Help

[Red Dragon]

Can anybody tell me
if x = x +1 is x++
What is x = x + 2 ?

Thanks
Khoon.

 

[Giovanni]

Can anybody tell me
if x = x +1 is x++
What is x = x + 2 ?

If applicable is: x += 2;
Ciao
Giovanni

 

[Robert Gamble]

Can anybody tell me
if x = x +1 is x++

The effect may be equivalent but in the former x is evaluated twice
whereas in the latter x is evaluated once. Additionally, the result of
evaluating the former is the same as ++x, not x++.

What is x = x + 2 ?

x += 2.

Robert Gamble

 

[Jordan Abel]

The effect may be equivalent but in the former x is evaluated twice
whereas in the latter x is evaluated once.

actually, no. it's read once and modified once in both cases.

 

[Red Dragon]

Can anybody tell me

actually, no. it's read once and modified once in both cases.

I got it. Thanks for your help.

Regards,
Khoon.

 

[Robert Gamble]

actually, no. it's read once and modified once in both cases.

As I said, x is *evaluated* twice in the first expression, once in the
second.

Robert Gamble

 

[slebetman]

As I said, x is *evaluated* twice in the first expression, once in the
second.

Robert Gamble

*Evaluated * by the compiler? Or machine code? Most compilers I know
generate the same assembly in both cases, even with optimisation turned
off.

 

[Robert Gamble]

*Evaluated * by the compiler? Or machine code?

Evaluated by the abstract machine as per the C Standard.

Most compilers I know generate the same assembly in both cases, even with
optimisation turned off.

An implementation does not have to evaluate part of an expression if
the result of the expression can be deduced without doing so and if the
program cannot tell the difference, irrespective of optimizations. In
other words, x may be evaluated a _maximum_ of 2 times in the first
expression and once in the second. It is therefore obviously possible
that both operations result in x being evaluated the same number of
times, namely one.

Robert Gamble

 

[Jordan Abel]

Evaluated by the abstract machine as per the C Standard.


An implementation does not have to evaluate part of an expression if
the result of the expression can be deduced without doing so and if the
program cannot tell the difference, irrespective of optimizations. In
other words, x may be evaluated a _maximum_ of 2 times in the first
expression and once in the second. It is therefore obviously possible
that both operations result in x being evaluated the same number of
times, namely one.

Robert Gamble

The compiler can evaluate it as many or as few times as it pleases, as
long as it acts "as if" it only did it as many times as you said. I take
"evaluate" to mean "get the value of", and assigning to a variable does
not "evaluate" it, it "assigns" it.

 

[Robert Gamble]

The compiler can evaluate it as many or as few times as it pleases, as
long as it acts "as if" it only did it as many times as you said. I take
"evaluate" to mean "get the value of", and assigning to a variable does
not "evaluate" it, it "assigns" it.

Well, you are wrong on both counts and it doesn't matter what you
consider evaluate to mean. If you care to actually read what the
Standard has to say you should start by reading the following
paragraphs:

5.1.2.3p3
6.5.2.4p2
6.5.16.2p3

Robert Gamble.

 

[Mark McIntyre]

I take
"evaluate" to mean "get the value of", and assigning to a variable does
not "evaluate" it, it "assigns" it.

In order to assign the value to a new variable, it has to evaluate the
previous variable first. Computing isn't magic.

 

[Keith Thompson]

In order to assign the value to a new variable, it has to evaluate the
previous variable first. Computing isn't magic.

Given

x = y;

y is evaluated, but x is not. (Specifically, the previous value of x
is not evaluted.)

 

[Mark McIntyre]

Given

x = y;

y is evaluated, but x is not. (Specifically, the previous value of x
is not evaluted.)

I agree.

 

[Jordan Abel]

I agree.

....that's what i said. why's he right and i'm not?

 

[SM Ryan]

# >>Given
# >>
# >> x = y;
# >>
# >>y is evaluated, but x is not. (Specifically, the previous value of x
# >>is not evaluted.)
# >
# > I agree.
#
# ...that's what i said. why's he right and i'm not?

The lvalue x and rvalue y are evaluated.

For example if an lvalue is a[i++], i will be incremented. Evaluation of
an lvalue does not include loading whatever is at the address into a
register or whatever.

 

[Tim Rentsch]

Given

x = y;

y is evaluated, but x is not. (Specifically, the previous value of x
is not evaluted.)

In the technical language of the Standard, I think it's more
accurate to say that both x and y are evaluated, but only the
evaluation of y results in a conversion of the lvalue to the
value stored (which results in a read access of y's object).

(In case someone wants a reference - try 6.3.2.1 p1,p2.)

 

[Mark McIntyre]

...that's what i said. why's he right and i'm not?

Because it wasn't how what you wrote, read.

Your words, quoted above, implied to me that assigning variable X to
variable Y didn't evaluate X, it only assigned it. I believe we can
agree that this is wrong.

 

[Charles Richmond]

Can anybody tell me
if x = x +1 is x++
What is x = x + 2 ?

x++; x++;


Question: If the king sits on gold, who sits on silver???



Answer: The Lone Ranger