CODEWITHSUNDEEP
javaintegerlong-integerinteger-overflow
aandis
aandis

Reputation: 4212

Java parseInt vs parseLong

Why are they different?

String a = "576055795";
long b = 10 * Integer.parseInt(a);
long c = 10 * Long.parseLong(a);

System.out.println(b); // Prints 1465590654
System.out.println(c); // Prints 5760557950

Upvotes: 1

Views: 26219

Answers (1)

user3553031
user3553031

Reputation: 6214

Integer.parseInt() returns an int, which is a signed 32-bit integer. 10 is also an int; multiplying 576055795 by 10 as ints overflows and yields an int, which is then promoted to a long.

Long.parseLong() returns a long, which is a signed 64-bit integer. Multiplying it by 10 yields a long with no overflow.

Upvotes: 15

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