Struct - Explain the output:

Question

I have the following C code.

struct values{ int a:3; int b:3; int c:2; }; void main(){ struct values v={2,-6,5}; printf("%d %d %d",v.a,v.b,v.c); }

When I execute the code, I am getting the following output:

2 2 1.

But output should be 2 -6 5, right?

If I'm wrong please explain.

The Clang compiler on OS X gives me a nice warning: "implicit truncation from 'int' to bitfield changes value from -6 to 2 [-Wconstant-conversion]". — Martin R
– Martin R, Commented Sep 1, 2013 at 15:53

Oliver Charlesworth · Accepted Answer · 2013-09-01 15:40:33Z

5

-6 exceeds the range of a 3-bit signed int. Therefore you're observing an artifact of ~~undefined~~ implementation-defined behaviour (in practice, the most-significant bits of your value are being thrown away).

edited Sep 1, 2013 at 15:40

answered Sep 1, 2013 at 15:34

Oliver Charlesworth

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1 Comment

ouah Over a year ago

yes but implementation defined behavior (c99, 6.3.1.3p3) not undefined

P0W · Accepted Answer · 2013-09-01 15:39:34Z

4

{2, | -6, | 5 } 010 last 3 bits | 010 last 3 bits | 01 last 2 bits 2 2 1

answered Sep 1, 2013 at 15:39

P0W

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2 Comments

Strawberry Over a year ago

Consider elaborating a little on this answer.

0decimal0 Over a year ago

@P0W Was simply looking through the answers and was in no mood to vote but could not resist after looking at your explanation ..... brief but explains everything except if one understands something .. +1 :) and for others too :P

Gokul Nath KP · Accepted Answer · 2013-09-01 15:34:33Z

No. Output is 2 2 1.

The C compiler converts the values to Binary, and stores in the memory.

Binary value of 2 : 00000010

Binary value of -6: 11111010 (11111001+1)

Binary value of 5 : 00000101

While storing in memory:

For 2, 010 will be stored.

For -6, 010 will be stored.

For 5, 01 will be stored.

When you access these variables from your main method, for v.a "010" will be returned, here left most bit is for sign.

So v.a is 2. Similarly v.b is 2 and v.c is 1.

Hope it helps.

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