Java Overflow Compiler Error
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Compiler Error Java
Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 4.7 examples of runtime errors in java million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up Compile time vs run time errors [duplicate] up vote 3 down vote favorite 1 Possible Duplicate: Runtime vs Compile compile time error java time How should I know whether a specific line of code in Java may throw a compile time or run-time error? Assuming that the specific line of code anyway throws and error. java runtime compile-time share|improve this question edited May 6 '13 at 22:09 Undo♦ 18.5k157399 asked Oct 14 '12 at 12:58 uml 4503921 marked as duplicate by home, Junuxx, Denys Séguret, S.L. Barth, dSquared Oct 14 '12 at 14:35 This question has
Runtime Error Vs Compile Error Java
been asked before and already has an answer. If those answers do not fully address your question, please ask a new question. 1 stackoverflow.com/questions/846103/runtime-vs-compile-time –Matt Whipple Oct 14 '12 at 13:01 add a comment| 5 Answers 5 active oldest votes up vote 11 down vote accepted In Eclipse, compile time errors will be underlined in red. A compile time error is an error that is detected by the compiler. Common causes for compile time errors include: Syntax errors such as missing semi-colon or use of a reserved keyword (such as 'class'). When you try and access a variable that is not in scope. When you declare multiple objects with the same name. If the compiler detects any errors during compilation it will fail to build a new assembly (or class file in Java). Even if your code has no compile time errors, errors can still occur on run-time. Errors such as 'logic errors' and 'runtime errors'. A good example of a runtime error is as followed: Pretend you're going to store an item in an array at index 5 but the array's size is only 4. The compiler won't detect an error here because it understands that the array size is subject to change but on run-time you'll be thrown an exception. To detect which li
here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the java compilation error cannot find symbol workings and policies of this site About Us Learn more about compile time error and runtime error in c Stack Overflow the company Business Learn more about hiring developers or posting ads with us Stack Overflow Questions
Compile Time Error In C
Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 4.7 million programmers, just like you, helping each other. http://stackoverflow.com/questions/12882448/compile-time-vs-run-time-errors Join them; it only takes a minute: Sign up Why does Java have an “unreachable statement” compiler error? up vote 71 down vote favorite 7 I often find when debugging a program it is convenient, (although arguably bad practice) to insert a return statement inside a block of code. I might try something like this in Java .... class http://stackoverflow.com/questions/3795585/why-does-java-have-an-unreachable-statement-compiler-error Test { public static void main(String args[]) { System.out.println("hello world"); return; System.out.println("i think this line might cause a problem"); } } of course, this would yield the compiler error. Test.java:7: unreachable statement I could understand why a warning might be justified as having unused code is bad practice. But I don't understand why this needs to generate an error. Is this just Java trying to be a Nanny, or is there a good reason to make this a complier error? java language-design unreachable-statement share|improve this question edited Sep 25 '10 at 21:25 asked Sep 25 '10 at 21:15 Mike 15.5k50138177 9 Java isn't entirely consistent about this either. For some control flows that cause dead code, but Java doesn't complain. For others it does. Determining what code is dead is an uncomputable problem. I don't know why Java decided to start something it couldn't finish. –Paul Draper Mar 21 '13 at 20:11 add a comment| 8 Answers 8 active oldest votes up vote 53 down vote accepted Because unreachable code is meaningless to
may consist of a limited amount of address space, often determined at the start of the program. The size of the https://en.wikipedia.org/wiki/Stack_overflow call stack depends on many factors, including the programming language, machine architecture, multi-threading, and amount of available memory. When a program attempts to use more space than is https://books.google.com/books?id=Uv7WBQAAQBAJ&pg=PA61&lpg=PA61&dq=java+overflow+compiler+error&source=bl&ots=vKEWVA3af-&sig=BNPS3TYNFHgNV2nQJp3mXhiYwUs&hl=en&sa=X&ved=0ahUKEwj54--V3d_PAhUH9WMKHcOWBBQQ6AEIYTAJ available on the call stack (that is, when it attempts to access memory beyond the call stack's bounds, which is essentially a buffer overflow), the stack is said time error to overflow, typically resulting in a program crash.[1] Contents 1 Infinite recursion 2 Very deep recursion 3 Very large stack variables 4 See also 5 References 6 External links Infinite recursion[edit] Main article: Infinite recursion The most common cause of stack overflow is excessively deep or infinite recursion, in which a function calls itself so many compile time error times that the space needed to store the variables and information associated with each call is more than can fit on the stack.[2] An example of infinite recursion in C. int foo() { return foo(); } The function foo, when it is invoked, continues to invoke itself, allocating additional space on the stack each time, until the stack overflows resulting in a segmentation fault.[2] However, some compilers implement tail-call optimization, allowing infinite recursion of a specific sort—tail recursion—to occur without stack overflow. This works because tail-recursion calls do not take up additional stack space.[3] C compiler options will effectively enable tail-call optimization; compiling the above simple program using gcc with -O1 will result in a segmentation fault, but not when using -O2 or -O3, since these optimization levels imply the -foptimize-sibling-calls compiler option. Other languages, such as Scheme, require all implementations to include tail-recursion as part of the language standard.[4] Very deep recursion[edit] A recursive function that terminates in theory but causes a call stack buffe