C++: Using 1 byte enum

By default, enum size in C++ is usually 4 bytes, both in Windows and Linux. If you need to use 1 byte enum, you have to declare it as follows:

enum e : unsigned char { a, b };

This requires at least C++ 11. printf("size = %ld\n", sizeof(e)) will output 1.

If you are using gcc (e.g. Eclipse), you can also use the compiler flag -fshort-enum (Eclipse properties - C/C++ Build - Settings - Tool Settings - GCC C++ Compiler - Miscellaneous) to convert all enums in code to 1 byte, but you won't have that option in Visual Studio. So it is better to use the unsigned char option.

Use memset only with zero

Consider the following C++ definitions:

typedef struct {

int i;

double d[3];

} A_STRUCT

A_STRUCT s;

When I use memset(&s, 0, sizeof(s)), i and d values are set to zero as expected. When I use memset(&s, -1, sizeof(s)), I would expect all values to be -1 but on inspection you will see that they have strange values, in my case i was -1 but d values were -nan. When I use memset(&s, 1, sizeof(s)), I get 16843009 for i and 7.74...e-304 for d values.

Reason: The memset() function writes bytes, not words. So writing 1 to sizeof(int)*100 bytes writes 00000001 to every set of 8-bits.Thus, each integer in binary looks like the following:

0000 0001 0000 0001 0000 0001 0000 0001 (1 int = 4 bytes)

which in decimal is, exactly, 16843009..

memset doesn't only work with 0. It also works with all numbers with identically repeating byte pattern. Like for example ~0.