What is the meaning of "l" in some variables initialization - C++ -

this question has answer here:

• what “l” mean @ end of integer literal? 1 answer

what meaning of "l" in variable initialisations? example:

#define maxpossible     (1000000000l) double = 1l; double b = 999999999l; 

is there difference between "l" , "l"?

this suffix type specifier, a , b can read more floating point literals here. short answer l , l both indicate long double. maxpossible can read integer literal here , l indicates long.

edit

as mike seymour kindly pointed out of literals integer literals. goes show times when not sanity check answer, wrong. simple sanity check have been follows:

#include <iostream> #include <typeinfo>  int main() {     std::cout << typeid( decltype( 1l ) ).name() << std::endl ;     std::cout << typeid( decltype( 999999999l ) ).name() << std::endl ;     std::cout << typeid( decltype( 1000000000l ) ).name() << std::endl ; } 

which gives me l each 1 , running through c++filt -t gives me long. have made literals floating point literals? either a:

• digits containing decimal point
• digits exponential notation, example 4e2

for example:

std::cout << typeid( decltype( .1l ) ).name() << std::endl ; std::cout << typeid( decltype( 1e2l ) ).name() << std::endl ; 

which gives me e both cases , running through c++filt -t gives me long double.