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I reduced this:
struct A
{
int * x;
A() : x( x = new int() )
{
}
};
to the following:
int m = m = 3;
//or
struct X;
//...
X x = x = X();
Seems legal to me. I don't see why you'd want to do it, but is it legal? Are there cases where you'd want to do this (not the int case, I realize that's completely useless)?
859
The Original Value of x is returned first then, x is incremented by 1. In the following example, x is incremented before the value it holds is used. Hope this helps! Originally Answered: What is the difference between X++ and ++X? X++ and ++x are different types of increment operators. 1 . X++ or in post increment operator,
Let’s make it pretty simple. Both x++; & ++x; are syntactic sugar for the expression, x=x+1; x++ is Post-Increment, which in simple terms means that “First Use/Execute Then Change (present value)”. ++x is Pre-Increment, which in simple terms means that “First Change (present value) Then Execute/Use”.
Obviously, b is first used then changed. This will produce an output 160. Here b is first changed then it is used to evaluate the expression. Hope it Helps. Cheers!! ++x changes the value of x before it is used (pre-increment). x++ changes the value of x after it is used (post-increment).
How to solve x plus x in algebra? To solve x plus x, imagine the variable x as a familiar object, for example an apple. So now instead of x plus x, you have one apple plus one apple which is obviously two apples. Therefore if we think of x instead of apples, it is clear that the answer x plus x is 2x.
It depends on how you define "legal". It will compile; that doesn't mean that it is guaranteed to work.
Until the full statement X x = ... executes, x is uninitialized. It is not an X yet. Therefore, performing x = X() means to create a temporary X and call X::operator=(const X&) on the uninitialized variable x.
Calling a function on a non-POD class instance that has not been initialized (who's constructor has not yet been called) yields undefined behavior. If X is a POD type (or trivial in C++11), then it will work. But otherwise no.
77
It's syntactically legal, but will result in undefined behavior at runtime. In a statement like:
X x = x = X();
the second = is assignment, and it assigns to an uninitialized variable. (The first = is simply the syntax for saying that what follows should be used for copy initialization; it isn't assignment.)
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