Why this statement int a= 012945; in C++ gives me an error like "invalid octal digit" ?

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I'm using visual studio 2017,so is it because of that it gives me an error "invalid octal digit"?

What I have tried:

if i give a integer variable value that starts with 0 then only this error comes something like this int e = 0129612;
but it doesn't gives any error if i just don't use 0 on the first place.

Posted 18-Jul-18 21:11pm

Member NFOC

Updated 19-Jul-18 7:05am

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KarstenK 19-Jul-18 4:56am

The language convention of C++ is saying that start with a "0" is an octet number. Compare to "0x" which is hexadecimal. :-O

4 solutions

Solution 3

C/C++ knows three types of integer literals: octal, decimal, and hex.
Since C++14 there is a fourth type: binary.

A literal beginning with a zero and followed by another digit is treated as octal value (base 8) which is not allowed to contain the digits 8 and 9.

See also integer literal - cppreference.com[^].

Posted 18-Jul-18 21:21pm

Jochen Arndt

Solution 4

Quote:
if i give a integer variable value that starts with 0 then only this error comes something like this int e = 0129612;
but it doesn't gives any error if i just don't use 0 on the first place.

Because it is the principle: integers starting with 0 are octal (base 8)

Posted 18-Jul-18 22:18pm

Patrice T

Solution 5

thank you everyone for responding it is of great help.

Posted 19-Jul-18 7:06am

Member NFOC

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Rick York · Accepted Answer · 2018-07-18T21:19:00

Solution 1

Octal digits can range from 0 to 7. This is because octal is base 8. A literal integer value starting with 0 is considered to be octal. If it starts with 0x it is a hexadecimal value.

Posted 18-Jul-18 21:19pm

Rick York

Updated 18-Jul-18 21:20pm

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