Doubt it; I believe ISO does NOT define a=...a++
|
| Author | Content |
|---|---|
| dwheeler Feb 28, 2005 8:44 AM EDT |
Just because compilers do something different does NOT mean that a standard has been violated. You need to cite WHERE in the standard it says something has to be done in a certain way. I don't have the ISO specs for C++, but I believe the construct a=...a++ is NOT defined, and thus, there IS NO VIOLATION. In the construct a=..a++..;, the post-increment happens sometime after the read of a, and before the statement is completed, but whether it happens before or after the assignment is, I believe, compiler-dependent. And intentionally so. It certainly varies in the real world, and has for a long time. Portable programs should NEVER use pre/post-increment and an assignment on the same variable in the same statement, and I believe that's really old advice. C# is a different language anyway. It's not at all unreasonable for C# to do one thing, and C++ to do something different. Having a C# example isn't convincing. |
| wwrafter Feb 28, 2005 11:04 AM EDT |
While I don't actually have the standard spec either, all of the sources that I've looked so far specify that the "++" operator has a higher precedence than the "=" operator. MS C++ is wrong. |
| JonPryor Feb 28, 2005 11:34 AM EDT |
Please do us a favor and consult Google. a = b++ * a++; is undefined behavior. Meaning that you can't portably rely on the result of this expression. Operator precedence has nothing to do with it; the C and C++ standards require that a variable can only be modified once within an expression. Attempts to modify it more than once are undefined. See: http://www.eskimo.com/~scs/C-faq/q3.1.html http://www.eskimo.com/~scs/C-faq/q3.2.html http://www.eskimo.com/~scs/C-faq/q3.8.html Remember, for your code to work `i = i++` must be valid, which would make `a[i] = i++` valid, and both of those ARE NOT VALID. C# is completely different, as the order of evaluation within an expression is fully defined. |
| chris Feb 28, 2005 12:33 PM EDT |
The standard does NOT forbid multiple modifications in the same expression. It forbids multiple modifications to the same scalar value without an intervening sequence point between the modifications. A sequence point is introduced after each complete expression, but they are also introduced by function calls and the builtin comma, &&, and || operators, among some other mechanisms. (Read section 1.9 and chapter 5 in the standard.) Thus it is possible to have a single expression with multiple sequence points, and therefore possible to safely have multiple modifications to the same value in the same expression. For example, even though there are 2 modifications to a, the comma operator makes the following expression valid and well-defined: a = ++a, a + 1; Chris |
| leimy Mar 01, 2005 10:03 AM EDT |
a = b++ * a++; Is not defined by the C++ standard. You are invoking undefined behavior. a = a++; alone is not defined. In short the C++ language standard basically says that '=' is not a sequence point. All side effects must be complete before the next sequence point in the code but it doesn't specify when each side effect's evaluation has to happen during the evaluation of the previous expression. As such the ONLY sequence point is the ';' and the incrementing of 'a' can happen before or after the assignment to 'a' . assume the following: a = 3; a = a++; Compiler 1 returns "4" for 'a'. Compiler 2 returns "3" for 'a'. Neither is incorrect but your code is not well formed. Does that clear it up a bit? I'd steer clear of all postfix operators because people really don't know how to use them properly and their value is questionable to begin with. Take the following: std::vector::iterator it = string.begin(); for (; it != string.end(); it++) { //... code } If you don't know why the above code is worse than: for (l it != string.end(); ++it) { //... code } then you probably shouldn't be using postfix ++ or --. Many C++ book authors choose to use i++ rather than ++i. I don't know why. I think it confuses people into thinking the wrong thing about for loops and side-effects in an expression are generally not well understood throughout the C++ and C communities. Dave |
Posting in this forum is limited to members of the group: [ForumMods, SITEADMINS, MEMBERS.]
Becoming a member of LXer is easy and free. Join Us!