gcc error operation on may be undefined Verbena Alabama

Address 1657 7th St S, Clanton, AL 35045
Phone (205) 755-4511
Website Link http://www.computerdoctoralabama.com

gcc error operation on may be undefined Verbena, Alabama

Compiler might translate it to: Code: 1. I don't know what other compilers would do. maybe you compiler set a prefix "_cdecl" for every standard function. (you could find that by opening a header file). For example: int i = 5; i = i * i; is perfectly legal, and the value of i is accessed twice (the compiler might optimize accesses and remember the value

You are wrong to think that. Code: i = (i + 1) % 10; should work. asked 2 years ago viewed 2004 times active 2 years ago Linked 2 overloading [][] operators in c++ 14 Why I got “operation may be undefined” in Statement Expression in C++? You should be careful about depending on such things, though, since they likely won't work the same way on other implementations. -- Keith Thompson (The_Other_Keith) ks***@mib.org San Diego Supercomputer Center

so I think the real reason is something about sequence point. You should consult the compiler's reference to figure out what semantics it is given (or not given, as the error message seems to suggest) to it. Hello, #include int main(int argc, char *argv[]) { int x = 1; printf("%d %d %d\n", ++x, x, x++); return 0; } Why does the above code when compiled with all Is there a possibility for an implementation to choose some behavior for a Standard-specified Undefined behaviour and document it?

There are 3 main tasks you should consider: 1. Therefore, we must do this. Browse other questions tagged c pointers gcc or ask your own question. Anyway, see the FAQ: http://www.eskimo.com/~scs/C-faq/q3.2.html -David Nov 15 '05 #2 P: n/a Antonio Contreras jimjim wrote: Hello, #include int main(int argc, char *argv[]) { int x = 1; printf("%d %d

This now makes sense to me. In previous post Why I got "operation may be undefined" in Statement Expression in C++? Why when run the output is: "3 2 1"? Therefore, we must do this.

Depending on the order, elems_[ptr] = data[ptr++] yields different results. Although my email address says spam, it is real and I read it. If you just use + 1 instead of ++, it will work fine. Last edited by mina86; 11-08-2013 at 11:07 AM.

operation on `x' may be undefined? Given that my compiler is the gcc, could you have a guess why is it 3 2 1 and not 1 2 3? ;-) If I were more familiar with the Introduction to Linux - A Hands on Guide This guide was created as an overview of the Linux Operating System, geared toward new users as an exploration tour and getting started In order to make sure that I understand what you are saying, is the sentence below equivalent with yours?

more stack exchange communities company blog Stack Exchange Inbox Reputation and Badges sign up log in tour help Tour Start here for a quick overview of the site Help Center Detailed Undefined behavior -- you are attempting to modify the value of an object more than once between sequence points, and the Standard imposes no requirement on how to handle that behavior. Word for someone who keeps a group in good shape? Appease Your Google Overlords: Draw the "G" Logo How to show hidden files in Nautilus 3.20.3 Ubuntu 16.10?

You probably meant ptr->count. Similar topics += operation insert() operation on vector, deque, list string operation?? Because then the issue could be solved upstream by the author and is available to everyone using it. If between sequence points an object's value is to be modified , its value should only be read once.

The code is broken because it does something that the C standard says it undefined behavior, a very specific term. If between sequence points an object's value is to be modified , its value should only be read once. Change your code to something like this: trackTail[nodeNumber - 1] = (trackTail[nodeNumber - 1] + 1) % 10; share|improve this answer edited May 16 '12 at 17:03 answered May 16 '12 The transformation T on the set of all continuous functions that is defined by T(f) = f (1) is a linear transformation.

It is C not C++ so I am the wrong person to comment on it at all (otherwise my opinion would be not to use unsafe pointer arithmetic at all but The code invokes undefined behavior, which means it could behave differently with the next release of gcc or with the phase of the moon. Compute the address to which the value must be written (&a + i * sizeof(a[0])) This things can be done in whatever order the compiler designers decide suits best their target Save reminder of dividing α by 10 to register β. 4.

Why doesn't ${@:-1} return the last element of [email protected]? Whatever the program actually does is irrelevant and off-topic here. It's like you're assigning i = ++i; which is also undefined behaviour. I think this is caused by the Function Calling Conventions used by your complier.

why no sequence points? and then the i++ would have been performed before proceeding to the next statement in the code. I have to patch it.