linux - [Solved-4 Solutions] What is “:-!!” in C code ? - ubuntu - red hat - debian - linux server - linux pc
Problem :
What is “:-!!” in C code ?
Linux - Solution 1:
Expression
sizeof(struct { int: -!!(e); }))click below button to copy the code. By - Linux tutorial - team
- (e): Compute expression e.
- !!(e): Logically negate twice: 0 if e == 0; otherwise 1.
- -!!(e): Numerically negate the expression from step 2: 0 if it was 0; otherwise -1.
- struct{int: -!!(0);} --> struct{int: 0;}: If it was zero, then we declare a struct with an anonymous integer bitfield that has width zero. Everything is fine and we proceed as normal.
- struct{int: -!!(1);} --> struct{int: -1;}: On the other hand, if it isn't zero, then it will be some negative number. Declaring any bitfield with negative width is a compilation error.
Linux - Solution 2:
- The : is a bitfield.
- As for !!, that is logical double negation and so returns 0 for false or 1 for true. And the - is a minus sign, i.e. arithmetic negation.
- It's all just a trick to get the compiler to barf on invalid inputs.
- Consider BUILD_BUG_ON_ZERO.
- When -!!(e) evaluates to a negative value, that produces a compile error.
- Otherwise -!!(e) evaluates to 0, and a 0 width bitfield has size of 0. Hence the macro evaluates to a size_t with value 0.
- BUILD_BUG_ON_NULL is very similar, but yields a pointer rather than an int.
Linux - Solution 3:
Some people have seem to be confusing these macros with assert().
- These macros are implement a compile-time test, while assert() is a runtime test.
Linux - Solution 4:
- It's creating a size 0 bitfield if the condition is false, but a size -1 (-!!1) bitfield if the condition is true/non-zero.
- In the former case, there is no error and the struct is initialized with an int member.