C++ - Legal Function Naming

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Legal Function Naming

Which of the following function name is legal in C++?
A. $_foo()
B. !_foo()
C. 1_foo()
D. -_foo()
E. #_foo()

Total Answers and Comments: 3

Last Update: February 16, 2009 Asked by: prettyfox

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Best Rated Answer

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Submitted by: OSaienni

A is the only one that will work.

The reason are :

!, - are operators,

# is the preprocessor operator.
#define, #pragma, also use #9 -> stringify value to char.

1 - numeric value (can be used at end of function / method)

$ isn't used in C++ (I may be wrong here but I've never used $) - So legal naming as its not a number/operator

Above answer was rated as good by the following members:
santhosh.kanchanapally, yzesong

September 10, 2008 09:13:21

OSaienni

Member Since: September 2008 Contribution: 7

RE: Legal Function Naming

A is the only one that will work.

The reason are :

! - are operators

# is the preprocessor operator.
#define #pragma also use #9 -> stringify value to char.

1 - numeric value (can be used at end of function / method)

$ isn't used in C++ (I may be wrong here but I've never used $) - So legal naming as its not a number/operator

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December 07, 2008 12:33:54

burraganesh

Member Since: December 2008 Contribution: 20

RE: Legal Function Naming

$_foo()
$ symbol is used by compiler in a process called "name mangling" where each functions name and signature are preceded by $.
Thus $ is a valid symbol that a compiler can understand

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February 16, 2009 01:29:00

santhosh.kanchanapally

Member Since: April 2008 Contribution: 10

RE: Legal Function Naming

void $_foo();

Can be used as function name

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