Infinite loop

[dmoran21@cox.net]

Hi all, I am a mathematician and I'm trying to write a program to try
out a formula that I've derived. However, it seems that I've got an
infinite loop and I don't quite understand why. I was hoping someone
could point me in the right direction.

Code:

#include <stdio.h>
#include <math.h>

int main()
{
float start = 0;
float end = 0;
float step = 0;
float x = 0;
float y = 0;

printf("Enter the starting value\n");
scanf("%f", &start);
printf("Enter the ending value\n");
scanf("%f", &end);
printf("Enter the step\n");
scanf("%f", &step);

for(x=start;x=e nd;x+=step) {
y = (pow(260,.5)/130)*cos(2*x-1.463) + (67/20)*exp(x) +
(37/52)*exp(-3*x);
printf("%3.3f %3.3f\n",x,y);
}
}

Thanks,
Dave

[osmium]

Re: Infinite loop

<dmoran21@cox.n etwrote:

Hi all, I am a mathematician and I'm trying to write a program to try
out a formula that I've derived. However, it seems that I've got an
infinite loop and I don't quite understand why. I was hoping someone
could point me in the right direction.
>
Code:
>
#include <stdio.h>
#include <math.h>
>
int main()
{
float start = 0;
float end = 0;
float step = 0;
float x = 0;
float y = 0;
>
printf("Enter the starting value\n");
scanf("%f", &start);
printf("Enter the ending value\n");
scanf("%f", &end);
printf("Enter the step\n");
scanf("%f", &step);
>
for(x=start;x=e nd;x+=step) {

Did you mean x==end ?

y = (pow(260,.5)/130)*cos(2*x-1.463) + (67/20)*exp(x) +
(37/52)*exp(-3*x);
printf("%3.3f %3.3f\n",x,y);
}
}
>
Thanks,
Dave
>

[dmoran21@cox.net]

Re: Infinite loop

On Mar 17, 8:39 pm, "osmium" <r124c4u...@com cast.netwrote:

<dmora...@cox.n etwrote:

Hi all, I am a mathematician and I'm trying to write a program to try
out a formula that I've derived. However, it seems that I've got an
infinite loop and I don't quite understand why. I was hoping someone
could point me in the right direction.

>

Code:

>

#include <stdio.h>
#include <math.h>

>

int main()
{
float start = 0;
float end = 0;
float step = 0;
float x = 0;
float y = 0;

>

printf("Enter the starting value\n");
scanf("%f", &start);
printf("Enter the ending value\n");
scanf("%f", &end);
printf("Enter the step\n");
scanf("%f", &step);

>

for(x=start;x=e nd;x+=step) {

>
Did you mean x==end ?
>

y = (pow(260,.5)/130)*cos(2*x-1.463) + (67/20)*exp(x) +
(37/52)*exp(-3*x);
printf("%3.3f %3.3f\n",x,y);
}
}

>

Thanks,
Dave

That stopped the infinite loop, but now, it's stopping after the input
is entered.

Dave

[santosh]

Re: Infinite loop

dmoran21@cox.ne t wrote:

Hi all, I am a mathematician

Didn't you start out with this exact same phrase in a post some time
ago. As I recall, it start a flame thread.

and I'm trying to write a program to try
out a formula that I've derived. However, it seems that I've got an
infinite loop and I don't quite understand why. I was hoping someone
could point me in the right direction.
>
Code:
>
#include <stdio.h>
#include <math.h>
>
int main()
{
float start = 0;
float end = 0;
float step = 0;
float x = 0;
float y = 0;

Unless you've good reasons, you might consider using double for these
objects. It provides greater default precision.

printf("Enter the starting value\n");
scanf("%f", &start);
printf("Enter the ending value\n");
scanf("%f", &end);
printf("Enter the step\n");
scanf("%f", &step);
>
for(x=start;x=e nd;x+=step) {

The = operator is the assignment operator. You probably meant <= or <
or some other relational or logical relation. Otherwise the test
simply assigns end to x and evaluates x. If x is other than zero, the
loop will continue. So if the user had entered a positive or negative
value for end, the program will enter an infinite loop.

y = (pow(260,.5)/130)*cos(2*x-1.463) + (67/20)*exp(x) + (37/52)*exp(-3*x);

Mismatched parenthesis. If you don't cut and paste the exact code that
failed to compile or run, then it's guessing all the way.

You should split up the above statement into multiple steps, with a
temporary or two. By reordering it you can preserve more accuracy and
enhance readability.

printf("%3.3f %3.3f\n",x,y);
}
}

[Thad Smith]

Re: Infinite loop

dmoran21@cox.ne t wrote:

for(x=start;x=e nd;x+=step) {
y = (pow(260,.5)/130)*cos(2*x-1.463) + (67/20)*exp(x) +
(37/52)*exp(-3*x);
printf("%3.3f %3.3f\n",x,y);
}

Besides the problem with the for loop, the expressions (67/20) and
(37/52) are integer expressions that yield 3 and 0, respectively. To
get floating point values, insert a decimal point: (67./27.). The other
integers in the expression, 260, 130, 2, and 3, are already converted to
type double because of
1) use as a parameter of type double and
2) operation with an expression of type double.

In the case of (-3*x), the integer 3 is negated, yielding -3 as an
integer, then converted to double (probably all at compile time) before
multiplying by x.

You can add decimal points to all these constants if you don't want to
bother remembering the rules for when conversion to floating point is done.

--
Thad

[osmium]

Re: Infinite loop

<dmoran21@cox.n etwrote:

On Mar 17, 8:39 pm, "osmium" <r124c4u...@com cast.netwrote:

><dmora...@cox. netwrote:

Hi all, I am a mathematician and I'm trying to write a program to try
out a formula that I've derived. However, it seems that I've got an
infinite loop and I don't quite understand why. I was hoping someone
could point me in the right direction.

>>

Code:

>>

#include <stdio.h>
#include <math.h>

>>

int main()
{
float start = 0;
float end = 0;
float step = 0;
float x = 0;
float y = 0;

>>

printf("Enter the starting value\n");
scanf("%f", &start);
printf("Enter the ending value\n");
scanf("%f", &end);
printf("Enter the step\n");
scanf("%f", &step);

>>

for(x=start;x=e nd;x+=step) {

>>
>Did you mean x==end ?
>>

y = (pow(260,.5)/130)*cos(2*x-1.463) + (67/20)*exp(x) +
(37/52)*exp(-3*x);
printf("%3.3f %3.3f\n",x,y);
}
}

>>

Thanks,
Dave

>
That stopped the infinite loop, but now, it's stopping after the input
is entered.

In general, you should never test any floating point type for equality, such
numbers are represented as approximations to the real underlying number.
Testing for x==end is like trying to balance a dime on the rim. Changing to
x<= end yields a short table of numbers for me.

[dmoran21@cox.net]

Re: Infinite loop

On Mar 17, 9:34 pm, "osmium" <r124c4u...@com cast.netwrote:

<dmora...@cox.n etwrote:

On Mar 17, 8:39 pm, "osmium" <r124c4u...@com cast.netwrote:

<dmora...@cox.n etwrote:
Hi all, I am a mathematician and I'm trying to write a program to try
out a formula that I've derived. However, it seems that I've got an
infinite loop and I don't quite understand why. I was hoping someone
could point me in the right direction.

>

Code:

>

#include <stdio.h>
#include <math.h>

>

int main()
{
float start = 0;
float end = 0;
float step = 0;
float x = 0;
float y = 0;

>

printf("Enter the starting value\n");
scanf("%f", &start);
printf("Enter the ending value\n");
scanf("%f", &end);
printf("Enter the step\n");
scanf("%f", &step);

>

for(x=start;x=e nd;x+=step) {

>

Did you mean x==end ?

>

y = (pow(260,.5)/130)*cos(2*x-1.463) + (67/20)*exp(x) +
(37/52)*exp(-3*x);
printf("%3.3f %3.3f\n",x,y);
}
}

>

Thanks,
Dave

>

That stopped the infinite loop, but now, it's stopping after the input
is entered.

>
In general, you should never test any floating point type for equality, such
numbers are represented as approximations to the real underlying number.
Testing for x==end is like trying to balance a dime on the rim. Changing to
x<= end yields a short table of numbers for me.

Would it be better to use a do loop? I thought that I remember hearing
something that you should use a for loop only if you are using
integers to count.

Dave

[Barry Schwarz]

Re: Infinite loop

On 17 Mar 2007 18:45:31 -0700, "dmoran21@cox.n et" <dmoran21@cox.n et>
wrote:

>On Mar 17, 8:39 pm, "osmium" <r124c4u...@com cast.netwrote:

><dmora...@cox. netwrote:

Hi all, I am a mathematician and I'm trying to write a program to try
out a formula that I've derived. However, it seems that I've got an
infinite loop and I don't quite understand why. I was hoping someone
could point me in the right direction.

>>

Code:

>>

#include <stdio.h>
#include <math.h>

>>

int main()
{
float start = 0;
float end = 0;
float step = 0;
float x = 0;
float y = 0;

>>

printf("Enter the starting value\n");
scanf("%f", &start);
printf("Enter the ending value\n");
scanf("%f", &end);
printf("Enter the step\n");
scanf("%f", &step);

>>

for(x=start;x=e nd;x+=step) {

>>
>Did you mean x==end ?
>>

y = (pow(260,.5)/130)*cos(2*x-1.463) + (67/20)*exp(x) +
(37/52)*exp(-3*x);
printf("%3.3f %3.3f\n",x,y);
}
}

>>

Thanks,
Dave

>
>That stopped the infinite loop, but now, it's stopping after the input
>is entered.
>

The conditional expression is evaluated before the first iteration.
Since x is not equal to end, the expression evaluates to zero and
looping is terminated even before the first iteration.


Remove del for email

[Richard Heathfield]

Re: Infinite loop

dmoran21@cox.ne t said:

<snip>

Would it be better to use a do loop? I thought that I remember hearing
something that you should use a for loop only if you are using
integers to count.

No, there is no such rule or guideline, or at least not amongst those
who know the language well!

If anything, your for(counter = start; counter < end; counter += step)
example is actually a rather good example of when a for-loop is
appropriate - but of course a while or a do would get the job done just
as well.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999

http://www.cpax.org.uk

email: rjh at the above domain, - www.

[Malcolm McLean]

Re: Infinite loop


<dmoran21@cox.n etwrote in message

#include <stdio.h>
#include <math.h>

>>

int main()
{
float start = 0;
float end = 0;
float step = 0;
float x = 0;
float y = 0;

>>

printf("Enter the starting value\n");
scanf("%f", &start);
printf("Enter the ending value\n");
scanf("%f", &end);
printf("Enter the step\n");
scanf("%f", &step);

>>

for(x=start;x=e nd;x+=step) {

>>

>Did you mean x==end ?

>>

y = (pow(260,.5)/130)*cos(2*x-1.463) + (67/20)*exp(x) +
(37/52)*exp(-3*x);
printf("%3.3f %3.3f\n",x,y);
}
}

>>

Thanks,
Dave

>>

That stopped the infinite loop, but now, it's stopping after the input
is entered.

>>
>In general, you should never test any floating point type for equality,
>such
>numbers are represented as approximations to the real underlying number.
>Testing for x==end is like trying to balance a dime on the rim. Changing
>to
>x<= end yields a short table of numbers for me.

>
Would it be better to use a do loop? I thought that I remember hearing
something that you should use a for loop only if you are using
integers to count.
>

There are certain difficulties in using floats as the counter within a for
loop.
One is that, if the numbers are not integers, the test x != end may fail
because of numerical precision. This can be fixed with x <= end + epsilon.

Another problem is that programmers are so used to seeing for loops used to
index arrays that it can be confusing to see one used for another purpose.

The final problem in mathematics is that the funny-looking E notation used
for sums, conventionally, takes integral indices.

However if a for() loop most naturally expresses the logic of your
calculation, then use one.
--
Free games and programming goodies.

http://www.personal.leeds.ac.uk/~bgy1mm

[Chris Johnson]

Re: Infinite loop

On Mar 17, 8:45 pm, "dmora...@cox.n et" <dmora...@cox.n etwrote:

On Mar 17, 8:39 pm, "osmium" <r124c4u...@com cast.netwrote:
>
>
>

<dmora...@cox.n etwrote:

Hi all, I am a mathematician and I'm trying to write a program to try
out a formula that I've derived. However, it seems that I've got an
infinite loop and I don't quite understand why. I was hoping someone
could point me in the right direction.

>

Code:

>

#include <stdio.h>
#include <math.h>

>

int main()
{
float start = 0;
float end = 0;
float step = 0;
float x = 0;
float y = 0;

>

printf("Enter the starting value\n");
scanf("%f", &start);
printf("Enter the ending value\n");
scanf("%f", &end);
printf("Enter the step\n");
scanf("%f", &step);

>

for(x=start;x=e nd;x+=step) {

>

Did you mean x==end ?

>

y = (pow(260,.5)/130)*cos(2*x-1.463) + (67/20)*exp(x) +
(37/52)*exp(-3*x);
printf("%3.3f %3.3f\n",x,y);
}
}

>

Thanks,
Dave

>
That stopped the infinite loop, but now, it's stopping after the input
is entered.

I think you're expecting the second part of the for loop to be an end
condition, i.e. You want to stop when x is equal to end. But the
second part is a boolean condition, i.e. the loop will continue until
that condition is false. You probably want x < end instead of x ==
end, so the loop will stop when x >= end.

[Mark L Pappin]

Re: Infinite loop

"Malcolm McLean" <regniztar@btin ternet.comwrite s:

in mathematics ... the funny-looking E notation used for sums

<OT>
By "funny-looking E" would you mean "Sigma", perchance?
</OT>

mlp

[Old Wolf]

Re: Infinite loop

On Mar 18, 1:32 pm, "dmora...@cox.n et" <dmora...@cox.n etwrote:

Hi all, I am a mathematician and I'm trying to write a program to try
out a formula that I've derived.
>
int main()
{
printf("Enter the starting value\n");
scanf("%f", &start);
printf("Enter the ending value\n");
scanf("%f", &end);
printf("Enter the step\n");
scanf("%f", &step);

What happens if the person types "x", or just presses Enter?

for(x=start;x=e nd;x+=step) {

'=' is the assignment operator. The equality test is '=='.

Also, as others have pointed out, testing for equality with
floating point types is unreliable.

y = (pow(260,.5)/130)*cos(2*x-1.463) + (67/20)*exp(x) +
(37/52)*exp(-3*x);

int divided by int gives int, in C, so 67/20 gives 3 and 37/52 gives
0.
You can fix it by writing 67./20 and 37./52

printf("%3.3f %3.3f\n",x,y);
}

Consider using 'double' instead of 'float'. You will also have
to change your scanf modifiers to '%lf' (or preferably use
fgets and strtod instead of scanf!).

[Norm Mann]

Re: Infinite loop


"Thad Smith" <ThadSmith@acm. orgwrote in message
news:45fca28d$0 $27529$892e0abb @auth.newsreade r.octanews.com. ..

dmoran21@cox.ne t wrote:
>

>for(x=start;x= end;x+=step) {
>y = (pow(260,.5)/130)*cos(2*x-1.463) + (67/20)*exp(x) +
>(37/52)*exp(-3*x);
>printf("%3.3 f %3.3f\n",x,y);
>}

>
Besides the problem with the for loop, the expressions (67/20) and (37/52)
are integer expressions that yield 3 and 0, respectively. To get floating
point values, insert a decimal point: (67./27.). The other integers in
the expression, 260, 130, 2, and 3, are already converted to type double
because of
1) use as a parameter of type double and
2) operation with an expression of type double.
>
In the case of (-3*x), the integer 3 is negated, yielding -3 as an
integer, then converted to double (probably all at compile time) before
multiplying by x.
>

Since (pow(260,.5)/130), (67/20) and (37/52) are all constants, if you
precompute them and insert them as actual values, you can save yourself some
processing time.

-NM

[Thad Smith]

Re: Infinite loop

Norm Mann wrote:

"Thad Smith" <ThadSmith@acm. orgwrote in message
news:45fca28d$0 $27529$892e0abb @auth.newsreade r.octanews.com. ..
>

>>dmoran21@cox. net wrote:
>>
>>

>>>for(x=start; x=end;x+=step) {
>>>y = (pow(260,.5)/130)*cos(2*x-1.463) + (67/20)*exp(x) +
>>>(37/52)*exp(-3*x);
>>>printf("%3.3 f %3.3f\n",x,y);
>>>}

>>
>>Besides the problem with the for loop, the expressions (67/20) and (37/52)
>>are integer expressions that yield 3 and 0, respectively. To get floating
>>point values, insert a decimal point: (67./27.). The other integers in
>>the expression, 260, 130, 2, and 3, are already converted to type double
>>because of
>>1) use as a parameter of type double and
>>2) operation with an expression of type double.

>
Since (pow(260,.5)/130), (67/20) and (37/52) are all constants, if you
precompute them and insert them as actual values, you can save yourself some
processing time.

True, but I prefer having the program compute them to show the
derivation. If speed is an issue, you can compute the constants once in
the code and assign to variables.

--
Thad