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C++17 post increment operation

By : Wael Al Khatib
Date : September 17 2020, 01:00 AM
Any of those help Before P0145 was adopted (in C++17), the first example had undefined behaviour. Anything could happen.
Before C++11, both had undefined behaviour.
code :

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Why does PHP post increment not behave as expected when in an array [ ] operation?

By : user3455708
Date : March 29 2020, 07:55 AM
This might help you You should not use a variable that gets incremented more than once in a line, else you risk this kind of unexpected behaviour.
code :
$EventArray[$lessonIndex]['Start'] = $TimeArray[$EventArray[$lessonIndex++]['Title']]['Start'];
$lessonIndex = 0;
$EventArray[$lessonIndex]['Title'] = 'Lesson 1';
$somevalue = $TimeArray[$EventArray[$lessonIndex]['Title']]['Start'];
$EventArray[$lessonIndex]['Start'] = $somevalue;

In Java, when exactly does the post-increment operation occur?

By : Shonali Peiris
Date : March 29 2020, 07:55 AM
With these it helps The n-- operation happens each time the while condition is evaluated; specifically when the left-hand side of the != operator is evaluated. It has nothing to do with the if statement. The while condition is evaluated at the start of each loop iteration; if it evaluates to true, then the body of the while loop (the if statement) is executed, but that's well after the n-- operation has finished. (But of course, then there's the next loop iteration.)

Not understanding the operation pre and post increment

By : Priyank Agarwal
Date : March 29 2020, 07:55 AM
it fixes the issue a++ is postfix notation, the value is first incremented, but the previous value is returned, which is then passed into prinf. ++a is prefix notation, the value is first incremented before it's passed into printf.
Your code is roughly equivalent to this:
code :
int a = 8;
printf("%d\n", a);

printf("%d\n", a);
printf("%d\n", a*a);
return 0;
int a = 8; int aTemp;

// a++
printf("%d\n", (aTemp = a, a = a + 1, aTemp) );

// ++a
printf("%d\n", (a = a + 1) );

post increment operation in while loop (java)

By : Wild Billa
Date : March 29 2020, 07:55 AM
Hope this helps This is one of the alternative ways I could wrap my head around this. Let f(ref i) be a function which takes in i by reference and increment it it's value by 1. So f(ref i) = i + 1
Now that we have f(ref i), the above code can be written as
code :
int i = 10
while( (f(ref i) -1) <=10 )
   f(ref i);
while(11 - 1 <= 10) {12}
while (13 -1 <= 10) -> break;

Post-increment vs Assignment in C++ operation precedence table

By : user2513388
Date : March 29 2020, 07:55 AM
I wish did fix the issue. "Precedence" is misleading. It has little to do in general with evaluation order (what happens first), but instead determines what is the operand of each operator for the purpose of evaluation. But let's examine your example.
code :
*(b++) = 5;
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