The provided code in the video does not seem to be working. https://github.com/total-typescript/typescript-generics-workshop/blob/main/src/02-passing-type-arguments/11-data-fetcher.solution.ts If you take the code and paste it into **Typescript playground** ([I have done it for you](https://www.typescriptlang.org/play?#code/JYWwDg9gTgLgBAbzgUwB5mQYxgGjseAXzgDMoIQ4AiANwOQGcYqBuAKFElkTgFEBHAK4BDADZ5e6LEVLlKVAAIwAnhgC0mABZjRyAHYBzRgHpBMYKIas2bTBD1NSyGFoAiwmMLgBeOMIbKephwADwAKu6eAHwAFIJQogBccExQwIYAlD5RiGxwcLrwACYewskRpT5+AO7CBE4umnEJGQB0MJr6MTFQjJAOyFneOb0M-QzIrQBWDPYxGRnseXC9MPF6cCWe7IRLBDFUAMqaEIKiRQ1am5VkFH4bAIIACgCSVHj+gcHz2bn5dg5ipVfMJavUSM43KUQkg9MIQMhkql0gY4IRYst8lRNDAYGAGIljMYGLUwMBWkVkDRjMIycYMBAwLpjABGKjLRbLNAYbAxLbCVpwhFtGAQAQiUQHAAyggA1sg4IdZcpaqJ5VAqJzlioMHAYIwYAwqgBtSQ8mAhcViEI65AQEjXTx4WHwxEpGBpQxoqJRAC6O0WQA)) - you will get a bunch of errors. The solution is as following: ```Typescript const fetchData = async <T>(url: string): Promise<T> => await fetch(url).then(resp => resp.json as T) ```