apache / apache/openwhisk-wskdeploy
Why default version to "master" inside manifest parser
- Dominant language
- Go
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- 75
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Description
in manifest_parser.yaml we default to version = "master"; however, this has no meaning in the specification. Are we assuming only GitHub and overloading a concept of version with "branch"? We need to examine why this was done and see if we need to introduce a new concept of "branch"; however, by default we should automatically use code from master branch (or equivalent concept) in other repo. systems. Perhaps we should see how we do this generically for Subversion, etc.
for reference, it is set as follows:
```
func (dm *YAMLParser) ComposeDependencies(mani *ManifestYAML, projectPath string) (map[string]utils.DependencyRecord, error) {
var errorParser error
depMap := make(map[string]utils.DependencyRecord)
for key, dependency := range mani.Package.Dependencies {
version := dependency.Version
if version == "" {
version = "master"
}
```
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