C Programming Language :: list = list->Next = malloc(sizeof(node));


	list = list->Next = malloc(sizeof(node)); Given: #include <stdlib.h> typedef struct Node { unsigned char Data[SIZE]; struct Node Next } node_t, list_t; list_t list; Does list = list->Next = malloc(sizeof(node)); cause UB? Is it any different (neglecting style) from list->Next = malloc(sizeof(node)); list = list->Next; ? In the former, apparently I only write to l itself once, and only read l to determine what l->next is. So apparently it is OK to do that within one sequence point. Am I wrong? "Army1987" <please.@for.it> writes: > Does > list = list->Next = malloc(sizeof(node)); > cause UB? Is it any different (neglecting style) from > list->Next = malloc(sizeof(node)); > list = list->Next; > ? We've had several extensive threads on this over the years. I started one of them. I'd suggest trying to find the earlier discussions instead of starting a new one. -- Ben Pfaff http://benpfaff.org On May 25, 11:39 am, "Army1987" <please.@for.it> wrote: > In the former, apparently I only write to l itself once, and only > read l to determine what l->next is. So apparently it is OK to do > that within one sequence point. Am I wrong? You are not sure whether your code invokes undefined behavior. It is trivial to change it so that it will clearly not invoke undefined behavior. If you don't change it, then yes, you are wrong. On May 25, 10:39 pm, "Army1987" <please.@for.it> wrote: > Does > list = list->Next = malloc(sizeof(node)); > cause UB? Yes. 'list' is both written, and read for a purpose other than to determine the value to be written, without an intervening sequence point. (That purpose is for determining the value of list->Next). Although '=' is right-left associative, that doesn't mean right-left order of evaluation. The associativity means that A=B=C is read as A=(B=C) rather than (A=B)=C. BTW. the latter would always cause undefined behaviour because A would be written twice without a sequence point. The code is like: void x = malloc( sizeof(node) ); list = x, list->Next = x; except that there is no sequence point where I have put the comma operator. So you could end up with the result of malloc having "->Next" applied to it. (Or any other result, since this means the behaviour is undefined). In article <1180313678.936468.323@n15g2000prd.googlegroups.com>, Old Wolf <oldw@inspire.net.nz> wrote: >> Does >> list = list->Next = malloc(sizeof(node)); >> cause UB? >Yes. 'list' is both written, and read for a purpose >other than to determine the value to be written, >without an intervening sequence point. (That purpose >is for determining the value of list->Next). That's not quite right, because the same is true of list = list->Next; in which list is read to determine the value of list->Next, but that is done in order to determine the value to be written, so there is no problem. The problem with the quoted example is that list is read in order to determine the location* to be written in the right-hand assignment. -- Richard -- "Consideration shall be given to the need for as many as 32 characters in some alphabets" - X3.4, 1963.