Weird 'for' loops in C

gregorian · 06-09-2007, 01:47 AM

Here are some for loops which compile and run properly but I have no idea how they work.

Code:

#include <stdio.h>

int main()
{
int i=0;

for(;i;)
 printf("\n Sentence");

return 0;
}

output: Nothing.

No error even though the second statement in the for loop is not a conditional statement?

Here's another:

Code:

#include <stdio.h>

int main()
{
int i;
for(i=1; i<=5; printf("\n %d",i));
i++;

return 0;
}

Output: Infinite 1's

How do these programs work? I only know how common for loops work.

Nylex · 06-09-2007, 01:54 AM

The second statement in the first for loop is a conditional statement - it's just false since i = 0. If i was initialised with any other number, it would be true.

gregorian · 06-09-2007, 02:00 AM

You are correct, I put i = 1 and got an infinite loop. May I know from where did you learn this? I never learned that you could substitute true with any number but zero and false with 0 in a C program.

What about the second loop??

IBall · 06-09-2007, 02:31 AM

This is just part of C. As you learn C, you will pick these things up. Of course, they would not be considered "good programming practice".

--Ian

Valkyrie_of_valhalla · 06-09-2007, 03:01 AM

about the second loop.
Try it like this:

Code:

#include <stdio.h>

int main()
{
int i;
for(i=1; i<=5; i++)
printf("\n %d",i);

return 0;
}

EDIT: don't put ";" after a for statement, as everything that follows after that for doesn't get included in the for statement. It also works how you mentioned if you remove that ";", but it would print out 2 3 4 5 6. So I would reccomend initialising i with 0.

gregorian · 06-09-2007, 03:28 AM

Thank you Valkyrie.

I have now understood how the programs work:

The for loop works by initialising the counter, checking for the validity of the second statement, executing the body of the loop, and then executing whatever is there after the second semi colon. Since the only body of the for loop in the original program (second one) was the semi colon, the value of the counter never got incremented and thus it broke into an infinite loop, continuously executing the statement after the second semi colon - the printf statement which gave the unincremented value of i i.e 1 continuously.

Thank you all for your help.

One more question: Why can't I declare the for loop variable in the for loop itself i.e.

Code:

for(int i=1; i<5; i++)

instead of

Code:

int i;
for(i=1; i<5; i++)

I get this error with gcc:

test.c: In function ‘main’:
test.c:6: error: ‘for’ loop initial declaration used outside C99 mode

IBall · 06-09-2007, 04:13 AM

Exactly as the error says - this is not complient with the ANSI C99 standard. There are gcc flags that will turn this off, but it is not really a problem.

Older versions of C, as well as some other languages permit this shortcut.

It is generally considered better programming practice to declare the variables separately from the initializations.

--Ian

gregorian · 06-09-2007, 05:14 AM

Ok, thank you!

gregorian · 06-09-2007, 05:48 AM

I was supposed to write a program that adds the first 7 terms of

1/1! + 2/2! + ...

I wrote this:

Code:

#include <stdio.h>

main()
{

int i,m,fact;
float sum,div;

sum=i=m=0;

/* First seven times */

for(i=1; i<=7 ; i++)
 {
  fact =1; 

/* Factorial function */

  for(m=1; m<=i; m++) 
    fact*=m;
    
   div = i/fact; 
      
  sum += div;

printf("%d / %d = %f sum = %f\n",i,fact,div,sum);

 }



}

This is the output:

Quote:

1 / 1 = 1.000000 sum = 1.000000
2 / 2 = 1.000000 sum = 2.000000
3 / 6 = 0.000000 sum = 2.000000
4 / 24 = 0.000000 sum = 2.000000
5 / 120 = 0.000000 sum = 2.000000
6 / 720 = 0.000000 sum = 2.000000
7 / 5040 = 0.000000 sum = 2.000000

3/6 = 0.000000

Nylex · 06-09-2007, 06:07 AM

You're doing integer division. You need to cast either the numerator or the denominator to a float.

gregorian · 06-09-2007, 06:22 AM

Thanks for that!

jschiwal · 06-09-2007, 06:42 AM

Quote:

Originally Posted by gregorian

I was supposed to write a program that adds the first 7 terms of

1/1! + 2/2! + ...

You can reduce the amount of round off error if you decrement from 7 to 1 instead.

Nylex · 06-09-2007, 06:43 AM

Out of interest, does that series have a name?

gregorian · 06-09-2007, 08:51 AM

Yes, it's called 'e' as in the 'e' of natural logarithms if you sum the terms till infinity.

Could you explain jschiwal? How's the error minimised if I add the series backwards?

Output:

Quote:

1 / 1 = 1.000000 sum = 1.000000
2 / 2 = 1.000000 sum = 2.000000
3 / 6 = 0.500000 sum = 2.500000
4 / 24 = 0.166667 sum = 2.666667
5 / 120 = 0.041667 sum = 2.708333
6 / 720 = 0.008333 sum = 2.716667
7 / 5040 = 0.001389 sum = 2.718056

The final result here is correct uptil 3 decimal places.

jschiwal · 06-11-2007, 02:49 AM

Quote:

Originally Posted by gregorian

Yes, it's called 'e' as in the 'e' of natural logarithms if you sum the terms till infinity.

Could you explain jschiwal? How's the error minimised if I add the series backwards?

Output:

The final result here is correct uptil 3 decimal places.

Each term is smaller than the last one. Your series is so short that you may not notice much of a problem, but for some series, the lost of significant digits from adding to 1.0 instead of a bunch of terms like 1.2234567e-13 being totaled can effect the final result.
For the smallest terms in a longer series, you will be adding 0 to the total.

By the way, you might want to look at the manpage of "bc" for the library function they use for e(x).

Try this on for size!
echo 'scale=1024; print e(1) ' | bc -l