int i = 0;
i = ++i;
preincrement and assignment
Author
Let's talk about this snippet of code:
No matter how you look at it I'm not sure how you can get a i == 2 after doing a ++ on a i of 0...
Unless you mean it assigns a ++'ed i to i then does the ++ on the already assigned i, but it's hardly logical
Unless you mean it assigns a ++'ed i to i then does the ++ on the already assigned i, but it's hardly logical
the ++ operator when placed before the variable does not return a value unless the variable itself has been increased by one. So I dont think saying the expression's value is anything if it isnt already evaluated, and by that point the registers already have the pointer to i waiting for the right side of the assignment to be caclulated.
im not sure if that answered the question/statement, maybe I missed something lol
im not sure if that answered the question/statement, maybe I missed something lol
++i is evaluated first and returns a reference to itself which it is assigned to...?
i dont see what the problem is?
i dont see what the problem is?
According to the standard, this is undefined, as you have pointed out correctly.
However, other than with the similar example of post-increment, I think that the result is nevertheless well-defined here, since the compiler really has no other choice but to evaluate ++i first.
If it was decided to do the assignment first, it would still have to evaluate the increment before that, as it needs the value to assign, which it can't get without evaluating the pre-increment first.
However, other than with the similar example of post-increment, I think that the result is nevertheless well-defined here, since the compiler really has no other choice but to evaluate ++i first.
If it was decided to do the assignment first, it would still have to evaluate the increment before that, as it needs the value to assign, which it can't get without evaluating the pre-increment first.
Quote:Original post by DevFred
4. If we further assume that the side effect of assignment happens before the side effect of increment, we get 2 as a value of i at the semicolon.
Nobody I have discussed this issue with seems to agree with me on point 4. Share your thoughts :)
Using your code as a starting point and producing this sample program:
int main(int, char **) { int i = 0; i = ++i; std::cout << i; return 0;}MSVC 2008 spits out 1 for the value of i. So given the fact that a compiler has disagreed with your logic, I think you're going to have to live with the fact that 4 doesn't hold true.
Author
Quote:Original post by SiCrane
So given the fact that a compiler has disagreed with your logic, I think you're going to have to live with the fact that 4 doesn't hold true.
So the output of one compiler defines undefined behavior? Hmmm... :)
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