64bit Integer has the same limit as 32bit integer

Question

This is just a short question but I'm still curious. When I initialize a int variable with the highest value 2147483647 it's allright. but when I want to initialize a long and assign it to it's max value, it gives me a "number too large" error

int i = 2147483647;
long j = 9223372036854775807;

Almost certainly you forgot to add the L at the end of the literal. Check the dupe link. — markspace, Jul 20 '18 at 20:38

score 1 · Accepted Answer · answered Jul 20 '18 at 20:38

1

9223372036854775807 is an int literal, but is too large to fit into an int, hence the error. You can use a long literal instead by adding L to the end of it:

long j = 9223372036854775807L;
// Here --------------------^

answered Jul 20 '18 at 20:38

Mureinik

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64bit Integer has the same limit as 32bit integer

1 Answers1