How to call a synchronous sub function from a Cloud Function - node.js

Question

I'd like to run all of my functions synchronously. But if I execute something like the code below the response is sent back before the return value is assigned.
index.js
exports.cfTest = (req, res) => {
try {
result = columnCount(
'project','dataset','table'
);
console.log('sending response');
res.send('<OUTPUT>' + result + '</OUTPUT>');
} catch (err) {
res.status(500).send('<ERROR>' + err + '</ERROR>');
}
};
function columnCount(projectId, bqDataset, bqTable) {
const BigQuery = require('#google-cloud/bigquery');
const bigquery = new BigQuery({projectId: projectId});
var result;
bigquery
.dataset(bqDataset)
.table(bqTable)
.get()
.then(results => {
result = results[1].schema.fields.length;
})
.catch(err => {
result = null;
});
console.log('returning ' + result);
return result;
}
I.e the console output would be something like:
<OUTPUT>undefined</OUTPUT>
and in the Stackdriver logs would show an execution order like this:
2018-07-25 10:00:00 - Function execution started
2018-07-25 10:00:01 - sending response
2018-07-25 10:00:02 - Function execution took 2000 ms, finished with status code: 200
2018-07-25 10:00:03 - returning 5
This is any elegant way to lock down execution into synchronous only behaviour?
Solution
The async/await approach suggested below by Alexandru works! Thanks Alexandru.
NB: But not with standard/GA Cloud Functions. You'll have to use the Google Cloud Functions Node.js v8 Beta.

Answer 1

Yep, there is you can look at async/await
Your code will look something like this one in the end:
async function columnCount(projectId, bqDataset, bqTable) {
const BigQuery = require('#google-cloud/bigquery');
const bigquery = new BigQuery({ projectId: projectId });
try {
let results = await bigquery
.dataset(bqDataset)
.table(bqTable)
.get();
return results[1].schema.fields.length;
} catch (err) {
return null;
}
}
And in the second one
exports.cfTest = async (req, res) => {
try {
result = await columnCount(
'project','dataset','table'
);
console.log('sending response');
res.send('<OUTPUT>' + result + '</OUTPUT>');
} catch (err) {
res.status(500).send('<ERROR>' + err + '</ERROR>');
}
};