sizeof(Structure) Confusion

In the following piece of code,

#include<stdio.h>
typedef struct {
    int bit1:1;
    int bit3:4;
    int bit4:4;
} node;

int main(){
    node n,n1,n2,ff[10];

    printf("%d\n",sizeof(node));
    return 0;
}

How do I predict the size of the开发者_JAVA技巧 structure?

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You cannot predict it without knowing the compiler and the target platform it compiles to.

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It depends on the platform and the compiler settings (packing, alignment, 32/64 machine)

According to comp.lang.c FAQ list

"Bit-fields are thought to be nonportable, although they are no less portable than other parts of the language."

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Unpredictable in general, but practically speaking it'll come out sizeof(int) more often than not. Which itself is very often 4; less commonly 2 and surely 8 at times.

Most of the time the bit fields will be packed and most of the time the int type will have 9 or more bits of storage.

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You will find that the size of your structure changes based on compiler optimization settings. I'd predict anywhere between 2 and 12 bytes for this structure.

Even when using bit-fields like you do, you can't always predict what the size of a struct is going to be. The compiler may have every bit-field take up the full space of an int, or possibly just the 1 or 4 bits that you specify. Using bit-fields, while it is great on memory storage space, is often bad for running time and executable size.

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Usually every compiler decides how to pack the union so you can't make many assumptions on the final size. They can decide a different layout according to its parameters.

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Add the bit-field sizes, divide by 8*sizeof(int), and take the ceiling of that value. In your example, it'll be 4.