The following code prints a value of 9. Why? Here return(i++) will return a value of 11 and due to --i the value should be 10 itself, can anyone explain how this works?
#include<stdio.h>
main()
{
int i= fun(10);
printf("%d\n",--i);
}
int fun (int i)
{
return(i++);
}
There is a big difference between postfix and prefix versions of ++.
In the prefix version (i.e., ++i), the value of i is incremented, and the value of the expression is the new value of i.
In the postfix version (i.e., i++), the value of i is incremented, but the value of the expression is the original value of i.
Let's analyze the following code line by line:
int i = 10; // (1)
int j = ++i; // (2)
int k = i++; // (3)
i is set to 10 (easy).
Two things on this line:
i is incremented to 11.
The new value of i is copied into j. So j now equals 11.
Two things on this line as well:
i is incremented to 12.
The original value of i (which is 11) is copied into k. So k now equals 11.
So after running the code, i will be 12 but both j and k will be 11.
The same stuff holds for postfix and prefix versions of --.
Prefix:
int a=0;
int b=++a; // b=1,a=1
before assignment the value of will be incremented.
Postfix:
int a=0;
int b=a++; // a=1,b=0
first assign the value of 'a' to 'b' then increment the value of 'a'
The function returns before i is incremented because you are using a post-fix operator (++). At any rate, the increment of i is not global - only to respective function. If you had used a pre-fix operator, it would be 11 and then decremented to 10.
So you then return i as 10 and decrement it in the printf function, which shows 9 not 10 as you think.
In fact return (i++) will only return 10.
The ++ and -- operators can be placed before or after the variable, with different effects. If they are before, then they will be processed and returned and essentially treated just like (i-1) or (i+1), but if you place the ++ or -- after the i, then the return is essentailly
return i;
i + 1;
So it will return 10 and never increment it.
There are two examples illustrates difference
int a , b , c = 0 ;
a = ++c ;
b = c++ ;
printf (" %d %d %d " , a , b , c++);
Here c has value 0 c increment by 1 then assign value 1 to a so value
of a = 1 and value of c = 1
next statement assiagn value of c = 1 to b then increment c by 1 so
value of b = 1 and value of c = 2
in printf statement we have c++ this mean that orginal value of c
which is 2 will printed then increment c by 1 so printf statement
will print 1 1 2 and value of c now is 3
you can use http://pythontutor.com/c.html
int a , b , c = 0 ;
a = ++c ;
b = c++ ;
printf (" %d %d %d " , a , b , ++c);
Here in printf statement ++c will increment value of c by 1 first then
assign new value 3 to c so printf statement will print 1 1 3
The postfix increment ++ does not increase the value of its operand until after it has been evaluated. The value of i++ is i.
The prefix decrement increases the value of its operand before it has been evaluated. The value of --i is i - 1.
Prefix increment/decrement change the value before the expression is evaluated. Postfix increment/decrement change the value after.
So, in your case, fun(10) returns 10, and printing --i prints i - 1, which is 9.
i++ is post increment. The increment takes place after the value is returned.
First, note that the function parameter named i and the variable named i in main() are two different variables. I think that doesn't matter that much to the present discussion, but it's important to know.
Second, you use the postincrement operator in fun(). That means the result of the expression is the value before i is incremented; the final value 11 of i is simply discarded, and the function returns 10. The variable i back in main, being a different variable, is assigned the value 10, which you then decrement to get 9.
Explanation:
Step 1: int fun(int); Here we declare the prototype of the function fun().
Step 2: int i = fun(10); The variable i is declared as an integer type and the result of the fun(10) will be stored in the variable i.
Step 3: int fun(int i){ return (i++); } Inside the fun() we are returning a value return(i++). It returns 10. because i++ is the post-increement operator.
Step 4: Then the control back to the main function and the value 10 is assigned to variable i.
Step 5: printf("%d\n", --i); Here --i denoted pre-increement. Hence it prints the value 9.
Let's keep this as simple as possible.
let i = 1
console.log('A', i) // 1
console.log('B', ++i) // 2
console.log('C', i++) // 2
console.log('D', i) // 3
A) Prints the value of I.
B) First i is incremented then the console.log is run with i as it's the new value.
C) Console.log is run with i at its current value, then i will get incremented.
D) Prints the value of i.
In short, if you use the pre-shorthand i.e(++i) I will get updated before the line is executed. If you use the post-shorthand i.e(i++) the current line will run as if I had not been updated yet then i get increased so the next time your interpreter comes across i it will have been increased.
It has to do with the way the post-increment operator works. It returns the value of i and then increments the value.
Actually what happens is when you use postfix i.e. i++, the initial value of i is used for returning rather than the incremented one. After this the value of i is increased by 1. And this happens with any statement that uses i++, i.e. first initial value of i is used in the expression and then it is incremented.
And the exact opposite happens in prefix. If you would have returned ++i, then the incremented value i.e. 11 is returned, which is because adding 1 is performed first and then it is returned.
fun(10) returns 10. If you want it to return 11 then you need to use ++i as opposed to i++.
int fun(int i)
{
return ++i;
}
Related
The semicolon has been added after the first while loop, but why is the value of the i variable 3 here, where j is 2?
#include<stdio.h>
int main()
{
int i=1;
while(i++<=1);
printf("%d",i);
int j=1;
while(j++<=1)
printf("%d",j);
return 0;
}
Both while loops run once (and once only); the difference is that, in the second case (the j loop) you are printing the 'control variable' inside the loop but, for the first case, you are printing it after the loop condition has evaluated to false. Also note that, in the first case, the semicolon immediately following the while statement defines the body of that loop as empty1.
Let's break down the first loop into steps:
On the first test of the condition, i++ evaluates to 1 and then i is incremented – so the loop runs.
On the second test, i++ evaluates to 2 (so the loop doesn't run) but i is still (post-)incremented, leaving it with the value of 3 (as shown in the output).
The same thing happens with j in the second loop but, in that case, as previously mentioned, you are displaying the value in the body of the loop (on its only run), so you see the value after the first (post-)increment.
As noted in the comments, if you add another printf("%d", j); after the body of the loop (which, in that case, consists of a single statement), you will see that j, too, has the value 3 when that loop has finished.
1 More precisely, the semicolon (on its own) defines a null statement, which forms the body of the while loop.
It is often helpful to clarify such 'null loops' by putting the semicolon on a line by itself (some compilers, with full warnings or static analysis enabled, may even suggest you do this):
#include<stdio.h>
int main()
{
int i = 1;
while (i++ <= 1)
; // Body of loop - Null Statement
printf("%d", i); // Loop has finished
int j = 1;
while (j++ <= 1)
printf("%d", j); // Body of loop
// Loop has finished
return 0;
}
For starters let;s consider how the postfix increment operator works. From the C Standard (6.5.2.4 Postfix increment and decrement operators)
2 The result of the postfix ++ operator is the value of the
operand. As a side effect, the value of the operand object is
incremented (that is, the value 1 of the appropriate type is added to
it).
Now consider this while loop
int i=1;
while(i++<=1);
In the first iteration of the loop the value of the expression i++ as the value of its operand that is 1. So the loop will iterate a second time, Due to applying the side effect to the variable i it will be equal already to 2 when the expression i++<=1 will evaluate the second time.
Now the value of i is equal to 2 and is greater than 1. so the loop will be interrupted. Again due to applying the side effect of the postfix increment operator to the variable i it will be equal to 3. This value is outputted in the following call of printf.
In simplest terms,
consider three things for each iteration of first while loop:
first iteration:
int i = 1;//initialization
while(i++<=1); //1. i evaluated as 1,
//2. loop condition is true causing loop to iterate again.
//3. ...but not before ++ causes i==2.
second iteration:
while(i++<=1); //1. i evaluated as 2,
//2. loop condition is false causing loop to eventually exit
//3. ...but not before ++ causes i==3.
printf is not included in the while loop (because of the semicolon) but
when program flow finally reaches printf("%d",i); the value of i is output as 3
In the second loop, because printf is included in the while construct, it will output the value of j for each iteration. For the same reasons as in loop one, it will also iterate only twice, and its values at time of output will be 2 & 3.
Using a debugger to set break points, and a watch on i, you can step through code such as this to see the sequence of these effects as they happen.
For explanation porpouses look at this example:
int main()
{
int i=1;
while(i++<=1)
printf("%d",i);
printf("%d",i);
}
The Output is:
2
3
Ok let's dive into the programm procedure:
Declaring i (i = 1)
while checking Condition and it's true because i++ returns 1(old Value). That's because the ++ are after the i. If you would code it like this ++i than it returns 2(new Value) (i = 2)
execute printf (i = 2)
while checking Condition and it's false because i++ returns 2 and 2 is not <= 1 (i = 3)
execute printf (i = 3)
Do you understand?
If it solved your thoughts jumble ;) mark it as answer.
You've got many answers already so I won't try to explain it in words but with an illustration in code. I've added two functions:
One that acts like a prefix increment operator (++i)
One that acts like a postfix increment operator (i++)
I'm only using the postfix version in the program though. I've added logging to the function so you can see how it works.
#include <stdio.h>
// this acts as if you put ++ before the variable
int preinc(int* p) {
++*p;
return *p;
}
// this acts as if you put ++ after the variable
int postinc(int* p) {
int rv = *p;
++*p;
printf("returning %d but incremented to %d\n", rv, *p);
return rv;
}
int main() {
int i=1;
while(postinc(&i) <= 1);
printf("end result: %d\n---\n", i);
int j=1;
while(postinc(&j) <= 1)
printf("in loop: %d\n", j);
printf("end result: %d\n", j);
}
Output:
returning 1 but incremented to 2
returning 2 but incremented to 3
end result: 3
---
returning 1 but incremented to 2
in loop: 2
returning 2 but incremented to 3
end result: 3
If you add the semicolon in the end of the while you will take: i = 3 and j = 3(IN THE OUTPUT).
If you add the semicolon only in the end of the printf and not to the while you will take i = 2 and j = 2(IN THE OUTPUT), but in the end you will, also, have the values
i = 3 and j = 3.
This happens because the "i" variable is incremented only after the first iteration of the loop, in which the empty statement (terminated by the semicolon) is evaluated.
that is why it is highly recommended to do "++i" rather than "i++", because then the incrementation is performed prior to the evaluation.
if you flip it here, you will see that i will be equal to 2 as well, regardless of the presence of the semicolon
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Can someone please explain this C code to me. This was one of the questions asked in my exam to find the output.
#include<stdio.h>
int r()
{
static int num = 7;
return num--;
}
int main()
{
for(r(); r(); r())
printf("%d\n",r() );
}
I can't understand how the for loop is working (condition and increment/decrement statement) and how this code gives this output.
Output:
5
2
Let's consider the loop
for(r(); r(); r())
the function r is called in the init expression
for(r(); r(); r())
^^^
Its static variable num is decreased and becames equal to 6.
Then the condition of the loop is checked
for(r(); r(); r())
^^^
Again the function is called and its static variable becames equal to 5.
Within the call of printf
printf("%d\n",r() );
there is called the function r. Because within the function there is used the post-decrement operator then the function returns 5 but the value of num becames equal to 4. So the returned value 5 is outputted.
Then the third expression of the loop is evaluated.
for(r(); r(); r())
^^^
num becames equal to 3.
Again the condition is checked and num becames equal to 2.
for(r(); r(); r())
^^^
In the call of printf the returned value 2 is outputted but within the function r the value of num is decremented and becames equal to 1.
After that the third expression of the loop is evaluated.
for(r(); r(); r())
^^^
num becames equal to 0.
This value is returned in the condition by the function r. Within the function the static variable now is equal to -1.
As returned value is 0 the loop stops its iterations.
From the C Standard (6.5.2.4 Postfix increment and decrement operators)
2 The result of the postfix ++ operator is the value of the operand.
As a side effect, the value of the operand object is incremented (that
is, the value 1 of the appropriate type is added to it)...
and
3 The postfix -- operator is analogous to the postfix ++ operator,
except that the value of the operand is decremented (that is, the
value 1 of the appropriate type is subtracted from it).
A for loop of the form
for(init; condition; repeat) {
body;
}
is roughly equivalent to
init;
while (condition) {
body;
repeat;
}
So your given statement is equivalent to:
r();
while (r()) {
printf("%d\n", r());
r();
}
The first call to r initializes the static variable num to 7. It then returns this value while also decrementing it to 6. This return value is not used.
Then the while condition is tested. This call to r() returns the value 6 while decrementing num to 5. Since this is non-zero, the condition succeeds, so we go into the loop body.
Then it executes printf("%d\n", r());. This call to r() returns the value 5 while decrementing the variable to 4. The return value is printed by printf(), so we see 5.
Then we call r() as the repetition operation. This decrements num to 3, and the return value is ignored.
Then we go back to the while condition. This decrements num to 2 while returning 3. This is non-zero, so we go into the body.
Then we execute printf("%d\n", r());. As on the previous iteration, r() returns the current value of num while decrementing it to 1, so 2 is printed.
Then we call r() as the repetition operation. This decrements num() to 0 and returns 1, but the return value is ignored.
We go back to the while condition. This time, r() returns the current value 0 while decrementing it to -1. This time, the return value is zero, so the loop stops.
The following code prints a value of 9. Why? Here return(i++) will return a value of 11 and due to --i the value should be 10 itself, can anyone explain how this works?
#include<stdio.h>
main()
{
int i= fun(10);
printf("%d\n",--i);
}
int fun (int i)
{
return(i++);
}
There is a big difference between postfix and prefix versions of ++.
In the prefix version (i.e., ++i), the value of i is incremented, and the value of the expression is the new value of i.
In the postfix version (i.e., i++), the value of i is incremented, but the value of the expression is the original value of i.
Let's analyze the following code line by line:
int i = 10; // (1)
int j = ++i; // (2)
int k = i++; // (3)
i is set to 10 (easy).
Two things on this line:
i is incremented to 11.
The new value of i is copied into j. So j now equals 11.
Two things on this line as well:
i is incremented to 12.
The original value of i (which is 11) is copied into k. So k now equals 11.
So after running the code, i will be 12 but both j and k will be 11.
The same stuff holds for postfix and prefix versions of --.
Prefix:
int a=0;
int b=++a; // b=1,a=1
before assignment the value of will be incremented.
Postfix:
int a=0;
int b=a++; // a=1,b=0
first assign the value of 'a' to 'b' then increment the value of 'a'
The function returns before i is incremented because you are using a post-fix operator (++). At any rate, the increment of i is not global - only to respective function. If you had used a pre-fix operator, it would be 11 and then decremented to 10.
So you then return i as 10 and decrement it in the printf function, which shows 9 not 10 as you think.
In fact return (i++) will only return 10.
The ++ and -- operators can be placed before or after the variable, with different effects. If they are before, then they will be processed and returned and essentially treated just like (i-1) or (i+1), but if you place the ++ or -- after the i, then the return is essentailly
return i;
i + 1;
So it will return 10 and never increment it.
There are two examples illustrates difference
int a , b , c = 0 ;
a = ++c ;
b = c++ ;
printf (" %d %d %d " , a , b , c++);
Here c has value 0 c increment by 1 then assign value 1 to a so value
of a = 1 and value of c = 1
next statement assiagn value of c = 1 to b then increment c by 1 so
value of b = 1 and value of c = 2
in printf statement we have c++ this mean that orginal value of c
which is 2 will printed then increment c by 1 so printf statement
will print 1 1 2 and value of c now is 3
you can use http://pythontutor.com/c.html
int a , b , c = 0 ;
a = ++c ;
b = c++ ;
printf (" %d %d %d " , a , b , ++c);
Here in printf statement ++c will increment value of c by 1 first then
assign new value 3 to c so printf statement will print 1 1 3
The postfix increment ++ does not increase the value of its operand until after it has been evaluated. The value of i++ is i.
The prefix decrement increases the value of its operand before it has been evaluated. The value of --i is i - 1.
Prefix increment/decrement change the value before the expression is evaluated. Postfix increment/decrement change the value after.
So, in your case, fun(10) returns 10, and printing --i prints i - 1, which is 9.
i++ is post increment. The increment takes place after the value is returned.
First, note that the function parameter named i and the variable named i in main() are two different variables. I think that doesn't matter that much to the present discussion, but it's important to know.
Second, you use the postincrement operator in fun(). That means the result of the expression is the value before i is incremented; the final value 11 of i is simply discarded, and the function returns 10. The variable i back in main, being a different variable, is assigned the value 10, which you then decrement to get 9.
Explanation:
Step 1: int fun(int); Here we declare the prototype of the function fun().
Step 2: int i = fun(10); The variable i is declared as an integer type and the result of the fun(10) will be stored in the variable i.
Step 3: int fun(int i){ return (i++); } Inside the fun() we are returning a value return(i++). It returns 10. because i++ is the post-increement operator.
Step 4: Then the control back to the main function and the value 10 is assigned to variable i.
Step 5: printf("%d\n", --i); Here --i denoted pre-increement. Hence it prints the value 9.
Let's keep this as simple as possible.
let i = 1
console.log('A', i) // 1
console.log('B', ++i) // 2
console.log('C', i++) // 2
console.log('D', i) // 3
A) Prints the value of I.
B) First i is incremented then the console.log is run with i as it's the new value.
C) Console.log is run with i at its current value, then i will get incremented.
D) Prints the value of i.
In short, if you use the pre-shorthand i.e(++i) I will get updated before the line is executed. If you use the post-shorthand i.e(i++) the current line will run as if I had not been updated yet then i get increased so the next time your interpreter comes across i it will have been increased.
It has to do with the way the post-increment operator works. It returns the value of i and then increments the value.
Actually what happens is when you use postfix i.e. i++, the initial value of i is used for returning rather than the incremented one. After this the value of i is increased by 1. And this happens with any statement that uses i++, i.e. first initial value of i is used in the expression and then it is incremented.
And the exact opposite happens in prefix. If you would have returned ++i, then the incremented value i.e. 11 is returned, which is because adding 1 is performed first and then it is returned.
fun(10) returns 10. If you want it to return 11 then you need to use ++i as opposed to i++.
int fun(int i)
{
return ++i;
}
#include <stdio.h>
#include <string.h>
main()
{
int a;
a=1;
while(a++<=2)
;
printf("%d",a);
}
This code gives the output as 4.
How is a getting incremented?
How does a++ work?
First things first, let's understand what happens when you type a++. What you are using is called a post increment operator. To explain its behavior, one can look at the "verbose equivalent":
prev_a = a;
a = a + 1;
return prev_a;
Note that the value returned by the a++ expression is the value it had before the increment.
On the other hand, if you had written ++a, you would be using a pre increment operator, which returns the value after increment. Kind of like:
a = a + 1;
return a;
So when you write a++ <= 2, you are comparing the current value of a, and incrementing it "after comparison", for the next iteration.
How does while work?
Now when you write something like while (expr) { /* ... */ }, what happens is that expr is evaluated before every iteration, and the iteration actually runs only if the result of the expression is true.
So what happens?
First iteration
a is equal to 1;
it is smaller or equal to 2: we decide go for another iteration;
increment a;
Second iteration
a is equal to 2;
it is smaller or equal to 2: we decide go for another iteration;
increment a;
Second iteration
a is equal to 3;
it is not smaller or equal to 2: we decide to stop after this iteration;
still increment a, as part of "evaluating the whole expression a++ <= 2";
In the end, a is equal to 4.
I guess what tricked you is the fact that the previous value of a (before increment) is used for comparison, but that the increment does always happen as part of the expression. a is being incremented, even if the comparison returns false!
In the end, your code is equivalent to:
int a;
a = 1;
int prev_a;
while(true) {
prev_a = a; // the previous value is saved for the comparison
a = a + 1; // the increment ALWAYS happen
if (prev_a > 2) {
break;
}
}
When a=3 we have a++<=2 i.e. 3++<=2 - hence it becomes 4 at the end of the statement even though the condition fails. You could have simply put a printf to check this:
while(a++<=2)
{
printf("%d ",a);
}
After 2 iteration the value of a is 2. So the condition is true and the value is incremented again i,e a=3.
Now the while condition is checked again and it returns false. But a is incremented again as (3++<=2) so a=4
Because the condition of the loop (a++<=2) evaluated by 3 times.
1st time: a is 1, and a <= 2, and then a -> 2 by operator++
2nd time: a is 2, and a <= 2, and then a -> 3 by operator++
3rd time: a is 3, and a not <= 2, and then a-> 4 by operator++
How is a getting incremented?
a is increased by the a++ in while(a++<=2);.
a++ will return the value of a and add one to it, so when a is 3, a++<=2 will be false, but a still get added by one.
init_stat execute status result
a =1 while(a++<=2) success a =2
a =2 while(a++<=2) success a =3
a =3 while(a++<=2) fail a =4
The following code prints a value of 9. Why? Here return(i++) will return a value of 11 and due to --i the value should be 10 itself, can anyone explain how this works?
#include<stdio.h>
main()
{
int i= fun(10);
printf("%d\n",--i);
}
int fun (int i)
{
return(i++);
}
There is a big difference between postfix and prefix versions of ++.
In the prefix version (i.e., ++i), the value of i is incremented, and the value of the expression is the new value of i.
In the postfix version (i.e., i++), the value of i is incremented, but the value of the expression is the original value of i.
Let's analyze the following code line by line:
int i = 10; // (1)
int j = ++i; // (2)
int k = i++; // (3)
i is set to 10 (easy).
Two things on this line:
i is incremented to 11.
The new value of i is copied into j. So j now equals 11.
Two things on this line as well:
i is incremented to 12.
The original value of i (which is 11) is copied into k. So k now equals 11.
So after running the code, i will be 12 but both j and k will be 11.
The same stuff holds for postfix and prefix versions of --.
Prefix:
int a=0;
int b=++a; // b=1,a=1
before assignment the value of will be incremented.
Postfix:
int a=0;
int b=a++; // a=1,b=0
first assign the value of 'a' to 'b' then increment the value of 'a'
The function returns before i is incremented because you are using a post-fix operator (++). At any rate, the increment of i is not global - only to respective function. If you had used a pre-fix operator, it would be 11 and then decremented to 10.
So you then return i as 10 and decrement it in the printf function, which shows 9 not 10 as you think.
In fact return (i++) will only return 10.
The ++ and -- operators can be placed before or after the variable, with different effects. If they are before, then they will be processed and returned and essentially treated just like (i-1) or (i+1), but if you place the ++ or -- after the i, then the return is essentailly
return i;
i + 1;
So it will return 10 and never increment it.
There are two examples illustrates difference
int a , b , c = 0 ;
a = ++c ;
b = c++ ;
printf (" %d %d %d " , a , b , c++);
Here c has value 0 c increment by 1 then assign value 1 to a so value
of a = 1 and value of c = 1
next statement assiagn value of c = 1 to b then increment c by 1 so
value of b = 1 and value of c = 2
in printf statement we have c++ this mean that orginal value of c
which is 2 will printed then increment c by 1 so printf statement
will print 1 1 2 and value of c now is 3
you can use http://pythontutor.com/c.html
int a , b , c = 0 ;
a = ++c ;
b = c++ ;
printf (" %d %d %d " , a , b , ++c);
Here in printf statement ++c will increment value of c by 1 first then
assign new value 3 to c so printf statement will print 1 1 3
The postfix increment ++ does not increase the value of its operand until after it has been evaluated. The value of i++ is i.
The prefix decrement increases the value of its operand before it has been evaluated. The value of --i is i - 1.
Prefix increment/decrement change the value before the expression is evaluated. Postfix increment/decrement change the value after.
So, in your case, fun(10) returns 10, and printing --i prints i - 1, which is 9.
i++ is post increment. The increment takes place after the value is returned.
First, note that the function parameter named i and the variable named i in main() are two different variables. I think that doesn't matter that much to the present discussion, but it's important to know.
Second, you use the postincrement operator in fun(). That means the result of the expression is the value before i is incremented; the final value 11 of i is simply discarded, and the function returns 10. The variable i back in main, being a different variable, is assigned the value 10, which you then decrement to get 9.
Explanation:
Step 1: int fun(int); Here we declare the prototype of the function fun().
Step 2: int i = fun(10); The variable i is declared as an integer type and the result of the fun(10) will be stored in the variable i.
Step 3: int fun(int i){ return (i++); } Inside the fun() we are returning a value return(i++). It returns 10. because i++ is the post-increement operator.
Step 4: Then the control back to the main function and the value 10 is assigned to variable i.
Step 5: printf("%d\n", --i); Here --i denoted pre-increement. Hence it prints the value 9.
Let's keep this as simple as possible.
let i = 1
console.log('A', i) // 1
console.log('B', ++i) // 2
console.log('C', i++) // 2
console.log('D', i) // 3
A) Prints the value of I.
B) First i is incremented then the console.log is run with i as it's the new value.
C) Console.log is run with i at its current value, then i will get incremented.
D) Prints the value of i.
In short, if you use the pre-shorthand i.e(++i) I will get updated before the line is executed. If you use the post-shorthand i.e(i++) the current line will run as if I had not been updated yet then i get increased so the next time your interpreter comes across i it will have been increased.
It has to do with the way the post-increment operator works. It returns the value of i and then increments the value.
Actually what happens is when you use postfix i.e. i++, the initial value of i is used for returning rather than the incremented one. After this the value of i is increased by 1. And this happens with any statement that uses i++, i.e. first initial value of i is used in the expression and then it is incremented.
And the exact opposite happens in prefix. If you would have returned ++i, then the incremented value i.e. 11 is returned, which is because adding 1 is performed first and then it is returned.
fun(10) returns 10. If you want it to return 11 then you need to use ++i as opposed to i++.
int fun(int i)
{
return ++i;
}