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# define side effects

 P: n/a This is one of the posts that i got. ------------------------------ A "side effect" of an operation is something that *happens*, not something that *is produced*. Examples: In the expression 2+2, the value 4 *is produced*. Nothing *happens*. Thus, 4 is the value of the expression, and it has no side effects. In the expression g=2.0, the value 2.0 is produced. What *happens* is that 2.0 is assigned to g. Thus, 2.0 is the value of the expression, and its side effect is to assign 2.0 to g. In the expression (int)g, the value 2 is produced. Nothing happens. Thus, 2 is the value of the expression (int)g, and it has no side effects. In the expression (a=1,++a), the value 2 is produced. What happens is that first 1 is assigned to a, and then a is incremented; those are the side effects of the expression. ----------------------------- This seems to be a good definition in most of the cases. I argue that in k=(int)2.0 , 2 is produced and the truncation of 2.0 -> 2 is that happens. I am really confused. Can some one give me a more clear definition of side effects ? If i apply the same definition to i=j++ where j=3 then i=3 and j=4 are the side effects and value 3,4 are produced. Am i correct ? Can someone throw more light on how to define side effects without ambiguity ? - Nik Nov 14 '05 #1
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 P: n/a ni*****@gamebox.net (Niklaus) wrote: This is one of the posts that i got. ------------------------------ A "side effect" of an operation is something that *happens*, not something that *is produced*. Examples: In the expression 2+2, the value 4 *is produced*. Nothing *happens*. Thus, 4 is the value of the expression, and it has no side effects. In the expression g=2.0, the value 2.0 is produced. What *happens* is that 2.0 is assigned to g. Thus, 2.0 is the value of the expression, and its side effect is to assign 2.0 to g. In the expression (int)g, the value 2 is produced. Nothing happens. Thus, 2 is the value of the expression (int)g, and it has no side effects. In the expression (a=1,++a), the value 2 is produced. What happens is that first 1 is assigned to a, and then a is incremented; those are the side effects of the expression. ----------------------------- This seems to be a good definition in most of the cases. I argue that in k=(int)2.0 , 2 is produced and the truncation of 2.0 -> 2 is that happens. No; the 2.0 is not changed. Nothing happens "behind the scenes"; the value 2.0 does get truncated, but only for the direct reason of calculating the value of the expression. Assigning this truncated value to k _is_ a side effect. I am really confused. Can some one give me a more clear definition of side effects ? Well, according to the Standard, # [#2] Accessing a volatile object, modifying an object, # modifying a file, or calling a function that does any of # those operations are all side effects, which are changes # in the state of the execution environment. Whether that is more clear, well... it's unambiguous, anyway. If i apply the same definition to i=j++ where j=3 then i=3 and j=4 are the side effects and value 3,4 are produced. Am i correct ? No; 4 is never produced. j is increased to 4, but that value is never passed on to any other sub-expression; its previous value, 3, is. Richard Nov 14 '05 #2

 P: n/a rl*@hoekstra-uitgeverij.nl (Richard Bos) wrote in message news:<40***************@news.individual.net>... ni*****@gamebox.net (Niklaus) wrote: This is one of the posts that i got. ------------------------------ A "side effect" of an operation is something that *happens*, not something that *is produced*. Examples: In the expression 2+2, the value 4 *is produced*. Nothing *happens*. Thus, 4 is the value of the expression, and it has no side effects. In the expression g=2.0, the value 2.0 is produced. What *happens* is that 2.0 is assigned to g. Thus, 2.0 is the value of the expression, and its side effect is to assign 2.0 to g. In the expression (int)g, the value 2 is produced. Nothing happens. Thus, 2 is the value of the expression (int)g, and it has no side effects. In the expression (a=1,++a), the value 2 is produced. What happens is that first 1 is assigned to a, and then a is incremented; those are the side effects of the expression. ----------------------------- This seems to be a good definition in most of the cases. I argue that in k=(int)2.0 , 2 is produced and the truncation of 2.0 -> 2 is that happens. No; the 2.0 is not changed. Nothing happens "behind the scenes"; the value 2.0 does get truncated, but only for the direct reason of calculating the value of the expression. Assigning this truncated value to k _is_ a side effect. I am really confused. Can some one give me a more clear definition of side effects ? Well, according to the Standard, # [#2] Accessing a volatile object, modifying an object, # modifying a file, or calling a function that does any of # those operations are all side effects, which are changes # in the state of the execution environment. Whether that is more clear, well... it's unambiguous, anyway. If i apply the same definition to i=j++ where j=3 then i=3 and j=4 are the side effects and value 3,4 are produced. Am i correct ? No; 4 is never produced. j is increased to 4, but that value is never passed on to any other sub-expression; its previous value, 3, is. Richard regarding side - effects what exactly does this sequence 'point mean' Nov 14 '05 #3

 P: n/a Gautam scribbled the following: rl*@hoekstra-uitgeverij.nl (Richard Bos) wrote in message news:<40***************@news.individual.net>... ni*****@gamebox.net (Niklaus) wrote: > This is one of the posts that i got. > ------------------------------ > A "side effect" of an operation is something that > *happens*, not something that *is produced*. Examples: > In the expression 2+2, the value 4 *is produced*. Nothing > *happens*. > Thus, 4 is the value of the expression, and it has no side effects. > In the expression g=2.0, the value 2.0 is produced. What *happens* > is that 2.0 is assigned to g. Thus, 2.0 is the value of the > expression, > and its side effect is to assign 2.0 to g. > In the expression (int)g, the value 2 is produced. Nothing happens. > Thus, 2 is the value of the expression (int)g, and it has no side > effects. > In the expression (a=1,++a), the value 2 is produced. What happens > is that first 1 is assigned to a, and then a is incremented; those are > the side effects of the expression. > ----------------------------- > > This seems to be a good definition in most of the cases. > > I argue that in k=(int)2.0 , 2 is produced and the truncation of 2.0 > -> 2 is that happens. No; the 2.0 is not changed. Nothing happens "behind the scenes"; the value 2.0 does get truncated, but only for the direct reason of calculating the value of the expression. Assigning this truncated value to k _is_ a side effect. > I am really confused. Can some one give me a more clear definition > of side effects ? Well, according to the Standard, # [#2] Accessing a volatile object, modifying an object, # modifying a file, or calling a function that does any of # those operations are all side effects, which are changes # in the state of the execution environment. Whether that is more clear, well... it's unambiguous, anyway. > If i apply the same definition to i=j++ where j=3 then i=3 and j=4 > are the side effects and value 3,4 are produced. Am i correct ? No; 4 is never produced. j is increased to 4, but that value is never passed on to any other sub-expression; its previous value, 3, is. regarding side - effects what exactly does this sequence 'point mean' It's a point during the evaluation of an expression, when all side effects are guaranteed to have taken place. Sequence points include: - The terminating ; in a statement - The && and || operators - The ?: operator - The , operator Also, AFAIK when a function is called, its entry point forms a sequence point for the expressions in its arguments. -- /-- Joona Palaste (pa*****@cc.helsinki.fi) ------------- Finland --------\ \-- http://www.helsinki.fi/~palaste --------------------- rules! --------/ "You could take his life and..." - Mirja Tolsa Nov 14 '05 #4

 P: n/a Simply The same as in medicine. A drug cures one problem, but causes another. In code you fix one bug, but cause another. Nov 14 '05 #5

 P: n/a On Sat, 8 May 2004, Neil Kurzman wrote only: Simply The same as in medicine. A drug cures one problem, but causes another. In code you fix one bug, but cause another. An interesting quotation, but why do you say so? If this is supposed to be a new topic of discussion (incidentally, one more suited to comp.programming than comp.lang.c, which is dedicated specifically to fixing bugs in *C* code :) then you might have considered starting a new thread rather than piggybacking on an existing one. We've seen people here before who apparently thought they should "conserve threads" by posting irrelevant replies to old threads; that's really not necessary. And if you *didn't* mean your reply to be irrelevant to the old thread, then you ought to have quoted some context so people could tell to what you were responding. Google "usenet faq" for more information. -Arthur Nov 14 '05 #6

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