I could not know why it happen! Want to know the reason.
{
int i=01;
printf("%d\n",i);
}
output: 1
but
{
int i=011;
printf("%d\n",i);
}
output: 9
Does anybody have the answer?
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3 Answers
011 is an octal constant. 11 (b8) = 9 (b10).
C11 (n1570), § 6.4.4.1 Integer constants
An octal constant consists of the prefix 0 optionally followed by a sequence of the digits 0 through 7 only.
4 Comments
Iqbal
why it will print octal number where as I have given a decimal integer value!
md5
Because your numeric constant starts with a
0. Read the quotation.
Daniel Fischer
@Iqbal Yes, of course,
1*8 + 2*1 = 10. If you start your literals with a 0, it's an octal literal. Try using the %o format in printf.
David Ranieri
Read the quotation ;) prefix 0 for an octal constant (base 8) (and prefix 0x or 0X or simply x or X for hexadecimal base 16)
011 = Octal, (1*8)+1=9 ........................
Comments
The numbers which are preceded by 0 is called octal numbers in c programming .
to evaluate such an expression we simply follow a conversion rule of converting octal to decimal number system
For conversion the following steps are to be proceed
such as 011
here 0 indicate the number is octal
and we are require to convert 11 which is (base 8) to decimal (base 10)
11= 1x8^1+1x8^0
=8+1
=9
