How is for loop parsed internally as in assembly language?

Question

How does a for loop work internally in terms of the machine language when the expressions are not written. For example

int i=0 , j=1;
for(;;)

@Osiris is there any other way of coming out of the loop other than break? — Harish Rajora, Aug 20 '18 at 18:29
With a `goto`for example, but thats really not good practice. — Osiris, Aug 20 '18 at 18:29
why are you trying to break out of an infinite loop without break? — DTSCode, Aug 20 '18 at 18:30
What about when the expressions are shuffled like for(i<2; i++; ). The loop still gets executed. Isn't there any kind of protocol? — Harish Rajora, Aug 20 '18 at 18:36
@HarishRajora: in that specific example, `i<2` will be evaluated once before the loop begins, then `i++` will be evaluated before each iteration - as long as the result is non-zero, the loop will execute. The position of each expression matters; you can't just shift everything left one position. — John Bode, Aug 20 '18 at 18:39

score 5 · Answer 1 · answered Aug 20 '18 at 18:33

5

A for loop with a missing middle expression is an infinite loop.

From section 6.8.5.3 of the C standard:

1 The statement
for ( clause-1 ; expression-2 ; expression-3 ) statement
behaves as follows: The expression expression-2 is the controlling expression that is evaluated before each execution of the loop body. The expression expression-3 is evaluated as a void expression after each execution of the loop body. If clause-1 is a declaration, the scope of any identifiers it declares is the remainder of the declaration and the entire loop, including the other two expressions; it is reached in the order of execution before the first evaluation of the controlling expression. If clause-1 is an expression, it is evaluated as a void expression before the first evaluation of the controlling expression.

2 Both clause-1 and expression-3 can be omitted. An omitted expression-2 is replaced by a nonzero constant.

Because a missing "expression-2" is replaced with a non-zero constant, and because a non-zero value evaluates to true, this gives you an infinite loop.

answered Aug 20 '18 at 18:33

dbush

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I see there can be shuffling of the expressions also like for (i<2; ;i++) and still the loop runs. – Harish Rajora Aug 20 '18 at 18:35
1

@HarishRajora It is still an infinite loop then, which increments `i` every turn. – Osiris Aug 20 '18 at 18:37
and if for(;i++;) is written then? Just last ques. – Harish Rajora Aug 20 '18 at 18:39
1

@HarishRajora In that case `i<2` is only executed at the start of the loop and its value is discarded, and as before the missing "expression-2" gives you an infinite loop. – dbush Aug 20 '18 at 18:39
1

@HarishRajora That will increment `i` until it evaluates to false which is (if starting value is positive) probably a signed int overflow and undefined behavior. – Osiris Aug 20 '18 at 18:39
@Osiris Thanks a lot. Understood. – Harish Rajora Aug 20 '18 at 18:41

score 3 · Answer 2 · answered Aug 20 '18 at 18:30

3

All the parts of the for-loop are optional:

initialization: you can do the initialization outside of the loop.
condition: will be true if empty
last part: you can also do nothing.

for(;;) is an infinite loop (if there are no breaks inside the loop).

answered Aug 20 '18 at 18:30

Maxime Chéramy

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score 2 · Answer 3 · answered Aug 20 '18 at 18:31

The expressions which are not present do nothing.

If the initialisation is absent, nothing is initialised.

If the test is absent, nothing is tested (and the body loops forever unless it contains abreak, return, etc.)

If the increment is absent, nothing is incremented.

score 2 · Answer 4 · edited Aug 20 '18 at 19:08

The for loop contains three clauses. They can be virtually any expression, but the standard way of using it is for(<init>; <loop condition>; <loop variable increment>). All of these are optional: if you omit the loop condition, it will be replaced by a constant that evaluates to true. You could for instance replace

for(int i=0; i<10; i++) {
   ...
}

with

for(int i=0; i<10; ) {
    ...
    i++;
}

for(;;) is an infinite loop. Any for loop that has second statement omitted is an infinite loop.

There are basically three ways to get out of an infinite loop, break, return and goto.

In most cases you should use break or return, but goto also has its usage.

If you want to just break out of the loop and continue after it, use break.
If you want to quit the function that contains the loop, use return. However, this can be bad practice since it will likely violate the "single exit" practice.
If you want to break out of a nested loop, goto may be appropriate.

And yes, you could exit a loop by calling exit() or simply make the program crash too, but I think you get the point.

Question from comments

for(i<2; i++; )

This does not make much sense. The first expression i<2 will not affect anything and is likely to be optimized away by the compiler. The second expression i++ will evaluate to false if i is 0:

if i has an initial value of 0, the loop body is not evaluated at all.
if i is initially a negative number. The loop iterates until i reaches the value 0, the last iteration occurs with i == 0 and the next test will end the loop.
if i has an unsigned type and is greater than 0. The loop iterates until i reaches the maximum value for its type and wraps to 0, the last iteration occurs with i == 0 and the next test stops the loop. If i is not unsigned, arithmetic overflow will cause undefined behavior, which is bad: the loop may stop or continue indefinitely, depending on the compiler choices, or anything else may occur.

@user3386109 Was adding it when you wrote the comment, but thanks. — klutt, Aug 20 '18 at 18:33
I took the liberty to fix some typos and refine the explanation for the `i++` condition expression. — chqrlie, Aug 20 '18 at 19:15

How is for loop parsed internally as in assembly language?

4 Answers4