Understanding sizeof(char) in 32 bit C compilers

Question

(sizeof) char always returns 1 in 32 bit GCC compiler.

But since the basic block size in 32 bit compiler is 4, How does char occup

Answer 1

sizeof(char) is always 1. Always. The 'block size' you're talking about is just the native word size of the machine - usually the size that will result in most efficient operation. Your computer can still address each byte individually - that's what the sizeof operator is telling you about. When you do sizeof(int), it returns 4 to tell you that an int is 4 bytes on your machine. Likewise, your structure is 8 bytes long. There is no information from sizeof about how many bits there are in a byte.

The reason your structure is 8 bytes long rather than 5 (as you might expect), is that the compiler is adding padding to the structure in order to keep everything nicely aligned to that native word length, again for greater efficiency. Most compilers give you the option to pack a structure, either with a #pragma directive or some other compiler extension, in which case you can force your structure to take minimum size, regardless of your machine's word length.

char is size 1, since that's the smallest access size your computer can handle - for most machines an 8-bit value. The sizeof operator gives you the size of all other quantities in units of how many char objects would be the same size as whatever you asked about. The padding (see link below) is added by the compiler to your data structure for performance reasons, so it is larger in practice than you might think from just looking at the structure definition.

There is a wikipedia article called Data structure alignment which has a good explanation and examples.