#include <stdio.h>
main()
{
int i;
for(i=0; i<0, 5; i++)
printf("%d\n", i);
}
I am unable to understand the i<0, 5 part in the condition of the for loop.
Even if I make it i>0, 5, there's no change in output.
How does this work?
On topic
The comma operator will always yield the last value in the comma separated list.
Basically it's a binary operator that evaluates the left hand value but discards it, then evaluates the right hand value and returns it.
If you chain multiple of these they will eventually yield the last value in the chain.
As per anatolyg's comment, this is useful if you want to evaluate the left hand value before the right hand value (if the left hand evaluation has a desirable side effect).
For example i < (x++, x/2) would be a sane way to use that operator because you're affecting the right hand value with the repercussions of the left hand value evaluation.
http://en.wikipedia.org/wiki/Comma_operator
Sidenote: did you ever hear of this curious operator?
int x = 100;
while(x --> 0) {
// do stuff with x
}
It's just another way of writing x-- > 0.
Comma operator evaluates i<0 Or i>0 and ignores. Hence, it's always the 5 that's present in the condition.
So it's equivalent to:
for(i=0;5;i++)
The coma operator is done to the initialization and to the increment part, to do something like for(i=0,j=20;i<j;i++,j--), if you do it in the comparation part it will evaluate the last one (as it was already answered before)
i<0,5 will always evaluate to 5, as always the right expression will be returned for ex1,ex2 .
The comma operator is intended for cases where the first operand has some side effects. It's just an idiom, meant to make your code more readable. It has no effect on the evaluation of the conditional.
For example,
for (i = 0; i<(i++, 5); i++) {
// something
}
will increment i, and then check if i<5.
Related
So, I understand that i++ increments post the condition is fulfilled, while ++i increments before the condition is fulfilled. That takes me to my question:
#include<stdio.h>
int main()
{
int i=0;
while(++i<10)
printf("%d\n",i);
return 0;
}
Now here we need the initialization of i from 0 as when it goes in the while loop, it will get incremented first, and thus it will be like while(1<10), and so it will print numbers from one to ten.
Second code:
#include<stdio.h>
int main()
{
int i=0;
while(i++<10)
printf("%d\n",i);
return 0;
}
Now since i++ increments after the value has been used, so why do we initialize i from i=0, as if it gets incremented after the comparision of value of i takes place, then why isn't 0 getting printed as well because the first loop should go like while(0<10), and not like while(1<10)? So, why is zero not getting printed?
Thank you for your time.
Yes, the first condition will evaluate to 0 < 10. You can verify that by changing it to while (i++ < 1) and see that the loop still runs (once), so clearly it's using 0 for the condition (1 < 1 would be false of course).
So why does it print 1? Because it doesn't print whichever value was used for the condition. It prints the current value of i. And the current value of i is 1 at that point because i was incremented right after it was used in the condition.
The difference of pre- and post- increment applies only to the value that the expression evaluates to.
After the evaluation of i++<10 is completed, the incrementation is also completed, and i has the value 1 in the first iteration.
printf("%d\n",i); is executed after the evaluation of i++<10, so i has the value 1 here in the first iteration.
Zero isn't printed because i is incremented right after the comparison, like this:
while(i<10) {
i += 1;
printf("%d\n",i);
}
The first sentence of your question is almost the answer:
i++ increments post the condition is fulfilled
It's incremented after the condition is checked, so i++ < 10 will increment i regardless of whether the condition ends up true or false.
The operation of ++ is not done before or after any “condition” it is in. It must be completed sometime in or around the full expression it is in. The answers that say i is updated “before” or “after” the ++i or i++ is evaluated are incorrect.
Prefix ++i does two separate things that may happen in any order:
It evaluates to the value of i after one is added.
It adds one to the stored value of i.
Postfix i++ does two separate things that may happen in any order:
It evaluates to the value of i before one is added.
It adds one to the stored value of i.
That change to the stored value of i is called a side effect. It is disconnected from the main evaluation of the expression. It can be performed before, during, or after the evaluation of i, but the evaluation still returns the pre- or post-increment value, as described above.
A full expression is one that is not contained inside another expression. So, in while (++i < 10), ++i < 10 is a full expression. The side effect can occur before any part of that is evaluated, after the value of i plus one is calculated, or after the < is evaluated. It can also occur in parts (such as updating the bytes of i one by one) during the evaluations. However, the side effect must occur after any previous full expression and before any later full expression. That is because the C standard says there is a sequence point between any two full expressions. (And there are some other rules about order of execution.)
Even if the stored value of i is updated before i++ or after ++i, the expression must still produce the value of i before or after the increment, respectively. For example, for i++, the compiler can fetch i, add one, store i, but then use the pre-add value in the expression.
The result of i++ is the value of i. As a side effect i is incremented.
The result of ++i is the value of i + 1. As a side effect i is incremented.
The statement
x = i++;
is logically equivalent to
tmp = i;
x = tmp;
i = i + 1;
with the caveat that the assignments to x and i can happen in any order, or even simultaneously (interleaved or in parallel).
Similarly, the statement
x = ++i;
is logically equivalent to
tmp = i + 1;
x = tmp;
i = i + 1;
with the same caveat as above. Again, these are logical equivalents, not what the compiler actually generates - depending on the compiler and the code involved there may not be a temporary.
In the second code, the increment will take place after the condition but before the printing statement.
Thus, the incremented value of i is printed.
This is your answer in simple language.
#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.
This question already has answers here:
C comma operator
(4 answers)
Closed 5 years ago.
Whats the difference between:
int i;
for( i=0;i<5 && i<3;i++)
{
printf("stackoverflow.com");
}
and
int i;
for( i=0;i<5 , i<3;i++)
{
printf("stackoverflow.com");
}
I mean use of && instead of ','
Regards
In the second code block, only i < 3 is actually used to evaluate whether the loop should exit. The i < 5 expression is evaluated, but its value is discarded. See the comma operator for more info.
There is functionally no difference between your examples because i < 3 is the limiting expression and appears second. Both loops will exit when i reaches 3. However, if you switched the terms so that you have i < 3, i < 5, then the second kicks out when it reaches 5 because only the value of i < 5 is considered.
In this case they will do exactly the same.
The thing that differs is that in the first case i<3 will not be evaluated every time. A statement on the form A && B will only execute B if the return value of A is true. The reason is that the logical value of && is true if both operands are true. If one of them are false, the whole expression is false, so if the left operand evaluates to false, there's no need to evaluate the right operand.
The comma operator on the other hand evaluates both operands, but the result of the left operand is discarded.
When using the comma operator , in a statement like the:
( <expression1>, <expression2> )
the <expression1> is evaluated and discarded, then the <expression2> is evaluated and its value is returned. In summary the right most expression is evaluated and its value returned. Other expressions are evaluated and discarded. So your statement of:
for(i=0; i<5 , i<3; i++)
is equivalent to:
for(i=0; i<3; i++)
As for the first statement, this expression:
(i<5 && i<3)
is a simple AND boolean evaluation. A funny one because it is enough to say:
(i<3)
Long story short, who ever made this example probably expects you to say "both conditions evaluate to second expression".
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.
I'm just learning some C, or rather, getting a sense of some of the arcane details. And I was using VTC advanced C programming in which I found that the sequence points are :
Semicolon
Comma
Logical OR / AND
Ternary operator
Function calls (Any expression used as an argument to a function call is finalized the call is made)
are all these correct ?. Regarding the last one I tried:
void foo (int bar) { printf("I received %d\n", bar); }
int main(void)
{
int i = 0;
foo(i++);
return 0;
}
And it didnt print 1, which according to what the VTC's guy said and if I undertood correctly, it should, right ?. Also, are these parens in the function call the same as the grouping parens ? (I mean, their precedence). Maybe it is because parens have higher precedence than ++ but I've also tried foo((i++)); and got the same result. Only doing foo(i = i + 1); yielded 1.
Thank you in advance. Please consider that I'm from South America so if I wasnt clear or nothing makes sense please, oh please, tell me.
Warmest regards,
Sebastian.
Your code is working like it should and it has nothing to do with sequence points. The point is that the postfix ++ operator returns the non-incremented value (but increments the variable by 1 nonetheless).
Basically:
i++ – Increment i by one and return the previous value
++i – Increment i by one and return the value after the increment
The position of the operator gives a slight hint for its semantics.
Sequence means i++ is evaluted before foo is invoked.
Consider this case (I am not printing bar!):
int i = 0;
void foo (int bar) { printf("i = %d\n", i); }
int main(void){
foo(i++);
return 0;
}
i = 1 must be printed.
C implements pass-by-value semantics. First i ++ is evaluated, and the value is kept, then i is modified (this may happen any time between the evaluation and the next sequence point), then the function is entered with the backup value as the argument.
The value passed into a function is always the same as the one you would see if using the argument expression in any other way. Other behavior would be fairly surprising, and make it difficult to refactor common subexpressions into functions.
When you do something like:
int i = 0, j;
j = i++;
the value of i is used first and then incremented. hence in your case the values of i which is 0 is used (hence passed to your function foo) and then incremented. the incremented values of i (now 1) will be available only for main as it is its local variable.
If you want to print 1 the do call foo this way:
foo(++i);
this will print 1. Rest you know, why!