How does +variable operate or +(+variable) operate?
int i=0;
while(+(+i--)!= 0){
// do
}
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1Just curious, is this your code ? If yes, why do you need it ? – Jagannath Mar 05 '15 at 06:44
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Is that Java or C, or C++? – ForceBru Mar 05 '15 at 06:44
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Is this [tag:c],[tag:c++] or [tag:java]? – Spikatrix Mar 05 '15 at 06:44
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I got it from a question paper of C – Q07 Mar 05 '15 at 06:48
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Then don't tag your question C++ and Java... – Lundin Mar 05 '15 at 07:18
4 Answers
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It's called the unary plus operator, and it has (almost) no effect on its argument.
By default, it only promotes its argument to an int. But since in your example i is an int already, +i is effectively a no-op.
Note that it can additionally be overloaded for custom classes in C++ (not in Java or C).
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It has an effect in an expression though: it applies integral promotion to the argument. – juanchopanza Mar 05 '15 at 06:51
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The value of the expression +variable is the same as the value of variable. The unary + operator changes neither the value of the expression nor the value of the variable.
R Sahu
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Both the + operators are unary operator, used for sign purpose and will not impact any functionality of -- (decrement operator).
Prateek Shukla
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The easiest way to see what it does is to run it and try it out.
Using i = 9, it printed 8, 7, 6, 5, 4, 3, 2, 1, 0.
Logically, what it does then is decrements i after performing the i != 0 check, but before the code runs.
This, you will find, is synonymous with while(i-- != 0) which fits the fact that + does not, in fact, affect your code.
Of course, when dealing with multiple operators on a single variable in a single expression you can quite often get undefined code; Code that may run differently on different compilers. For that sake, you probably shouldn't try using anything as confusing as that in your code.
Nonanon
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