What is the full "for" loop syntax in C? - c

I have seen some very weird for loops when reading other people's code. I have been trying to search for a full syntax explanation for the for loop in C but it is very hard because the word "for" appears in unrelated sentences making the search almost impossible to Google effectively.
This question came to my mind after reading this thread which made me curious again.
The for here:
for(p=0;p+=(a&1)*b,a!=1;a>>=1,b<<=1);
In the middle condition there is a comma separating the two pieces of code, what does this comma do? The comma on the right side I understand as it makes both a>>=1 and b<<=1.
But within a loop exit condition, what happens? Does it exit when p==0, when a==1 or when both happen?
It would be great if anyone could help me understand this and maybe point me in the direction of a full for loop syntax description.

The comma is not exclusive of for loops; it is the comma operator.
x = (a, b);
will do first a, then b, then set x to the value of b.
The for syntax is:
for (init; condition; increment)
...
Which is somewhat (ignoring continue and break for now) equivalent to:
init;
while (condition) {
...
increment;
}
So your for loop example is (again ignoring continue and break) equivalent to
p=0;
while (p+=(a&1)*b,a!=1) {
...
a>>=1,b<<=1;
}
Which acts as if it were (again ignoring continue and break):
p=0;
while (true) {
p+=(a&1)*b;
if (a == 1) break;
...
a>>=1;
b<<=1;
}
Two extra details of the for loop which were not in the simplified conversion to a while loop above:
If the condition is omitted, it is always true (resulting in an infinite loop unless a break, goto, or something else breaks the loop).
A continue acts as if it were a goto to a label just before the increment, unlike a continue in the while loop which would skip the increment.
Also, an important detail about the comma operator: it is a sequence point, like && and || (which is why I can split it in separate statements and keep its meaning intact).
Changes in C99
The C99 standard introduces a couple of nuances not mentioned earlier in this explanation (which is very good for C89/C90).
First, all loops are blocks in their own right. Effectively,
for (...) { ... }
is itself wrapped in a pair of braces
{
for (...) { ... }
}
The standard sayeth:
ISO/IEC 9899:1999 §6.8.5 Iteration statements
¶5 An iteration statement is a block whose scope is a strict subset of the scope of its
enclosing block. The loop body is also a block whose scope is a strict subset of the scope
of the iteration statement.
This is also described in the Rationale in terms of the extra set of braces.
Secondly, the init portion in C99 can be a (single) declaration, as in
for (int i = 0; i < sizeof(something); i++) { ... }
Now the 'block wrapped around the loop' comes into its own; it explains why the variable i cannot be accessed outside the loop. You can declare more than one variable, but they must all be of the same type:
for (int i = 0, j = sizeof(something); i < j; i++, j--) { ... }
The standard sayeth:
ISO/IEC 9899:1999 §6.8.5.3 The for statement
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 variables 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.133)
Both clause-1 and expression-3 can be omitted. An omitted expression-2 is replaced by a
nonzero constant.
133) Thus, clause-1 specifies initialization for the loop, possibly declaring one or more variables for use in
the loop; the controlling expression, expression-2, specifies an evaluation made before each iteration,
such that execution of the loop continues until the expression compares equal to 0; and expression-3
specifies an operation (such as incrementing) that is performed after each iteration.

The comma simply separates two expressions and is valid anywhere in C where a normal expression is allowed. These are executed in order from left to right. The value of the rightmost expression is the value of the overall expression.
for loops consist of three parts, any of which may also be empty; one (the first) is executed at the beginning, and one (the third) at the end of each iteration. These parts usually initialize and increment a counter, respectively; but they may do anything.
The second part is a test that is executed at the beginning of each execution. If the test yields false, the loop is aborted. That's all there is to it.

The C style for loop consists of three expressions:
for (initializer; condition; counter) statement_or_statement_block;
The initializer runs once, when the loop starts.
The condition is checked before each iteration. The loop runs as long it evaluates to true.
The counter runs once after each iteration.
Each of these parts can be an expression valid in the language you write the loop in. That means they can be used more creatively. Anything you want to do beforehand can go into the initializer, anything you want to do in between can go into the condition or the counter, up to the point where the loop has no body anymore.
To achieve that, the comma operator comes in very handy. It allows you to chain expressions together to form a single new expression. Most of the time it is used that way in a for loop, the other implications of the comma operator (e.g. value assignment considerations) play a minor role.
Even though you can do clever things by using syntax creatively - I would stay clear of it until I find a really good reason to do so. Playing code golf with for loops makes code harder to read and understand (and maintain).
The wikipedia has a nice article on the for loop as well.

Everything is optional in a for loop. We can initialize more than one variable, we can check for more than one condition, we can iterate more than one variable using the comma operator.
The following for loop will take you into an infinite loop. Be careful by checking the condition.
for(;;)

Konrad mentioned the key point that I'd like to repeat: The value of the rightmost expression is the value of the overall expression.
A Gnu compiler stated this warning when I put two tests in the "condition" section of the for loop
warning: left-hand operand of comma expression has no effect
What I really intended for the "condition" was two tests with an "&&" between. Per Konrad's statement, only the test on to the right of the comma would affect the condition.

Formal syntax of the for loop:
for (init-statement condition; iteration-expression)
statement
Corresponding flowchart:
Equivalent program:
{
init-statement
while (condition) {
statement
iteration-expression;
}
}
Except that
Names declared by init-statement (if init-statement is a declaration) and names declared by condition (if condition is a declaration) are in the same scope (which is also the scope of statement).
continue in statement will execute iteration-expression.
Empty condition is equivalent to while (true).

the for loop is execution for particular time for(;;)
the syntex for for loop
for(;;)
OR
for (initializer; condition; counter)
e.g (rmv=1;rmv<=15;rmv++)
execution to 15 times in for block
1.first initializ the value because start the value
(e.g)rmv=1 or rmv=2
2.second statement is test the condition is true or false ,the condition true no.of time execution the for loop and the condition is false terminate for block,
e.g i=5;i<=10 the condition is true
i=10;i<10 the condition is false terminate for block,
3.third is increment or decrement
(e.g)rmv++ or ++rmv

Related

Can we interchange the for loop parameters in C?

#include <stdio.h>
int main(void) {
char i=250;
for(i<0;i++;i=0,printf("%d", i));
return 0;
}
In this program, the output is 0. From what i understand is, a for loop should have the first parameter as initialisation, then condition, then increment. But in this question the initialisation is happening at the last and still the code is giving valid result. Can someone explain how?
The clauses or expressions in a for statement are always interpreted based on where they are (first, second, or third) in the for statement. In this code:
char i=250;
for(i<0;i++;i=0,printf("%d", i));
return 0;
i is set to either 250 (if char is unsigned) or −6 (if char is signed, eight-bit, and typical two’s complement wrapping is used for the conversion from 250 to char). (The C standard permits other possibilities, but they are unusual and are not discussed further in this answer.)
To start the loop, the initial clause i<0 is evaluated. Its result is inconsequential because it is ignored.
To decide whether the loop ends, the test clause i++ is evaluated. This produces either 250 or −6, per the above, and, separately, increments i to either 251 or −5. In either case, the result of the expression is non-zero, so the loop continues.
The body of the loop, ; is evaluated. Since this is an empty statement, it has no effect.
The post-iteration clause, i=0,printf("%d", i) is evaluated. This sets i to 0 and prints i, resulting in output of “0”.
The test clause i++ is evaluated again. Since i is zero, this produces 0, and, separately, increments i to 1. Since the result of the expression is zero, the loop terminates.
return 0; is executed, cause the program to end with a success status.
Short answer to your question "Can we interchange the for loop parameters in C?" is: No .
This code, if not meant to confuse the reader, must be a badly written one.
Walk-through
i is initialized with a value 250
i<0 as the initialization step is ignored
i++ is executed, since the old i is not 0, the loop continues
nothing in the loop body
set i=0 and print its value 0 out
come back to the i++, noting that the previous i value is zero, so loop stops
Actually for loop works in the following way:
for (step 0; step 1; step 3) {
step 2;
}
Here, step 0 is executed only once. Then step 1 -> step 2 -> step 3 -> step 1 and the loop goes on.
a for loop should have the first parameter as initialisation, then condition, then increment.
this is more like a convention. And explaining the for loop in this way makes perfect sense (especially because in step 1, the program executes the statement and continues to step 2 only when it returns true). And so we utilize for loop in this way.
"Can we interchange the for loop parameters in C?"
We can interchange the expressions itself (for whatever reason, for example as experiment because it usually makes no sense to do so), but we can't change the syntax (how an expression is evaluated at a certain place).
A for loop has a fixed syntax following conventionally the form:
for (initializations; condition; in-/decrements)
You can use expressions where you want to but it has quite different effects.
If you place for example the initialization expressions at the second place they are used as condition.
Same goes if you would take the in-/decrements and place them at the second place. Then they would be evaluated as condition, too.
Equivalently, if you would place the expression used as condition at the first or third place, this expression will not be used as the condition anymore.
That's what the C standard says to this topic:
6.8.5.3 The for statement
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. Ifclause-1is 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.161)
2 Both clause-1 and expression-3 can be omitted. An omitted expression-2 is replaced by a non zero constant.
Source: ISO/IEC 9899:2018 (C18), §6.8.5.3
A for loop for(init; cond; inc) { body } Is basically executed like this:
init;
while(cond) {
body;
inc;
}
For some cases it is possible to switch things, but in the general case it is not.

How is for loop parsed internally as in assembly language? [duplicate]

This question already has answers here:
Two semicolons inside a for-loop parentheses
(4 answers)
Closed 4 years ago.
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(;;)
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.
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).
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.
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.

C Programming - comma operator within while loop [duplicate]

This question already has answers here:
What does the comma operator , do?
(8 answers)
Closed 6 years ago.
Prog 1:
#include<stdio.h>
int main()
{
int i=0;
while(i<=8,i++);
printf("%d",i);
return 0;
}
Prog 2:
#include<stdio.h>
int main()
{
int i=0;
while(i++,i<=8);
printf("%d",i);
return 0;
}
The output of Prog 1 is 1 and that of Prog 2 is 9.
Can someone explain whats going here. How the two codes are different?
The comma operator evaluates both of its arguments in turn, throwing away the result, except for the last. The last evaluated expression determines the result of the entire expression.
i<=8,i++ - here the value of the expression is the value of i++, which is the value of i before being incremented. It's 0 so the loop immediately terminates.
i++,i<=8 - here the value of the expression is the value of i<=8 which is 0 only when i is incremented to 9.
On a personal note: I think the second form, while somewhat comparable to a for loop, is less clear to the reader of the code than an actual for loop.
1 while ( condition )
2 statement;
3 more_code();
In the above code snippet, the statement can be executed repeatedly as long as condition is true. On each iteration of the while loop, condition is evaluated to either true or false. If it's false, the while loop ends and execution continues beyond it's scope (in this case, line 4 with more_code().
We are usually accustomed to enclosing parts of code that we want to be executed in loop with curly brackets { and }, but that is not mandatory. If we do not do so, the looping code will consist of single statement, the one immediately following the while part.
It could actually be argued that the more common situation, where we combine while with curly brackets enclosed block of code could be interpreted as providing this block of code in place of a single statement, with braces providing information that the block should be treated (by compiler analysing it's relation with preceding and following code) as if it was a single statement.
However, as it is perfectly valid to provide a single statement, not a usual block of code, it's worth to understand that there is a valid statement that is empty. We get an empty statement by typing a semicolon without preceding it with a code causing anything. So following is perfectly valid:
1 code;
2 ; // empty statement
3 ; // another empty statement
or in fact this:
1 code;; // a "code" statement followed by empty statement in the same line
The while( condition ) part is not terminated with a semicolon, so if it's supposed to control some actual code (apart from condition), it should not be followed by a semicolon. If it is immediately followed by a semicolon, that semicolon will constitute (and be so interpreted by compiler) an empty statement, so the looping code will be empty. If that's unintended, then the code we wanted to be looped, whether a block of code or a statement, will not be looped, but rather executed once, after (and if) loop ends.
1 int a = 0;
2 while ( a < 3 ) ; // Next line is not part of loop - only the empty statement this semicolon creates is. This loop is infinite, or in other words it will never end.
3 a++; // This won't be executed even once.
4 printf("This never happens.");
(It's worth realizing that lines are only important for us, humans, in C. Lines and indentation can be misleading if they represent the intentions of programmer, when he failed to write the code functioning as he wanted it to.)
Therefore what happens in both snippets from the question, is we get condition evaluated continuously until it yields false. To understand what's going on we need to examine the way comma operator works.
(Note, while comma as a character can be used with a completely different meaning in various places in C - I can think of function declarations, definitions and calls - in this case comma character is part of condition, therefore it acts as an operator - something akin to + or % operators.)
expression1 , expression2
Comma operator causes expression1 to be evaluated first, then expression2, and returns the value of expression2.
Upon every evaluation of condition, we will thus evaluate both expressions, (in this case both being operands, i++ and i<=8), then consider value of the right one as result of comma operand itself, and thus as value of our condition. So the loop will keep repeating as long as the right operand resolves as true.
While usually we use condition to control the execution of loop, often, as in this case, condition may have "side" effects (intentional or unintended). In our case variable i is affected by every evaluation of condition: it is increased by one.
Our example differs only in order of operands of condition, therefore pay attention to the right operand which really controls the execution of the loop.
Let's examine the second example first. In this case we have condition i++, i<=8. This means upon every evaluation we first increase i, then check if it's less than or equal to 8. So on first evaluation of condition we will increase i from 0 to 1 and conclude that 1<=8, so the loop continues. The loop so constructed will break when i becomes 9, ie. on the 9th iteration.
Now as for the first example, the condition is i<=8, ++i. Since comparison has no side effects, that is we could perform any number of comparisons in any order and if that's the only thing we did, that is if we did not perform any other action in a way or order dependent of results of the comparisons, those comparisons would do absolutely nothing. As is in our case, we evaluate i<=8 which evaluates to true or false, but we make no use of this result, just proceed to evaluating the right operand. So left operand absolutely doesn't matter. Right operand, on the other hand, has both a side effect and it's value becomes value of entire condition. Before each loop iteration we check if i++ evaluates to true or false.
i++ is a unary operator of post-incrementation. It returns value of i then increases it by one (the difference between i++ and ++i is subtle but crucial in cases like this one). So what happens is we first check whether i is true or false, then i is increased by one.
In C there is no boolean type. Integers are considered to be true if they have a non-zero value.
So upon first evaluation of i++ we get 0, that is false. This means the loop is broken without even a single iteration. However it does not break evaluation of i++, which causes i to increase by one before we're done with the loop and execution proceeds beyond it. So once we're done with the while loop, i is already 1.
If we want to be very precise in our understanding, the part where we take the result of evaluating entire condition happens, chronologically, after we are finished executing any code involved in this evaluation. So we first memorize that i was 0 at the point we got toward i++ part, then we increase i by one, and then we are finished executing condition, so we provide value of 0 to the code that decides if we should do another (in this case first) iteration or jump beyond looping part and move on. This is exact reason why everything within condition will actually happen even though the fact that loop will end was already determined: it was determined, but it was not checked and acted upon until condition finishes executing.
The expression separator operator , forces evaluation from left to right, and is also a sequencing point.
Prog 1: consider i <= 8, i++. i <= 8 is evaluated and discarded then i++ is evaluated. The entire expression has the unincremented value of i. Since i is initially 0, the while loop terminates on the first iteration. The output will be that singly incremented value of i, i.e. 1.
Prog 2: i++ is evaluated and the result discarded, then i <= 8 is evaluated with the new value of i since , is a sequencing point. So the while loop runs until i <= 8 is no longer true with the incremented value of i. The output will be 9.

Blank 'For' Loop Giving Output

I encountered a strange FOR LOOP question in the book. Here is the code for the loop -
#include<stdio.h>
int main()
{
int i=1,j=1;
for(;;)
{
if(i>5)
break;
else
j+=i;
printf("%d\n",j);
i+=j;
}
return 0;
}
The program prints 2 & 5 as the output. Now could anyone please explain how this for loop is executing?
That for loop is a classic idiom for a non-terminating loop. It's what you use when you want a loop that never terminates, or whose termination is controlled by control flow statements inside the loop. The latter is the case here.
for(;;)
All parts of the for loop are empty. They key to understanding why it is a non-terminating loop is the empty controlling expression in the for statement.
The standard (6.8.5.3 The for statement) says, with my emphasis:
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.158)
Both clause-1 and expression-3 can be omitted. An omitted
expression-2 is replaced by a nonzero constant.
And this means that your loop will never terminate due to the controlling expression part of the for statement.
for(;;)
is a for loop which performs no initialisation, has no exit condition and performs no post-action.
It will loop forever unless code inside the loop contains a condition that can result in break being called.
Your loop is equivalent to
int i,j;
for(i=1, j=1 ;i<=5 ;i+=j) {
j+=i;
printf("%d\n",j);
}
for(;;) is equivalent to while(true) (or while(1) in old-school C), both of whose terminations are only controlled by the statements inside them.
Edit:
To rehash a couple of old jokes (borrowed from here):
Don't use for (;;) {} — it makes the statement cry.
Unless, of course, you #define EVER ;;.

a case of for loop in c

I came across the following question :
How many times will the following for loop run -
for(;0;)
printf("hello");
I executed and it runs 1 time . I am not able to understand how?
This will not execute even for 1 time. I guess you have a bad compiler?
Ok. I think you are using Turbo C ;-)
EDIT:
From C99 standard:
6.8.5.3 The for statement 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 variables 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.134)
It clearly states that condition is evaluated first before executing the loop. Any standard conforming compiler should not execute the loop for(;0;) {} even once.
Either the code you copied here is not really what is in your .c file or you have a buggy compiler.
Maybe you have an additional semicolon?: for(;0;); printf("!"); will print once.
for loops are defined as:
for(startExpression; testExpression; countExpression)
{
block of code;
}
startExpression is evaluated before the code;
testExpression is evaluated before the code;
countExpression is evaluated after code;
Decoding:
for(;0;)
Means
no startExpression
testExpression evaluated to false, therefore loop exits.
Edited to show correct for loop decoding.
The code as written above will never enter the for loop.
Check the code on ideone link.
My be this not what you have in your souce code, you probably typed a ; after for without noticing it like this:
for(;0;);
printf("hello");
In that case your program will print "hello".
Since the expression is 0, it is taken to be false. So, in this case the loop runs 0 times.

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