How can I access the elements of a pointer-to-map?

Question

I've passed a map by pointer to a function, but now I'm unable to change the contents of the map because the code doesn't compile. I'm unsure whether I'm even allowed to pass the map as a pointer. For example:

#include <iostream>
#include <stdlib.h>
#include <map>
#include <string>
using namespace std;

void faa(map<int,string> *baa){
   *baa[1] = "somethingnew";
}

int main(){
    map<int,string> bar;
    bar[1] = "one";
    bar[2] = "two";
    bar[3] = "three";

    int ii;
    for(ii=1;ii<4;ii++){
        cout<<"bar["<<ii<<"]="<<bar[ii]<<endl;
    }

    faa(&bar);
    cout<<"after"<<endl;
    for(ii=1;ii<4;ii++){
        cout<<"bar["<<ii<<"]="<<bar[ii]<<endl;
    }
}

When I compile this I get the error:

error: no match for ‘operator*’ (operand type is ‘std::map >’)
     *baa[1] = "somethingnew";

Is a function like faa possible? What is the syntax?

@ddriver doesn't have the same meaning WRT constness and what to do with unfound keys though. — Quentin
– Quentin, Commented Aug 12, 2015 at 14:06
@ddriver at and operator[] are semantically different if the key is not found. — Csq
– Csq, Commented Aug 12, 2015 at 14:06

TartanLlama · Accepted Answer · 2015-08-12 14:00:28Z

8

Due to operator precedence rules, *baa[1] is parsed as *(baa[1]) instead of (*baa)[1]. You could just put in the parentheses to make it work. However, you'd be better off using references instead:

void faa(map<int,string> &baa){
   baa[1] = "somethingnew";
}

Then you just call the function without taking the address of the map:

faa(bar);

answered Aug 12, 2015 at 14:00

TartanLlama

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1 Comment

Paani Over a year ago

Plus1 for pointing out use of reference. That is the way to go with STL containers in general.

Csq · Accepted Answer · 2015-08-12 14:07:49Z

2

You can use (*baa)[1] instead.

The default precedence is the same as if you wrote *(baa[1])

To explain the error message: if baa is a pointer, you can use the [] operator to access the nth element pointed by the pointer, therefore the type of baa[1] is an std::map, and it has no * operator.

edited Aug 12, 2015 at 14:07

answered Aug 12, 2015 at 14:02

Csq

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