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************ *************** *******@x19g200 0prg.googlegrou ps.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] c-cnrc.gc.ca (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] c-cnrc.gc.ca (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************ *************** *******@s33g200 0pri.googlegrou ps.com>,
spasmous <sp******@gmail .comwrote:

>On Jun 20, 12:28=A0pm, [e-mail] c-cnrc.gc.ca (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
misunderstandin g 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 <spasm...@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?

{
// ...
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 <ben.use...@bsb .me.ukwrote:

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.

[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/>