mean ISO C++, which didn't even _exist_ until ten years later!! Is a thing still "real" if it doesn't work on platforms created within the 1990s, 2000s, 2010s? — Lightness Races in Orbit, Sep 23 '16 at 15:13
@LightnessRacesinOrbit Is anything real? ( no but really, the 2006 version doesn't ) — Hatted Rooster, Sep 23 '16 at 15:14

Rakete1111 · Accepted Answer · 2016-09-23T12:55:54.207

2

"Test string" is of type const char[12], not char[11] (you forgot '\0').

Your code is equivalent to

test("Test String");

_{Macros just do a find+replace, so it's not "converted".}

Arrays decay into pointers, to here, the char arrays decays into char*. And so test calls it. This however is has been removed as of C++, since const char* != char*.

edited Sep 23 '16 at 12:55

answered Sep 23 '16 at 12:38

Rakete1111

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Actually it is `const char[12]`, and this is illegal on more updated compilers. – Gabriel Vasconcelos Sep 23 '16 at 12:45
Hello my function I have an array, take my variable and give it play. It works I think without delay but woe is me we have decay. – wally Sep 23 '16 at 12:48
@paizza When a pointer is asked and an array is passed, yes, always. (except on sundays but that's personal). – Hatted Rooster Sep 23 '16 at 12:48
But If I do this `const char kSamplerDefaultPath[] = "Test String";` instead of #define, it doenst work anymore :( – markzzz Sep 23 '16 at 12:53
@paizza Because a `const char*` is not a `char*`. – Rakete1111 Sep 23 '16 at 12:57
arrays decay into pointers?: Pretty much always. The exceptions are 1) when passed to `sizeof`; 2) when initializing a value which is a reference to an array of the right size (and I always have to look up the syntax for defining one of those). – Martin Bonner supports Monica Sep 23 '16 at 13:14

score 2 · Answer 2 · answered Sep 23 '16 at 12:51

2

There is no explicit conversion at all in your code.

STRING_TEST is a macro, so the statement

test(STRING_TEST);

is preprocessed into

test("Test String");

which will then be compiled.

"Test String" is a string literal, so when passed to a function will be IMPLICITLY converted to a const char *.

So, assuming your compiler complies with the standard, the statement will not compile (since implicit conversion from const char * (or from a string literal to char *) is not permitted.

However, some compilers do allow this implicit conversion (for reasons of backward compatibility to older versions of C). Those compilers can often be made to complain by cranking up diagnostics (e.g. with Microsoft compilers adding the /Wall command line option).

answered Sep 23 '16 at 12:51

Peter

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I don't get an error on VS 2015 with `/Wall` for this. – wally Sep 23 '16 at 12:52
Ah, well, bad luck. Often != always. – Peter Sep 23 '16 at 12:53
I think you are mistaken. The standard specifies an implicit conversion from string literal to `char *` (no const). This conversion is deprecated, and obviously it is UB to modify the characters through the pointer. – Martin Bonner supports Monica Sep 23 '16 at 13:03
Looks like C++14 has removed that - but I'm pretty sure it used to be there. – Martin Bonner supports Monica Sep 23 '16 at 13:11
@MartinBonner: It was indeed. Deprecated in C++98 and C++03; removed in C++11. Non-existent therefore in C++14. – Lightness Races in Orbit Sep 23 '16 at 15:15

Which kind of explicit conversion is done here?

2 Answers2