Question: I tried to open pentaho spoon but it doesnt launch . The error is "System does not find the path"
I use "-Xms1024m" and C:\Program Files\Java\jre1.8.0\_261 path as an environment but still does not working. Can someone help me please?

[![enter image description here](https://i.stack.imgur.com/8hE1X.png)](https://i.stack.imgur.com/8hE1X.png)

Answer: 
The structure returned by `foo` is just a value (also called an rvalue). It is not an object and does not have any lifetime.

Consider a function `int foo(void) { return 3; }`. This returns an `int` value of 3, and we would not expect to be able to take its address, as in `printf("%p", (void *) &foo());`. The 3 is just a value used in the computer with no associated storage.

Similarly, given `struct S { int x, y; }`, `struct S foo(void) { return (struct S) { 3, 4 }; }` returns a `struct S` value containing 3 and 4. Although we often think of structures as layouts of memory, the C standard treats this return value as just a value. It is a compound value, having multiple parts, but it is just a value with no associated storage. It is not an object in the C model.

Also similarly, given `struct S { int x, y[1]; }`, `struct S foo(void) { return (struct S) { 3, { 4 } }; }` returns a `struct S` value containing 3 and an array containing 4. Here the C standard painted itself into a corner. It wanted to support returning structures from functions, but, when you access an array, as in `foo().y[0]`, the rule currently in C 2018 6.3.2.1 3 says the array is converted to a pointer to its first element. A pointer has to point to storage, so there has to be some object to point to. I suppose one solution might have been to say you cannot use arrays within such structure values individually. (You could use the return value by copying it into an object with `struct S x = foo();` and then using `x`.) However, the solution the C committee adopted was to define a temporary lifetime for such structures. Their definition for that, in C 2018 6.2.4 8, defines a temporary lifetime only for structures and unions that contain an array member.

However, in your code, this is not a concern. Because your `foo(2, 4)` returns a value, you can use that value as you desire; `foo(2, 4).x` works because it takes the `x` member of the value. It does not need to worry about the lifetime of any object because there is no object involved.
From the C Standard

> 
> 4 **A full expression is** an expression that is not part of another
> expression or of a declarator. Each of the following is a full
> expression: an initializer that is not part of a compound literal;
> **the expression in an expression statement**; the controlling expression of a selection statement (if or switch); the controlling
> expression of a while or do statement; each of the (optional)
> expressions of a for statement; the (optional) expression in a return
> statement. There is a sequence point between the evaluation of a full
> expression and the evaluation of the next full expression to be
> evaluated.
> 
> 
> 

The lifetime of a temporary object ends when the evaluation of the containing full expression ends.

So within this expression statement

```
a = foo(2, 4).x;

```

the returned object of the structure type is alive and the value of its data member `x` is assigned to the variable `a` that is declared in main and has the block scope of main.