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Why is sizeof an operator?
Why is sizeof supposed to be an operator in C, & not a function?
Its usage seems similar to that of a function call, as in sizeof(int).
Is it supposed to be some kind of a pseudo-function?
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2Probably because `sizeof foo++` does not look like a function call, does not evaluate `foo++`, and `int` in `sizeof(int)` is not an expression. – R.. GitHub STOP HELPING ICE Oct 17 '10 at 19:25
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Because it's not a function; it's not possible to write a function that can take a type as an argument and return its size! Note also that sizeof is implemented at compile-time, and that the parentheses aren't necessary if you're using it on a variable.
Oliver Charlesworth
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A function call is evaluated at runtime. sizeof must be evaluated at compile time (though C99 introduces some exceptions to this, I believe)
leonbloy
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A function in C cannot do what sizeof needs to do.
A function by definition is allocated space at runtime and can operate on data only.
While sizeof operates on datatypes as well, which are compiler specific. A function does not know how to interpret the datatype "int" and ask the compiler for its size, by design.
Before C99 sizeof was totally compile-time, but in the new standard, it can be used to get information at runtime as well, for variable-length arrays.
sarshad
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function is something you can call at runtime. sizeof is only understood by the compiler.
unbeli
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sizeof has its effects at complite time not execution time. You only need the ( ) when you enter a type such as sizeof(int) but if 'i' is of type int you could do sizeof i
JonH
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