Microsoft.VisualStudio.Project.ConfigProvider.GetDefaultOutputPath C# (CSharp) Method

        protected virtual string GetDefaultOutputPath(string configurationName, string platformName)
        {
            if (configurationName == null)
                throw new ArgumentNullException("configurationName");
            if (platformName == null)
                throw new ArgumentNullException("platformName");
            if (string.IsNullOrEmpty(configurationName))
                throw new ArgumentException("configurationName cannot be null or empty");
            if (string.IsNullOrEmpty(platformName))
                throw new ArgumentException("platformName cannot be null or empty");
            //Contract.Ensures(!string.IsNullOrEmpty(Contract.Result<string>()));

            string outputBasePath = ProjectManager.OutputBaseRelativePath;
            if (outputBasePath[outputBasePath.Length - 1] == Path.DirectorySeparatorChar || outputBasePath[outputBasePath.Length - 1] == Path.AltDirectorySeparatorChar)
                outputBasePath = Path.GetDirectoryName(outputBasePath);

            string platformProperty = GetPlatformPropertyFromPlatformName(platformName);
            if (!string.Equals(platformProperty, ProjectFileValues.AnyCPU, StringComparison.OrdinalIgnoreCase))
                outputBasePath = Path.Combine(outputBasePath, platformProperty);

            return Path.Combine(outputBasePath, configurationName) + Path.DirectorySeparatorChar;
        }