When displaying the value of variable “int a = 011”, I get 9. Why? [duplicate]

问题

This question already has answers here:

What does it mean when a numeric constant in C/C++ is prefixed with a 0? (7 answers)

printf with “%d” of numbers starting with 0 (ex “0102”) giving unexpected answer (ex '“66”) (3 answers)

Closed 4 years ago.

With this code snippet:

int a = 011;
printf("a = %d", a);

Why is the result

a = 9

回答1:

011 is an octal value and its decimal equivalent is 9. Preceding integer literal with 0 indicates octal value.
Use %o specifier in printf to print the value in octal.

回答2:

A leading 0, in an int literal or int constant, represents the octal value. It is called an octal constant.

Related: C11 standard, chapter 6.4.4.1, Integer constants, Paragraph 3,

An octal constant consists of the prefix 0 optionally followed by a sequence of the digits 0 through 7 only.

回答3:

With 0 at the beginning of of a numeric literal, you specify the octal system. And 11 in the octal system is 1*8 + 1 = 9.

来源：https://stackoverflow.com/questions/30615960/when-displaying-the-value-of-variable-int-a-011-i-get-9-why