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I am beginner in C and can't understood how this piece of code is working:

struct marks{ int p:3; int c:3; int m:2; }; void main(){ struct marks s={2,-6,5}; printf("%d %d %d", s.p, s.c, s.m); }

I found that the output is: 2 2 1

But after a lot of try I was unable to figure out how the output is like that. I have less knowledge about the struct. That's why may be I am feeling some problem here.

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  • Possible duplicate of When to use bit-fields in C? Commented Jun 23, 2017 at 10:32
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    The int c and p have only 3 bits. With 3 bits you have only 8 possible values: from 0 to 7 using unsigned int, from 0 to 3 and from -1 to -4 using (signed) int. You cannot insert the value -6 into a 3 bit int, if you assign it only the 3 less significant bits are set into the variable. Commented Jun 23, 2017 at 10:37
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    What did you expect? Why do you think you can put two buckets full of water into a single bucket? That's not even related to bit-fields, but simply using too small integer types. Did you enable compiler wanrnings? Why not? If yes: didn't the complier warn? Why did you ignore them? Your code relies on implementation defined conversions. Commented Jun 23, 2017 at 11:05

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Well, inside the struct you defined- p:3, c:2, m:2 are actually denoting the bit fields. To know more about bit fields in C go through this link in SO or this link in Wikipedia.

For simplicity, know that the 3 or 2 after the colon(:) sign is representing bit-fields of width 3 or 2. That means 3 or 2 adjacent computer memory locations which have been allocated to hold a sequence of bits.

Now, inside the main() function of your code:

struct marks s={2,-6,5};

Here,

Binary value of 2: 00000010

Binary value of -6: 11111010

Binary value of 5: 00000101

Now according to bit-fields, from binary value of 2, we will take last 3 digits which is 010 (in decimal it is 2) and from binary value of -6, we will take last 2 digits which is 10 (in decimal it is 2) and from binary value of 5, we will take last 3 digits which is 01 (in decimal it is 1) and finally assign them to p, c or m.

That's how the output comes as 2 2 1.

Hope, I could make you clear.

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3 Comments

C does not support methods. An main is not a method in C++ either. And your explanation is not quite correct. The result is implementation defined, there is no guarantee for this result.
Thanks.that was my mistake(it would be main() function).edited.
But how the result is not guaranteed. Can you make it more clear? It will be helpful for everyone. Thanks.
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first of all your declaration of struct is using bit fields. 

struct marks{ int p:3; int c:3; int m:2; };

here p:3 means we are defining P to be an int and we are preserving only 3 bit of the given number. For example , You need 3 bits to represent the value 5, which would be represented as 101. The lower 2 bits are the value 1. That's why you get m=1.

This link will give you more explanation.

Thanks.

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Thanks, this was helpful :)