function pointer definition on ternary operator

Question

When I define a function pointer in C++ using the ternary operator I get the compiler error overloaded function with no contextual type information. I'm confused. Could someone explain me the reason of this behavior?

#include <string>
#include <string.h>
#include <iostream>

const char *my_strstr1 (const char *__haystack, const char *__needle) {
    std::cout << "my_strstr" << std::endl;
    return strstr(__haystack, __needle);
}

const char *my_strstr2 (const char *__haystack, const char *__needle) {
    std::cout << "my_strstr2" << std::endl;
    return strstr(__haystack, __needle);
}

int main(int argc, char** argv) {
    std::cout << "argc:" << argc << std::endl;

    //ok
    // const char* (*StrStr)(const char*, const char*) = strstr;
    // const char* (*StrStr)(const char*, const char*) = (argc > 1) ? my_strstr2 : my_strstr1;

    // error: overloaded function with no contextual type information
    const char* (*StrStr)(const char*, const char*) = (argc > 1) ? my_strstr1 : strstr;

    StrStr("helloworld", "h");

    return 0;
}

Did you wish to write: (argc > 1) ? my_strstr1 : my_strstr2; — nishantsingh, Sep 21 '16 at 08:58
@user3286661 No, OP wants to use the standard library `strstr`. — Hatted Rooster, Sep 21 '16 at 08:59

Barry · Answer 1 · 2016-09-21T09:14:32.420

1

The overload of strstr defined in string.h is:

char *strstr( const char* str, const char* substr );

And in <cstring> there are two strstrs:

const char* strstr( const char* str, const char* target );
      char* strstr(       char* str, const char* target );

Either way, that doesn't match your other function. The former has the wrong return type, and the latter has two overloads and you're using it in a context that doesn't allow for that.

The simplest approach is just to "fix" which strstr you want with a lambda and just use that:

using F = const char*(*)(const char*, const char*);
F std_version = [](const char* s, const char* t) -> const char* {
    return strstr(s, t);
};

F StrStr = (argc > 1) ? my_strstr1 : std_version ;

edited Sep 21 '16 at 09:14

answered Sep 21 '16 at 08:59

Barry

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Won't this clash because `std` is a namespace? – Hatted Rooster Sep 21 '16 at 09:01
I think that does match, bcz the code could be compiled correctly. //ok // const char* (*StrStr)(const char*, const char*) = strstr; – izual Sep 21 '16 at 09:02
1

@GillBates Nope. But changed it anyway. – Barry Sep 21 '16 at 09:15

score 0 · Accepted Answer · edited May 23 '17 at 12:32

0

There are more than one version of strstr function. So compiler don't know which one should be used so you have to tell the compiler which one you want:

const char* (*StrStr)(const char*, const char*) = (argc > 1) ? my_strstr1 : (const char* (*)(const char*, const char*))strstr;

You can take a look to similar SO question: Return type of '?:' (ternary conditional operator).

And also section 5.16 in Working draft, Standard for Programming Language C++

edited May 23 '17 at 12:32

Community

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answered Sep 21 '16 at 09:00

KIIV

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then why this code knows how to choose the right version? //ok // const char* (*StrStr)(const char*, const char*) = strstr; – izual Sep 21 '16 at 09:02
Assigment operator knows, ternary operator don't. – KIIV Sep 21 '16 at 09:03
can u explain more or share a link about this? It's the 1st time I heard about this since using C++. 3q. – izual Sep 21 '16 at 09:09

function pointer definition on ternary operator

2 Answers2