[{"_1":2,"_3":-5,"_4":-5},"loaderData",{"_5":6,"_7":8},"actionData","errors","layouts/app-shell",{"_245":-5,"_9":10},"routes/subjects.$subject.topics.$topic",{"_9":10,"_11":12,"_13":14,"_15":16},"guest",true,"subjectTitle","C# for Beginners","topicTitle","Loops: for, while and do-while","questions",[17,18,19,20,21,22,23,24,25,26,27,28],{"_29":228,"_31":211,"_33":229,"_35":36,"_37":230,"_39":231,"_41":232},{"_29":210,"_31":211,"_33":212,"_35":36,"_37":213,"_39":214,"_41":215},{"_29":194,"_31":195,"_33":-5,"_35":-5,"_37":181,"_39":196,"_41":197},{"_29":179,"_31":32,"_33":180,"_35":36,"_37":181,"_39":182,"_41":183},{"_29":162,"_31":32,"_33":163,"_35":36,"_37":164,"_39":165,"_41":166},{"_29":144,"_31":145,"_33":146,"_35":36,"_37":147,"_39":148,"_41":149},{"_29":127,"_31":32,"_33":128,"_35":36,"_37":129,"_39":130,"_41":131},{"_29":109,"_31":110,"_33":111,"_35":36,"_37":112,"_39":113,"_41":114},{"_29":92,"_31":32,"_33":93,"_35":36,"_37":94,"_39":95,"_41":96},{"_29":75,"_31":32,"_33":76,"_35":36,"_37":77,"_39":78,"_41":79},{"_29":58,"_31":32,"_33":59,"_35":36,"_37":60,"_39":61,"_41":62},{"_29":30,"_31":32,"_33":34,"_35":36,"_37":38,"_39":40,"_41":42},"id","cmuo4yz3801da20jx034g5d6l","stem","What does the code below show on screen?","codeBlock","for (int i = 1; i <= 5; i++)\n{\n if (i % 2 == 0)\n {\n continue;\n }\n Console.WriteLine(i);\n}","codeLang","csharp","notes","`continue` skips the rest of the current pass. It jumps straight to checking the condition again (in `while` and `do-while`) or to the iterator step (in `for`).\n\n```csharp\nfor (int i = 0; i < 5; i++)\n{\n if (i == 2)\n {\n continue; // skips printing 2\n }\n Console.WriteLine(i); // prints 0, 1, 3, 4\n}\n```","explanation","The continue statement skips the rest of the current pass through the loop and jumps straight to the iterator step, then re-checks the condition. i % 2 == 0 (the remainder after dividing by 2 is 0) is true for even numbers, so continue skips printing them. Only the odd numbers 1, 3, and 5 get printed.","options",[43,44,45,46],{"_29":56,"_48":57,"_33":-5,"_50":53},{"_29":54,"_48":55,"_33":-5,"_50":53},{"_29":51,"_48":52,"_33":-5,"_50":53},{"_29":47,"_48":49,"_33":-5,"_50":10},"cmuo4yz3e01de20jxzed9xez9","text","1, then 3, then 5","isCorrect","cmuo4yz3d01dd20jxmz8knmfn","1, 2, 3, 4, 5, all of them",false,"cmuo4yz3b01dc20jxwtnmktal","1, 3, 5, but the program never stops","cmuo4yz3a01db20jxoj1m1579","2, then 4","cmuo4yz3001d520jx1z8ere3y","int count = 5;\ndo\n{\n Console.WriteLine(count);\n count++;\n} while (count < 3);","**`do-while`** checks its condition **after** running the body. This means the body always runs **at least one time**, even if the condition is `false` from the very start.\n\n```csharp\nint y = 10;\ndo\n{\n Console.WriteLine(\"runs once\"); // this runs even though y < 5 is false\n}\nwhile (y < 5);\n```\n\n**Watch out:** `do-while` needs a semicolon after `while (condition)` at the end. If you forget it, you get a compile error.","A do-while loop checks its condition after running the body, so the body always runs at least once no matter what. Here it prints 5 first. Then count becomes 6, and the check count < 3 is false, so the loop stops after only one pass.",[63,64,65,66],{"_29":73,"_48":74,"_33":-5,"_50":10},{"_29":71,"_48":72,"_33":-5,"_50":53},{"_29":69,"_48":70,"_33":-5,"_50":53},{"_29":67,"_48":68,"_33":-5,"_50":53},"cmuo4yz3701d920jxq2lf7mi1","Nothing shows up at all","cmuo4yz3501d820jx9elkv3z6","The code will not build, because the check has to come before the body in a do-while loop","cmuo4yz3401d720jxbz6i19zz","5, 6, 7, and so on forever, because the loop never stops","cmuo4yz3201d620jxgwmug3k9","Just the number 5","cmuo4yz2t01d020jxn17f9v4h","int count = 5;\nwhile (count < 3)\n{\n Console.WriteLine(count);\n count++;\n}\nConsole.WriteLine(\"Done\");","**`while`** checks its condition **before** running the body each time. If the condition is `false` right at the start, the body never runs at all (zero times).\n\n```csharp\nint x = 10;\nwhile (x < 5)\n{\n Console.WriteLine(\"never runs\");\n}\n```","A while loop checks its condition before running the body, even the first time. Here count starts at 5, so count < 3 is already false. The body never runs at all, so the program skips straight to printing \"Done\".",[80,81,82,83],{"_29":90,"_48":91,"_33":-5,"_50":53},{"_29":88,"_48":89,"_33":-5,"_50":53},{"_29":86,"_48":87,"_33":-5,"_50":10},{"_29":84,"_48":85,"_33":-5,"_50":53},"cmuo4yz2z01d420jxzg9mwnuq","The program never stops, and nothing gets shown","cmuo4yz2x01d320jxsiri72r8","Just the word Done","cmuo4yz2w01d220jxlu4cwptm","5, then the word Done","cmuo4yz2u01d120jx5eax9zdb","0, 1, 2, then the word Done","cmuo4yz2j01cv20jxbjvjonwk","for (int i = 0; i < 10; i++)\n{\n if (i == 3)\n {\n break;\n }\n Console.WriteLine(i);\n}","`break` makes the program leave the loop right away. It skips any passes that are left.\n\n```csharp\nfor (int i = 0; i < 10; i++)\n{\n if (i == 3)\n {\n break;\n }\n Console.WriteLine(i); // prints 0, 1, 2\n}\n```","The break statement immediately stops the loop and skips everything else, including any remaining passes. When i becomes 3, the if condition is true, so break runs before Console.WriteLine(3) has a chance to execute, ending the loop right there. So only 0, 1, and 2 get printed.",[97,98,99,100],{"_29":107,"_48":108,"_33":-5,"_50":53},{"_29":105,"_48":106,"_33":-5,"_50":53},{"_29":103,"_48":104,"_33":-5,"_50":10},{"_29":101,"_48":102,"_33":-5,"_50":53},"cmuo4yz2r01cz20jxz7n90kdn","Just the number 3","cmuo4yz2p01cy20jx6suyt8tk","0, then 1, then 2","cmuo4yz2n01cx20jxf4j3b61r","0, then 1, then 2, then 3","cmuo4yz2m01cw20jxyv083ty1","0 all the way up to 9","cmuo4yz2a01cq20jxwj5ubcgc","What happens when you run the code below, and why?","int[] numbers = { 1, 2, 3 };\nforeach (int n in numbers)\n{\n n = n * 10;\n}\nConsole.WriteLine(numbers[0]);","A `foreach` loop gives you a **copy** of each value. You cannot change the array through it. If you try to set a new value to the loop variable, the code will not even compile.\n\n```csharp\nforeach (int n in numbers)\n{\n // n = n * 2; // error: you cannot assign a new value to n\n}\n```","In C#, the variable you use in a foreach loop, like n, is read-only: you are not allowed to assign it a new value, so the compiler blocks this code before it can even run. This is exactly why foreach cannot be used to change elements in place. A for loop with an index avoids this problem, because you write to the array directly, like numbers[i] = ..., instead of writing to the loop variable.",[115,116,117,118],{"_29":125,"_48":126,"_33":-5,"_50":53},{"_29":123,"_48":124,"_33":-5,"_50":53},{"_29":121,"_48":122,"_33":-5,"_50":10},{"_29":119,"_48":120,"_33":-5,"_50":53},"cmuo4yz2i01cu20jxb1od0kdk","It shows 1, because foreach only gives you a copy of the value, so changing n does not affect the array","cmuo4yz2g01ct20jxhrwls64m","The code will not build, because n is a read-only value inside a foreach loop, so you cannot change it","cmuo4yz2f01cs20jxxup1k9ts","It shows 3, because the loop replaces every item with the last value it saw","cmuo4yz2d01cr20jx7ba3o3pf","It shows 10, because foreach changes the real array while it runs","cmuo4yz1y01cl20jxbcixkjpz","int[] numbers = { 1, 2, 3 };\nfor (int i = 0; i < numbers.Length; i++)\n{\n numbers[i] = numbers[i] * 10;\n}\nConsole.WriteLine(numbers[1]);","If you use `for` with an index number, you can **change** the values in the array. This works because `numbers[i] = ...` writes a new value straight into that spot in the array.\n\n```csharp\nint[] numbers = { 1, 2, 3 };\nfor (int i = 0; i < numbers.Length; i++)\n{\n numbers[i] = numbers[i] * 2; // this changes the array\n}\n```","The for loop uses the counter i as a position (index) to reach into the array and replace each item with itself multiplied by 10. numbers[1] starts as 2, so after the loop it becomes 2 times 10, which is 20. This is something a for loop can do that foreach cannot: change items in place using their index.",[132,133,134,135],{"_29":142,"_48":143,"_33":-5,"_50":53},{"_29":140,"_48":141,"_33":-5,"_50":10},{"_29":138,"_48":139,"_33":-5,"_50":53},{"_29":136,"_48":137,"_33":-5,"_50":53},"cmuo4yz2601cp20jxbzwg7x3x","2","cmuo4yz2401co20jx855clz3k","10","cmuo4yz2201cn20jxp9rsvqps","20","cmuo4yz2001cm20jx8ytgfdj7","The code will not build, because you are not allowed to change an item inside a for loop","cmuo4yz1m01cg20jxh96wf6lc","What happens when you run the code below?","int[] numbers = { 10, 20, 30 };\nfor (int i = 0; i <= numbers.Length; i++)\n{\n Console.WriteLine(numbers[i]);\n}","The condition decides exactly how many times the loop runs. Mixing up `<` and `<=` is one of the most common loop mistakes.\n\n- `for (int i = 0; i < 5; i++)` → runs 5 times: 0,1,2,3,4\n- `for (int i = 0; i <= 5; i++)` → runs 6 times: 0,1,2,3,4,5\n\n**Watch out:** When going through an array by index, use `<`, not `<=`, when comparing to `array.Length`. `array[array.Length]` is past the end of the array. This causes an `IndexOutOfRangeException` (a crash).\n\n```csharp\nint[] numbers = { 10, 20, 30 };\nfor (int i = 0; i < numbers.Length; i++) // right: use \"<\"\n{\n Console.WriteLine(numbers[i]);\n}\n```","numbers.Length is 3, so the valid positions (indexes) are 0, 1, and 2. Using <= instead of < makes the loop also try i = 3, but there is no item at position 3, so C# throws an IndexOutOfRangeException while the program is running. This is a classic off-by-one error; using i < numbers.Length would fix it.",[150,151,152,153],{"_29":160,"_48":161,"_33":-5,"_50":53},{"_29":158,"_48":159,"_33":-5,"_50":10},{"_29":156,"_48":157,"_33":-5,"_50":53},{"_29":154,"_48":155,"_33":-5,"_50":53},"cmuo4yz1x01ck20jxkygjrbav","The program shows 10, 20, 30, 0, because C# fills a missing spot with zero","cmuo4yz1u01cj20jxbll4fama","The program will not build, because you are not allowed to use <= in a for loop's check","cmuo4yz1q01ci20jx8fh0pzeq","The program crashes with an error, because the loop tries to read numbers[3], and there is no item at position 3","cmuo4yz1o01ch20jxry1shnwv","The program shows 10, 20, 30 and stops with no problem","cmuo4yz1e01cb20jx7pjcygom","for (int i = 5; i > 0; i--)\n{\n Console.WriteLine(i);\n}","Count down: `for (int i = n - 1; i >= 0; i--)`\n\n```csharp\nfor (int i = 5; i > 0; i--)\n{\n Console.WriteLine(i);\n}\n```","The counter i starts at 5 and decreases by 1 each time (i--). The loop keeps going as long as i > 0. So it prints 5, 4, 3, 2, 1, and stops before printing 0 because 0 > 0 is false.",[167,168,169,170],{"_29":177,"_48":178,"_33":-5,"_50":53},{"_29":175,"_48":176,"_33":-5,"_50":53},{"_29":173,"_48":174,"_33":-5,"_50":10},{"_29":171,"_48":172,"_33":-5,"_50":53},"cmuo4yz1l01cf20jx9lf3fy3u","4, 3, 2, 1","cmuo4yz1j01ce20jx21hnwr7w","5, 4, 3, 2, 1","cmuo4yz1i01cd20jxb9hn49q2","5, 4, 3, 2, 1, 0","cmuo4yz1g01cc20jxf8s639pt","1, 2, 3, 4, 5","cmuo4yz1401c620jx8eswbttl","for (int i = 0; i < 3; i++)\n{\n Console.WriteLine(i);\n}","A `for` loop repeats a block of code a set number of times. Its header has three parts, split by semicolons:\n\n- **Initializer** — runs one time, before the loop begins. It usually creates a counter variable.\n- **Condition** — checked before each run of the loop. The loop keeps going while this is `true`.\n- **Iterator** — runs after each pass. It usually adds to or takes from the counter.\n\n```csharp\nfor (int i = 0; i < 5; i++)\n{\n Console.WriteLine(i);\n}\n```","The counter i starts at 0. The loop keeps going as long as i < 3. After each print, i increases by 1. So it prints 0, then 1, then 2, and stops before printing 3 because 3 < 3 is false.",[184,185,186,187],{"_29":193,"_48":104,"_33":-5,"_50":10},{"_29":191,"_48":192,"_33":-5,"_50":53},{"_29":190,"_48":106,"_33":-5,"_50":53},{"_29":188,"_48":189,"_33":-5,"_50":53},"cmuo4yz1c01ca20jxd6v9ox0l","1, then 2, then 3","cmuo4yz1a01c920jxuqghubvf","cmuo4yz1801c820jxqsxaphma","3, then 2, then 1","cmuo4yz1701c720jx4tur15qt","cmuo4yz0w01c120jxuwnf0lpn","In a C# 'for' loop, what three parts go inside the round brackets, separated by semicolons?","A for statement's header has three parts separated by semicolons: the initializer (runs once at the start, usually creating a counter variable), the condition (checked before every pass through the loop), and the iterator (runs after every pass, usually changing the counter).",[198,199,200,201],{"_29":208,"_48":209,"_33":-5,"_50":53},{"_29":206,"_48":207,"_33":-5,"_50":53},{"_29":204,"_48":205,"_33":-5,"_50":53},{"_29":202,"_48":203,"_33":-5,"_50":10},"cmuo4yz1301c520jxc2ddxqqi","A starting step, a check, and an update step","cmuo4yz1101c420jxljboaygf","A variable, a data type, and a method name","cmuo4yz1001c320jxfbwhk9zo","A check, the loop's body, and a return statement","cmuo4yz0y01c220jxhmiyukph","A start number, an end number, and a name for the loop","cmumydl3v012u3ijxog0cl4vd","What is wrong with the code below?","for (int i = 10; i >= 0; i++)\n{\n Console.WriteLine(i);\n}","If a loop's condition never becomes `false`, the loop never stops. This often happens because you forgot to update the counter, or you updated it in a way that never makes the condition false, or there is a mistake in the condition itself.\n\n**Watch out:** Always check that the step that updates your variable (the iterator) actually moves it toward making the condition `false`.","The iterator i++ increases i on every pass, but the condition i >= 0 only becomes false if i drops below 0. Since i keeps growing larger and larger starting from 10, it never becomes negative, so i >= 0 stays true forever. This is an infinite loop caused by using the wrong direction of iterator (i++ instead of i--) for a countdown.",[216,217,218,219],{"_29":226,"_48":227,"_33":-5,"_50":53},{"_29":224,"_48":225,"_33":-5,"_50":53},{"_29":222,"_48":223,"_33":-5,"_50":53},{"_29":220,"_48":221,"_33":-5,"_50":10},"cmumydl43012y3ijxw93xdu2p","It never stops, because i starts at 10 and i++ keeps making it bigger, so the check i >= 0 stays true","cmumydl40012x3ijx14d3txkk","It crashes with an error the moment i goes above 10","cmumydl3y012w3ijxpmvxm5ag","It shows 10 down to 0 and then stops","cmumydl3x012v3ijx1w6slii9","It will not build, because you cannot use i++ together with a >= check","cmumydl3o012p3ijx9te9fxwg","int i = 0;\nwhile (i < 5)\n{\n Console.WriteLine(i);\n}","If a loop's condition never becomes `false`, the loop never stops. This often happens because you forgot to update the counter, or you updated it in a way that never makes the condition false, or there is a mistake in the condition itself.\n\n```csharp\nint i = 0;\nwhile (i < 5)\n{\n Console.WriteLine(i);\n // missing i++ here — the condition is always true, so the loop never stops\n}\n```","Nothing inside the loop body ever changes i, so the condition i < 5 stays true forever, since i stays 0. This creates an infinite loop that keeps printing 0 forever and never lets the program move on. Unlike a for loop, a while loop does not have a built-in iterator step, so you must remember to update the counter yourself inside the body.",[233,234,235,236],{"_29":243,"_48":244,"_33":-5,"_50":10},{"_29":241,"_48":242,"_33":-5,"_50":53},{"_29":239,"_48":240,"_33":-5,"_50":53},{"_29":237,"_48":238,"_33":-5,"_50":53},"cmumydl3u012t3ijxry70o86l","It will not build, because a while loop must have i++ written in its round brackets","cmumydl3s012s3ijx2opysb54","It crashes right away with a stack overflow error","cmumydl3r012r3ijxqyl89fwx","It shows 0, 1, 2, 3, 4 and then stops on its own","cmumydl3p012q3ijxoce5m15s","It never stops, because i never changes inside the loop, so the check i < 5 is always true","email"]