CODEWITHSUNDEEP
c++
Koz
Koz

Reputation: 151

struct cannot appear in a constant-expression; template arguments invalid

This function is giving me errors such as:

error: 'vertice' cannot appear in a constant-expression

error: template argument 1 is invalid

error: `iterator' does not name a type

list< list<vertice> >::iterator extraiLista(string vertice, list< list<vertice> >& listaVertices){
    list< list<vertice> >::iterator itVert;

    for(itVert = listaVertices.begin(); itVert != listaVertices.end(); itVert++){
        list<vertice>::iterator aux = itVert->begin();
        if(aux->nome == vertice)
            return itVert;
    }

    return NULL;
}

'vertice' is a struct that I created and I'm having no problems using it in other functions. It's just this one that is giving me trouble. I thought the problem might be with the iterator but I tried it in another function and it worked.

Upvotes: 0

Views: 416

Answers (2)

user31264
user31264

Reputation: 6737

You made at least two errors.

Error 1:

string vertice, list< list<vertice> >

"vertice" cannot be both a typename (as in list<list<vertice> >) and a variable (as in string vertice).

Error 2:

return NULL;

Your function returns list<list<iterator> >. In C++, NULL is int. There is no casting from int to list<list<iterator> > .

Upvotes: 1

Mike Seymour
Mike Seymour

Reputation: 254771

Within the function, vertice is the name of the function parameter, not the type which it hides.

To refer to the type, you can either elaborate it:

list<class vertice>

or qualify it:

list<::vertice>    // assuming it's in the global namespace

but it might be better to choose a different name for the parameter.

Upvotes: 2

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