In C, what does a colon mean inside a declaration? [duplicate]
It's a bitfield member. Your code means dumpable
occupies exactly 1 bit in the structure.
Bitfields are used when you want to pack members in bit-level. This can greatly reduce the size of memory used when there are a lot of flags in the structure. For example, if we define a struct having 4 members with known numeric constraint
0 < a < 20
b in [0, 1]
0 < c < 8
0 < d < 100
then the struct could be declared as
struct Foo {
unsigned a : 5; // 20 < 2^5 = 32
unsigned b : 1; //
unsigned c : 3; //
unsigned d : 7; // 100 < 2^7 = 128
};
then the bits of Foo may be arranged like
ddddddd c cc b aaaaa
--------- --------- --------- ----------
octet 1 octet 0
===========================================
uint32
instead of
struct Foo {
unsigned a;
unsigned b;
unsigned c;
unsigned d;
};
in which many bits are wasted because of the range of values
# wasted space which is not used by the program
# v v
ddddddd ccc
------------------------------------ ------------------------------------
uint32 uint32
b aaaaa
------------------------------------ ------------------------------------
uint32 uint32
so you can save space by packing many members together.
Note that the C standard doesn't specify how the bitfields are arranged or packed within an "addressable storage unit". Also, bitfields are slower compared with direct member access.
It means it's a bitfield - i.e. the size of dumpable is a single bit, and you can only assign 0 or 1 to it. Normally used in old code to save space, or in low-level code that interfaces with hardware (even though the packing is non-portable). See here for more information
If I remember correctly, when used inside of a struct the number after the colon signifies how many bits make up the variable (or a bitfield).
So unsigned dumpable:1;
is a single bit bitfield.