![]() Linker can do simple work to link regular static variables inside one translation unit. If static member is defined in class, and the class definition will be included into many translation unit, so that the linker has to do more work to decide which static member should be used as the only one through all the related translation unit.īut for regular static variables, they can only be used inside one translation unit, even in the case different static variables in different translation unit with same name, they will not affect each other. Static member has only one instance in the class scope, not like regular static variables used heavily in C, which has only one instatnce inside one translation unit. ![]() In order to make linker rules more simple, C++ requires that every object has a unique definition. So basically, this is not allowed because C++ do not allow this. That rule would be broken if C++ allowed in-classĭefinition of entities that needed to be stored in memory as objects. However, to avoidĬomplicated linker rules, C++ requires that every object has a uniqueĭefinition. Typically included into many translation units. Here is my understanding based on Stroustrup's statement about in-class definitionĪ class is typically declared in a header file and a header file is I suspect that floating point was left out of these types of constant expressions simply because they are not 'simple' enough. Note that he pretty much leaves out floating point as being able to play in 'constant expressions'. That can be evaluated by the compiler before the program Intuitively, constant expressions are simple expressions Functions, class objects, pointers, and references Floating literals (4.5.1)Ĭan be used only if explicitly converted to an integral Is composed of literals (4.3.1, 4.4.1, 4.5.1),Įnumerators (4.8), and consts initialized byĬonstant expressions. A constant expression evaluates toĪn integral or enumeration constant. In places such as array bounds (5.2), case labels (6.3.2),Īnd initializers for enumerators (4.8), C++ requires aĬonstant expression. In that way, no member definition is needed elsewhere, and you are notĪnd in Appendix C (Technicalities) in Section C.5 (Constant Expressions), Stroustrup has this to say about "constant expressions": I see no technical reason why struct type With no optimization, this will not hurt performance, since you're not optimizing anyways, but with optimizations enabled, this can hurt performance, since the compiler can no longer inline those constants into other source files, unless you enable "whole program optimization". Thus, the following code compiles at -O1 but NOT at -O0: // File a.hĬommand line: g++ a.cc b.cc -O0 -o a # Linker error: ld: undefined symbols: X::dįor maximal portability, you should declare your constants in header files and define them once in some source file. Using g++ with no optimization ( -O0), it automatically inlines constant integer variables but not constant double values. And in that case, you need to separate the declaration from the definition, since otherwise it would get defined in multiple translation units. ![]() ![]() In order to give it an address, it needs to be defined in some translation unit. It can still decide to give it a memory location - if its address is ever taken (or if it's passed by const reference to a function), that it must. It doesn't exist at runtime, and every use of it gets inlined into the surrounding code. I disagreed.The problem is that with an integer, the compiler usually doesn't have to ever create a memory address for the constant. $\int _0^\infty f(x)dx$ is finite but the following limit doesn't exist: $\lim_$ She was trying to say this functions integral will be 0, but still it won't be bounded. Find an example for a non-negative and continuous function s.t.
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