What does for(;;) mean?

Question

I am confused by the for(;;) construct. I think it is a form of shorthand for an unlimited for loop but I can't be sure.

Here is the code:

for(;;)
{
    //whatever statements
}

Answer 1

Your guess is correct; it's an infinite loop.^* This is a common C idiom, although many people (including me) believe the following to be less cryptic:

while (1) { whatever statements; }

---

_{* It's infinite assuming there are no break/return/etc. statements inside the loop body.}

Answer 2

It's an un-terminated loop. It is sometimes written with a while:

while (1)

or even better:

while (true)

I would expect to see a break or return inside any such loop, no matter whether it is written with for or while. There has to be some abnormal control flow or it really will be an infinite loop.

Answer 3

Yes, that's the for C syntax with blank fields for initialization expression, loop condition and increment expression.

The for statement can also use more than one value, like this sample :

for (i=0, j=100, k=1000; j < 500 || i<50 || k==5000; i++, j+=2, k*=6) {}; 

Maybe one step beyond in for understanding ? =)

Answer 4

Yes, the expressions in the for loop are just optional. if you omit them, you will get an infinite loop. The way to get out is break or exit or so.

Answer 5

As I understand it, for(;;) creates a deliberate non-exiting loop. Your code is expected to exit the loop based on one or more conditions. It was once provided to me as a purer way to have a do while false loop, which was not considered good syntax. Based on the exit condition, it is easier to dispatch to a function to handle the result, failure, warning, or success, for example.

My explanation may not be the reason someone used that construct, but I'll explain in greater detail what it means to me. This construct may be someone's "Pure C" way of having a loop in which you can serially perform multiple steps, whose completion mean something like your application has performed all steps of initialization.

#define GEN_FAILURE -99
#define SUCCESS 0

/* perform_init_step1() and perform_init_step2() are dummy 
   place-holder functions that provide a complete example. 
   You could at least have one of them return non-zero
   for testing. */

int perform_init_step1();
int perform_init_step2(); 

int perform_init_step1()
{ 
   return 0;
}

int perform_init_step2()
{
   return 0;
}   
int ret_code = GEN_FAILURE;
for(;;)
{
   if(SUCCESS != perform_init_step1())
   {
     ret_code = -1;
     break;
   }

   if(SUCCESS != perform_init_step2())
   {
     ret_code = -2;
     break;
   }

   break;
}

If part of the initialization fails, the loop bails out with a specific error code.

I arrived at using C having done a lot of firmware work, writing in assembly language. Good assembly language programmers taught me to have a single entry point and single exit. I took their advice to heart, because their creed helped them and me immensely when debugging.

Personally, I never liked the for(;;) construct, because you can have an infinite loop if you forget to break; out at the end.

Someone I worked with came up with do..until(FALSE), but the amount of proper C furvor this caused was not to be believed.

#define GEN_FAILURE -99
#define SUCCESS 0

/* perform_init_step1() and perform_init_step2() are dummy 
   place-holder functions that provide a complete example. 
   You could at least have one of them return non-zero
   for testing. */

int perform_init_step1();
int perform_init_step2(); 

int perform_init_step1()
{ 
   return 0;
}

int perform_init_step2()
{
   return 0;
}   

int ret_code = GEN_FAILURE;


do
{
   if(SUCCESS != perform_init_step1())
   {
     ret_code = -1;
     break;
   }

   if(SUCCESS != perform_init_step2())
   {
     ret_code = -2;
     break;
   }

}

until (FALSE);

This runs once, no matter what.

Answer 6

This statement is basically equal to:

while(1) {}

There is no start, no condition and no step statement.