Why is the size of a struct not the sum of the sizes of its variables?

Question

For example, the following code:

#include <stdio.h>

struct exemplo{
    char letra;
    int numero;
    float flutuante;
};

int main()
{
    printf("Tamanho do char: %u\n", sizeof(char));
    printf("Tamanho do int: %u\n", sizeof(int));
    printf("Tamanho do float: %u\n", sizeof(float));
    printf("Tamanho da struct: %u\n", sizeof(struct exemplo));
}

Show on console:

Tamanho do char: 1
Tamanho do int: 4
Tamanho do float: 4
Tamanho da struct: 12

That is, the struct size is 12 and not 9 (1+4+4) as expected. Why does this happen?

Answer 1

In fact the structure may be the same size as the sum of its members. It depends on the alignment. If the members allow to assemble the structure in an aligned way will have the same size.

The case presented really requires aligning the type char for a word, then there is a waste of space. This is called padding. The bytes used for the padding are not used, indicate nothing. And there is no problem in that, it is the normal one. It is done assom for performance issues. In most cases the compiler knows how to do better than the programmer.

If you need something different can be solved with #pragma pack(1) or __attribute__((packed)). More information can be obtained in the answer linked above. Rarely is this really necessary and it is better not to touch it if you do not understand all the implications it will have.