[Solved] Identify the error in this code snippet:using System.Text.Json; string json = File.ReadAllText("data.json"); var person = JsonSerializer.Deserialize(json); | C Sharp (C#)

C Sharp (C#) - File IO

Identify the error in this code snippet:

using System.Text.Json;

string json = File.ReadAllText("data.json");
var person = JsonSerializer.Deserialize(json);

AJsonSerializer.Deserialize cannot be used without JsonOptions

BMissing generic type parameter in Deserialize method

CJsonSerializer.Deserialize requires a stream, not string

DFile.ReadAllText cannot read JSON files

Step-by-Step Solution

Solution:

Step 1: Check Deserialize method usage
Deserialize requires a generic type parameter to know which object type to create from JSON string.
Step 2: Verify other statements
File.ReadAllText can read any text file including JSON. Deserialize accepts string input. JsonOptions are optional.
Final Answer:
Missing generic type parameter in Deserialize method -> Option B
Quick Check:
Deserialize needs generic type parameter [OK]

Quick Trick: Always specify type in Deserialize(string) [OK]

Common Mistakes:

MISTAKES

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