You can try like this : @echo off set Source=C:\TEMP\MyDoc.txt set Destination=E:\MyData\ set Filename=MyDoc set a=1 :loop if exist %Destination%\%Filename%(%a%).txt set /a a+=1 && goto :loop copy %Source% %Destination%\%Filename%(%a%).txt pause
No, it doesn’t, see: R Language Definition: Operators
It comes from the history of using integer values as booleans. If x is an int, but I am using it as a boolean as per if(x)… then incrementing will mean that whatever its truth value before the operation, it will have a truth-value of true after it (barring overflow). However, it’s impossible to predict … Read more
The results are undefined. You’re modifying a variable more than once in an expression (or sequence point to be more accurate). Modifying a variable more than once between sequence points is undefined, so don’t do it. It might be your compiler, for this particular case decides to evalate ++i + ++i as increment the last … Read more
Pre- or post-incrementing do not magically delay things until later. It’s simply inline shorthand. // pre-increment $var = 5; print(++$var); // increments first, then passes value (now 6) to print() // post-increment $var = 5; print($var++); // passes value (still 5) to print(), then increments Now let’s look at a loop. for ($i = 0; … Read more
No, the printout of 10 is correct. The key to understanding the reason behind the result is the difference between pre-increment ++x and post-increment x++ compound assignments. When you use pre-increment, the value of the expression is taken after performing the increment. When you use post-increment, though, the value of the expression is taken before … Read more
the answer to my question is not to use let (or shift, or…) but to use i=$((i+1)) when trying to check a bash script by setting ‘exit on non-zero status code‘ with set -e The bash manual states that set -e has the effect of ‘Exit immediately if a simple command exits with a non-zero … Read more
Simple, direct solution function nextChar(c) { return String.fromCharCode(c.charCodeAt(0) + 1); } nextChar(‘a’); As others have noted, the drawback is it may not handle cases like the letter ‘z’ as expected. But it depends on what you want out of it. The solution above will return ‘{‘ for the character after ‘z’, and this is the … Read more
These statements produce different results because of the way in which the operators bind. In particular, the prefix ++ operator has the same precedence as *, and they associate right-to-left. Thus ++*ptr is parsed as ++(*ptr) meaning “increment the value pointed at by ptr,”. On the other hand, the postfix ++ operator has higher precedence … Read more
No, this code is absolutely, demonstrably not threadsafe. import threading i = 0 def test(): global i for x in range(100000): i += 1 threads = [threading.Thread(target=test) for t in range(10)] for t in threads: t.start() for t in threads: t.join() assert i == 1000000, i fails consistently. i += 1 resolves to four opcodes: … Read more