Teach Start-DevPowerShell run CoreCLR version

This commit is contained in:
Sergei Vorobev
2016-07-27 10:49:46 -07:00
parent 1349372b39
commit 99e4b87cc6
2 changed files with 7 additions and 6 deletions
+4 -3
View File
@@ -884,10 +884,11 @@ function Publish-NuGetFeed
function Start-DevPowerShell {
param(
[switch]$FullCLR,
[switch]$ZapDisable,
[string[]]$ArgumentList = '',
[switch]$LoadProfile,
[string]$binDir = (Split-Path (New-PSOptions -FullCLR).Output),
[string]$binDir = (Split-Path (New-PSOptions -FullCLR:$FullCLR).Output),
[switch]$NoNewWindow,
[string]$Command,
[switch]$KeepPSModulePath
@@ -918,7 +919,7 @@ function Start-DevPowerShell {
$env:COMPLUS_ZapDisable = 1
}
if (-not (Test-Path $binDir\powershell.exe.config)) {
if ($FullCLR -and (-not (Test-Path $binDir\powershell.exe.config))) {
$configContents = @"
<?xml version="1.0" encoding="utf-8" ?>
<configuration>
@@ -932,7 +933,7 @@ function Start-DevPowerShell {
# splatting for the win
$startProcessArgs = @{
FilePath = "$binDir\powershell.exe"
FilePath = "$binDir\powershell"
ArgumentList = "$ArgumentList"
}
+3 -3
View File
@@ -92,20 +92,20 @@ Assembly Cache (GAC), not your output directory.
Use `Start-DevPowerShell` helper funciton, to workaround it with `$env:DEVPATH`
```powershell
Start-DevPowerShell
Start-DevPowerShell -FullCLR
```
This command has a reasonable default to run `powershell.exe` from the build output folder.
If you are building an unusual configuration (i.e. not `Debug`), you can explicitly specify path to the bin directory
```powershell
Start-DevPowerShell -binDir .\src\Microsoft.PowerShell.ConsoleHost\bin\Debug\net451
Start-DevPowerShell -FullCLR -binDir .\src\Microsoft.PowerShell.ConsoleHost\bin\Debug\net451
```
Or more programmatically:
```powershell
Start-DevPowerShell -binDir (Split-Path -Parent (Get-PSOutput))
Start-DevPowerShell -FullCLR -binDir (Split-Path -Parent (Get-PSOutput))
```
The default for produced `powershell.exe` is x64.