Unexpected Output For C Program For Int Datatype

Question

Can anyone explain why the output is not as expected ? I was expecting 16 but got 14.

#include<stdio.h>

int main()
{
    int x=016;
    printf("%d",x);
    return 0;
}

You should use %o to print octal numbers — not %d. Octal numbers start with a 0 on input. To get a leading zero on output, you need %#o. See scanf() and printf(). (And yes, this is a very tongue-in-cheek comment — it is accurate up to a point, but it is also very misleading. There's no way to decide whether x contains an octal or decimal value; it really contains a binary number and the formatted I/O functions control the appearance.) — Jonathan Leffler
– Jonathan Leffler, Commented Aug 5, 2017 at 6:11

BLUEPIXY · Accepted Answer · 2017-08-05 05:14:35Z

4

Numeric literals beginning with 0 are interpreted as octal numbers.

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

E.g 016(8 base) ==> 0 x 8^2 + 1 x 8^1 + 6 x 8^0 ==> 0 + 8 + 6 ==> 14(10 base)

%d of printf outputs the value of int in 10 bases, so the result is 14.

If you output int as an octal number using %o, 16 is obtained.

E.g

int x = 016;
printf("%#o", x);//016

If you want results with three digits include leading 0 using %d,

printf("%03d", 16);//016

Please refer to the reference of printf for details of format string.

answered Aug 5, 2017 at 4:57

BLUEPIXY

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1 Comment

Thanks for helping out.