Help avoid goto

[spasmous]

// return early if all points are the same
for(int i=1; i<n; i++)
if(y[i] != y[0]) goto SKIP;
return;
SKIP:

Can someone help me with an alternative to this snippet that avoids
goto? Without introducing a new variable?

[Walter Roberson]

In article <fc**********************************@x19g2000prg. googlegroups.com>,
spasmous <sp******@gmail.comwrote:

>// return early if all points are the same
for(int i=1; i<n; i++)
if(y[i] != y[0]) goto SKIP;
return;
SKIP:

>Can someone help me with an alternative to this snippet that avoids
goto? Without introducing a new variable?

int i = 1;
while( i<n && y[i] != y[0] ) i++;
if (i == n) return;

This can also be written as a for loop.
--
"Let me live in my house by the side of the road --
It's here the race of men go by.
They are good, they are bad, they are weak, they are strong
Wise, foolish -- so am I;" -- Sam Walter Foss

[pete]

spasmous wrote:

// return early if all points are the same
for(int i=1; i<n; i++)
if(y[i] != y[0]) goto SKIP;
return;
SKIP:

Can someone help me with an alternative to this snippet that avoids
goto? Without introducing a new variable?

int i;

for (i = 1; n i; ++i) {
if (y[i] != y[0]) {
break;
}
}
if (i == n || 1 n) {
return;
}
--
pete

[Ben Bacarisse]

spasmous <sp******@gmail.comwrites:

// return early if all points are the same
for(int i=1; i<n; i++)
if(y[i] != y[0]) goto SKIP;
return;
SKIP:

Can someone help me with an alternative to this snippet that avoids
goto? Without introducing a new variable?

9 times out of 10, this sort of situation is a reminder that you need
another function. Testing if an array has all elements equal is the
job of a separate function. There are lots of ways to write it:

bool all_equal(T *array, size_t n)
{
for (size_t i = 1; i < n; i++)
if (array[i] != array[0])
return false;
return true;
}

May people prefer this style (I think I do):

bool all_equal(T *array, size_t n)
{
for (size_t i = 1; i < n && array[y] == array[0]; i++)
continue;
return i == n;
}

(Note that they are not the same when n == 0 -- you need to decide what
you mean by that case.)

I had to use T because I don't know the type of the elements of y. If
you don't like using C99isms, move the declaration of i out of the
loop and return int rather than bool (or declare bool yourself).

--
Ben.

[spasmous]

On Jun 20, 12:28*pm, [e-mail] (Walter Roberson)
wrote:

>
int i = 1;
while( i<n && y[i] != y[0] ) i++;
if (i == n) return;

This can also be written as a for loop.

Very nice Walter. I went with a for loop variant.

// return early if all points are the same
for(int i=1; y[i]!=y[0]; i++)
if(i == n) return;

[Eric Sosman]

spasmous wrote:

On Jun 20, 12:28 pm, [e-mail] (Walter Roberson)
wrote:

>int i = 1;
while( i<n && y[i] != y[0] ) i++;
if (i == n) return;

This can also be written as a for loop.

Very nice Walter. I went with a for loop variant.

// return early if all points are the same
for(int i=1; y[i]!=y[0]; i++)
if(i == n) return;

Note that this tests y[n], the (n+1)st array element.

--
Er*********@sun.com

[Walter Roberson]

In article <10**********************************@s33g2000pri. googlegroups.com>,
spasmous <sp******@gmail.comwrote:

>On Jun 20, 12:28=A0pm, [e-mail] (Walter Roberson)
wrote:

>>
int i = 1;
while( i<n && y[i] != y[0] ) i++;
if (i == n) return;

This can also be written as a for loop.

Very nice Walter. I went with a for loop variant.

// return early if all points are the same
for(int i=1; y[i]!=y[0]; i++)
if(i == n) return;

If y is defined from index 0 to n-1 then your code will access
y[n] in the termination test, which would be undefined behaviour
under that sizing assumption.

--
"When a scientist is ahead of his times, it is often through
misunderstanding of current, rather than intuition of future truth.
In science there is never any error so gross that it won't one day,
from some perspective, appear prophetic." -- Jean Rostand

[Kaz Kylheku]

On Jun 20, 12:19*pm, spasmous <[e-mail]:

// return early if all points are the same
for(int i=1; i<n; i++)
* * if(y[i] != y[0]) goto SKIP;
return;
SKIP:

Can someone help me with an alternative to this snippet that avoids
goto? Without introducing a new variable?

{
// ...
for (int i = 1; i < n; i++) {
if (y[i] != y[0]) {
// entire ``SKIP:'' section here
break;
}
}
}

Like, put the conditional code into the body of the conditional
statement which tests that condition? Doh?

The reason you have a goto is that you relocated that logic away from
that construct. In general, you can't arbitrarily relocate control
without using GOTO or additional state variables.

[Kaz Kylheku]

On Jun 20, 1:12*pm, Ben Bacarisse <[e-mail]:

9 times out of 10, this sort of situation is a reminder that you need
another function. *Testing if an array has all elements equal is the
job of a separate function. *There are lots of ways to write it:

bool all_equal(T *array, size_t n)
{
* * for (size_t i = 1; i < n; i++)
* * * * if (array[i] != array[0])
* * * * * * return false;
* * return true;

}

May people prefer this style (I think I do):

bool all_equal(T *array, size_t n)
{
* * for (size_t i = 1; i < n && array[y] == array[0]; i++)
* * * * continue;
* * return i == n;

This algorithm is not the same as the first version with the early
return. What if the array size is zero? The ``all equal'' condition is
true then: all elements of an empty sequence are equal to each other,
because it is not the case that there exist two distinct positions x
and y such that the x-th element is equal to the y-th element.

Moreover, the variable i is not in scope of the i == n expression.

[Kaz Kylheku]

On Jun 20, 1:12*pm, Ben Bacarisse <[e-mail]:

bool all_equal(T *array, size_t n)
{
* * for (size_t i = 1; i < n && array[y] == array[0]; i++)
* * * * continue;
* * return i == n;

The name i is not in scope here, oops! :)

}

(Note that they are not the same when n == 0 -- you need to decide what
you mean by that case.)

That case means that all elements are equal. For an empty sequence, it
is not the case that there can exist two distinct indices x and y such
that the x-th element is equal to the y-th element. Since you can't
find two unequal elements, you cannot declare that the sequence does
not have all elements equal.

Here is another view. The all-equal property, if true of a sequence S,
should also be true of a subsequence of S, because how could it be the
case that all elements of a large sequence are equal, yet elements of
a subsequence of that sequence are not equal? But the empty sequence E
is a subsequence of every sequence; if the property is declared not
true of E by definition, then this represents a troublesome special
case; there now exists a subsequence of all-equal sequence S for which
the property does not hold.

[Ben Bacarisse]

Kaz Kylheku <kk******@gmail.comwrites:

On Jun 20, 1:12Â*pm, Ben Bacarisse <[e-mail]:

>9 times out of 10, this sort of situation is a reminder that you need
another function. Â*Testing if an array has all elements equal is the
job of a separate function. Â*There are lots of ways to write it:

bool all_equal(T *array, size_t n)
{
Â* Â* for (size_t i = 1; i < n; i++)
Â* Â* Â* Â* if (array[i] != array[0])
Â* Â* Â* Â* Â* Â* return false;
Â* Â* return true;

}

May people prefer this style (I think I do):

bool all_equal(T *array, size_t n)
{
Â* Â* for (size_t i = 1; i < n && array[y] == array[0]; i++)
Â* Â* Â* Â* continue;
Â* Â* return i == n;

This algorithm is not the same as the first version with the early
return. What if the array size is zero?

I know. Why did you clip the bit where I said that? I explained they
were not the same and when.

Moreover, the variable i is not in scope of the i == n expression.

That is a good point. Thank you.

--
Ben.

[Daniel Pitts]

spasmous wrote:

// return early if all points are the same
for(int i=1; i<n; i++)
if(y[i] != y[0]) goto SKIP;
return;
SKIP:

Can someone help me with an alternative to this snippet that avoids
goto? Without introducing a new variable?

if (isAllEqual(y, n)) return;
....
int isAllEqual(foo *y, int n) {
for (int i = 0; i <n; ++i) {
if (y[n] != y[0]) {
return false;
}
}
return true;
}

--
Daniel Pitts' Tech Blog: <http://virtualinfinity.net/wordpress/>

[Mike Wahler]

"spasmous" <sp******@gmail.comwrote in message
news:fc**********************************@x19g2000 prg.googlegroups.com...

// return early if all points are the same
for(int i=1; i<n; i++)
if(y[i] != y[0]) goto SKIP;

if(y[i] != y[0])
break;
-Mike

return;
SKIP:

Can someone help me with an alternative to this snippet that avoids
goto? Without introducing a new variable?

[CBFalconer]

spasmous wrote:

>
// return early if all points are the same
for(int i=1; i<n; i++)
if(y[i] != y[0]) goto SKIP;
return;
SKIP:

Can someone help me with an alternative to this snippet that
avoids goto? Without introducing a new variable?

for (i = 1; i < n; i++)
if (y[i] != y[0] break;
if (n == i) return;
else {
/* SKIP is here */
}

--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home.att.net>
Try the download section.

** Posted from http://www.teranews.com **

[Ben Bacarisse]

Kaz Kylheku <kk******@gmail.comwrites:

On Jun 20, 1:12Â*pm, Ben Bacarisse <[e-mail]:

>bool all_equal(T *array, size_t n)
{
Â* Â* for (size_t i = 1; i < n && array[y] == array[0]; i++)
Â* Â* Â* Â* continue;
Â* Â* return i == n;

The name i is not in scope here, oops! :)

Yes, you said that already. Do you get paid for posts? :-)

>}

(Note that they are not the same when n == 0 -- you need to decide what
you mean by that case.)

That case means that all elements are equal.

Yes.

For an empty sequence, it
is not the case that there can exist two distinct indices x and y such
that the x-th element is equal to the y-th element. Since you can't
find two unequal elements, you cannot declare that the sequence does
not have all elements equal.

I agree that there is only one logical return value for the empty
case, but I opted just to suggest that the OP think it over.

--
Ben.

[Kaz Kylheku]

On Jun 20, 4:04*pm, Ben Bacarisse <[e-mail]:

Kaz Kylheku <[e-mail]:

The name i is not in scope here, oops! :)

Yes, you said that already. *Do you get paid for posts? :-)

Alas, I deleted that post very soon after submitting it, literally
within a minute or two.

[Walter Banks]

spasmous wrote:

// return early if all points are the same
for(int i=1; i<n; i++)
if(y[i] != y[0]) goto SKIP;
return;
SKIP:

Can someone help me with an alternative to this snippet that avoids
goto? Without introducing a new variable?

for(int i=1; i<n; i++)
if(y[i] == y[0]) return;
w..

[=?iso-8859-1?Q?=D8yvind_R=F8tvold?=]

Walter Banks <wa****@bytecraft.comwrites:

spasmous wrote:

>// return early if all points are the same
for(int i=1; i<n; i++)
if(y[i] != y[0]) goto SKIP;
return;
SKIP:

Can someone help me with an alternative to this snippet that avoids
goto? Without introducing a new variable?

for(int i=1; i<n; i++)
if(y[i] == y[0]) return;

That's not equivalent, consider the case where y[1] matches while y[2]
doesn't.

This might work:

int i;
for(i=1; i<n; i++)
if(y[i] != y[0]) break;
if (i=n)
return;
--
... __o Øyvind
... _`\(, http://www.darkside.no/olr/
... (_)/(_) ... biciclare necesse est ...

[Bill Leary]

"Walter Banks" <wa****@bytecraft.comwrote in message
news:48***************@bytecraft.com...

spasmous wrote:

Assuming 'n' = 5 and y[0] = 9, y[1] = 9, y[2] = 9, y[3] = 99, y[4] = 9

>// return early if all points are the same
for(int i=1; i<n; i++)
if(y[i] != y[0]) goto SKIP;
return;
SKIP:

This will not return. It will goto SKIP when i = 3.

>Can someone help me with an alternative to this snippet that avoids
goto? Without introducing a new variable?

for(int i=1; i<n; i++)
if(y[i] == y[0]) return;

But this will return when i = 1.

- Bill

[Bill Leary]

"Øyvind Røtvold" <or******@gmail.comwrote in message
news:v6*************@jjj.jjj...

That's not equivalent, consider the case where y[1] matches while y[2]
doesn't.

This might work:

int i;
for(i=1; i<n; i++)
if(y[i] != y[0]) break;
if (i=n)
return;

Well, that's not equivalent either, though, since it changes the scope of
"i".

And I'm assuming "if (i=n)" is a typo for "if (i==n)". :)

- Bill

[CBFalconer]

Walter Banks wrote:

spasmous wrote:

>// return early if all points are the same
for(int i=1; i<n; i++)
if(y[i] != y[0]) goto SKIP;
return;
SKIP:

Can someone help me with an alternative to this snippet that
avoids goto? Without introducing a new variable?

for(int i=1; i<n; i++)
if(y[i] == y[0]) return;

No. This treats (y[1] == y[0]) as the non-SKIP case.

--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home.att.net>
Try the download section.
** Posted from http://www.teranews.com **

[Walter Banks]

All, quite correct missed the extent of the for loop. My bad.

w..
Bill Leary wrote:

"Walter Banks" <wa****@bytecraft.comwrote in message
news:48***************@bytecraft.com...

spasmous wrote:

Assuming 'n' = 5 and y[0] = 9, y[1] = 9, y[2] = 9, y[3] = 99, y[4] = 9

// return early if all points are the same
for(int i=1; i<n; i++)
if(y[i] != y[0]) goto SKIP;
return;
SKIP:

This will not return. It will goto SKIP when i = 3.

Can someone help me with an alternative to this snippet that avoids
goto? Without introducing a new variable?

for(int i=1; i<n; i++)
if(y[i] == y[0]) return;

But this will return when i = 1.

- Bill