When an integer is initialized as <code>int a = 010</code>, <code>a</code> is actually set to 8, but for <code>int a = 10</code>, <code>a</code> is set to 10. Can anyone tell me why <code>a</code> is not set to 10 for <code>int a = 010</code>?

Because it's interpreting <code>010</code> as a number in octal format. And in a base-8 system, the number <code>10</code> is equal to the number <code>8</code> in base-10 (our standard counting system). More generally, in the world of C++, prefixing an integer literal with <code>0</code> specifies an octal literal, so the compiler is behaving exactly as expected.

Why is initializing an integer in C++ to 010 different from initializing it to 10?

1 Answers

Because it's interpreting 010 as a number in octal format. And in a base-8 system, the number 10 is equal to the number 8 in base-10 (our standard counting system).

More generally, in the world of C++, prefixing an integer literal with 0 specifies an octal literal, so the compiler is behaving exactly as expected.