bit-fields

I want to have a typedef that is 1-bit integer, so I though of this typedef int:1 FLAG; but I'm getting errors with it, is there a way I can do so? Thanks No. The smallest addressable "thing" in a C Program is a byte or char . A char is at least 8 bits long. So you cannot have a type (or objects of any type) with less than 8 bits. What you can do is have a type for which objects occupy at least as many bits as a char and ignore most of the bits #include <limits.h> #include <stdio.h> struct OneBit { unsigned int value:1; }; typedef struct OneBit onebit; int main(void) { onebit x; x.value = 1; x

Here I have a code snippet. #include <stdio.h> int main() { struct value { int bit1 : 1; int bit2 : 4; int bit3 : 4; } bit; printf("%d",sizeof(bit)); return 0; } I'm getting the output as 4 (32 bit compiler). Can anyone explain me how? Why is it not 1+ 4 + 4 = 9? I've never worked with bit fields before so would love some help. Thank you. :) sampathsris When you tell the C compiler this: int bit1 : 1 It interprets it as, and allocates to it, an integer; but refers to it's first bit as bit1 . So if we consider your code: struct value { int bit1 : 1; int bit2 : 4; int bit3 : 4; } bit; What you

A typical use of bitfield is to declare a space efficient variable smaller than 8 bits. What i don't understand is the value of declaring those bits as short, int , long , bool etc. For example typedef struct{ int first:3, short second:3, char third:3 } somestruct; In above case, all 3 variables, i.e. first, second and third are 3 bit long. What is the value of declaring the variable first as int, second as short and third as char? Or, why is even a data type required? I should be able to declare the above as typedef struct{ first:3, second:3, third:3 } modifiedstruct; The modifiedstruct