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Given that x is a variable of type int with the number 5 as its value, consider the following statement:
int y = !!x;
This is what I think it happens: x is implicitly casted to a bool and the first negation is executed, after that the last negation is made, so a cast and two negations.
My question is, isn't just casting to bool (executing int y = (bool)x; instead of int y = !!x) faster than using double negation, as you are saving two negations from executing.
I might be wrong because I see the double negation a lot in the Linux kernel, but I don't understand where my intuition goes wrong, maybe you can help me out.
There was no bool type when Linux was first written. The C language treated everything that was not zero as true in Boolean expressions. So 7, -2 and 0xFF are all "true". No bool type to cast to. The double negation trick ensures the result is either zero or whatever bit pattern the compiler writers chose to represent true in Boolean expressions. When you're debugging code and looking at memory and register values, it's easier to recognize true values when they all have the same bit patterns.
Addendum: According the C89 draft standard, section 3.3.3.3:
The result of the logical negation operator ! is 0 if the value of its operand compares unequal to 0, 1 if the value of its operand compares equal to 0. The result has type int . The expression !E is equivalent to (0==E).
So while there was no Boolean type in the early days of the Linux OS, the double negation would have yielded either a 0 or a 1 (thanks to Gox for pointing this out), depending on the truthiness of the expression. In other words any bit pattern in the range of INT_MIN..-1 and 1..INT_MAX would have yielded a 1 and the zero bit pattern is self-explanatory.
C language unlike other languages does not have bool type. bool in C is actually defined in stdbool.h which is not included in many C projects. Linux kernel is one such projects, it would be a pain to go through Linux code and update everything to use bool now as well. That is reason why Linux kernel does not use bool in C.
why !!x? This is done to ensure that value of y is either 1 or 0. As an example if you have this cocd
x=5;
int y = !!x;
We know that everything that non-zero values in C mean true. So above code would brake down to y= !!(5) followed by y = !(0) and than y = 1.
EDIT:
One more thing, I just saw OP mentioned casting to bool. In C there is no bool as base type, bool is defined type, thus compilers do not cast integers to Boolean type.
EDIT 2:
To further explain, in C++, Java and other languages when you type bool a = false you do not have to use headers or compile some other libraries or define bool type for bool type to work, it is already incorporated into compilers, where as in c you have to.
EDIT 3:
bool is not the same as _Bool.
The only reason I can imagine is because this saves some typing (7 chars vs 2 chars).
As #jwdonahue and #Gox have already mentioned, this is not the correct reason. C did not have bool when the linux kernel was written therefore casting to bool was not an option.
As far as efficiency goes, both are equivalent because compilers can easily figure this out. See https://godbolt.org/g/ySo6K1
bool cast_to_bool_1(int x) {
return !!x;
}
bool cast_to_bool_2(int x) {
return (bool) x;
}
Both the functions compile to the same assembly which uses the test instruction to check if the argument is zero or not.
test edi, edi // checks if the passed argument is 0 or not
setne al // set al to 0 or 1 based on the previous comparison
ret // returns the result
I had the need to code a statement of the form
a = a || expr;
where expr should be evaluated and the result be assigned to a iff a is not set. this relies on the logical OR's short-circuiting capabilities.
The shorter way to write the above would, of course, be
a ||= expr;
but (to my surprise) C does not have logical assignment operators.
So my question is twofold. First, is there a shorter way to write the first statement in standard C (the ternary operator is even worse - a = a ? a : expr requires me to spell out a thrice).
Secondly, why aren't there logical assignments in C? The possible reasons I could think of are:
it makes the grammar harder to parse?
there is some subtlety in handling short-circuiting for these cases?
it was considered superfluous (but isn't that an argument against ALL the operator assignments?)
EDIT
Please unlock this question because:
The question it has been linked to (as a alleged duplicate of) HAS NOT BEEN ANSWERED. The (accepted) answer to that question states that ||= is not present because duplicates the functionality of |=. That is the wrong answer. |= does not short-circuit.
C and C++ are NOT the same languages. I wish to know why C doesn't have it. In fact, the fact that derived languages like C++ and, particularly, Java (which did not suffer from the problems of legacy code as has been suggested in Edmund's answer) makes the question even more interesting.
EDIT 2
It now seems like my original intent was wrong. In the statement a = a || expr (where a is integral and expr returns an integral value, first both a and expr will be implicitly converted to "booleans", and then the "boolean" value will be assigned to a. This will be incorrect — the integral value will be lost. Thanks, Jens and Edmund.
So for the first part of the question, the correct ways, not alternatives :), to code my intention would be:
if (!a) a = expr;
or
a = a ? a : expr;
they should be optimized the same (I think) though personally I would prefer the first one (because it has one less a to type).
However, the second part of the question still remains. The arguments that Jens and Edmund about have given about the ambiguity in a ||= expr apply equally well to a = a || expr. the assignment case can simply be treated as the normal one:
convert a to boolean
if it is true, the value of the entire expression becomes equal to the boolean value of a
otherwise evaluate expr, convert result to boolean, assign to a, and return it
The steps above seem to be the same for both the assignment and normal case.
a ||= expr is problematic due to short circuit evaluation of its equivalent a = a || expr.
To have a ||= expr function like a = a || expr consider OP's assertion:
"In the statement a = a || expr ..., first both a and expr will be implicitly converted to "booleans","
This is not quite correct. expr will not be converted if a evaluates to true. This would make a difference should expr be something like scanf() or rand() or some function that affected the state of the program.
Code such as a ||= scanf("%d", &i) != 1; would only attempt to scan data with a false value in a. Although it would be possible to extend the language this way, additional short-circuit operators to the current set of || and && would likely cause more coding problems than clear simplifications.
On the other hand: A quick, if obfuscated, way to write code where functions return non-zero codes on error.
// Perform functions until an error occurs.
bool error = foo1();
error &&= foo2(); // Only valid if C was extended with &&=
error &&= foo3();
Because the return type of operators || and && is not the same as type of their left argument.
The return type of || and && is always int1, while the left argument may be any integral, floating point or pointer type. The operands also don't have to be of the same type. Therefore defining x ||= y as x = x || y and x &&= y as x = x && y as would be consistent with other augmented assignments would not be able to store the result in the argument for most types.
You could come up with other definitions, e.g. x ||= y as if(!x) x = y and x &&= y as if(!y) x = y, but that would not be exactly obvious and it is not that useful, so it was not included.
1In C++ it is bool.
I guess the simple answer is that || is a boolean operator: and in C, a "boolean" is 0 or 1. The operands are implicitly converted to booleans (I have not checked that that's what the spec actually says, but it's how C behaves), and the result is a boolean.
Altering the semantics to support this pattern may well be feasible -- until someone relies on || doing what it's always done.
I cannot find any particular reason, why the operators don't exist (in C99).
So the only reason I can find is, that there was no boolean type in C89, and those boolean operators were intended to be solely used in if's.
Example:
int i = 5;
/* This should not make any difference,
since or'ing with false, shouldn't change
the value... dib di dib diddy...*/
i ||= 0; /* Actually: i = i || 0, which gives 'true' */
i is now '1'', which for most people is pretty counter intuitive.
This operator obviously doesn't bring any clearence or coding improvement without the boolean type, that would make sence being or'd with another one.
In my opinion, the implementation of a ||= b; as if(!a) a = b; would be pretty straightforward and has aleardy been implemented by e.g. Lua.
So you're question seems to be a bit, why C has been designed the way it has been designed.
If this question was about C++, you could for example ask Bjarne Stroustrup and ask him, what had went into him. Since this is not the case, this seems to me to be kind of a dead end, because the standard has been written quite some time ago and you cannot really ask people anymore, why the h***.
On the other hand, this incomplete operator set should (in my opinion) aleardy have been made whole using a similar notation than yours, since in my opinion, there is no reason against it.
I hope I could help a little.
One simple explanation is this.
bool resultsComputeAll = false;
bool resultsUntilFirst = false;
for (int i = 0; i < 10; ++i) {
resultsComputeAll = compute(i) || resultsComputeAll;
resultsUntilFirst = resultsUntilFirst || compute(i);
}
Which one would be result ||= compute(i)? It's ambiguous so it's better to not define.
We know that any numbers that are not equal to 0 are viewed as true in C, so we can write:
int a = 16;
while (a--)
printf("%d\n", a); // prints numbers from 15 to 0
However, I was wondering whether true / false are defined as 1/0 in C, so I tried the code below:
printf("True = %d, False = %d\n", (0 == 0), (0 != 0)); // prints: True = 1, False = 0
Does C standard explicitly indicate the truth values of true and false as 1 and 0 respectively?
Does the C standard explicitly indicate the truth values of true and false as 0 and 1 respectively?
The C standard defines true and false as macros in stdbool.h which expand to 1 and 0 respectively.
C11-§7.18:
The remaining three macros are suitable for use in #if preprocessing directives. They are
true
which expands to the integer constant 1,
false
which expands to the integer constant 0 [...]
For the operators == and != , standard says
C11-§6.5.9/3:
The == (equal to) and != (not equal to) operators are analogous to the relational operators except for their lower precedence.108) Each of the operators yields 1 if the specified relation is true and 0 if it is false. The result has type int. For any pair of operands, exactly one of the relations is true.
It is not explicitly indicated in C11. All language-level operations will return 1 as truthy (and accept any nonzero including NaN as true).
If you concern about _Bool, then true must be 1 because the standard only require it to hold 0 and 1. (§6.2.5/2).
Also in <stdbool.h> the macro true expands to 1 (§7.18/3)
==, !=, <, >, <= and >= return 0 or 1 (§6.5.8/6, §6.5.9/3).
!, && and || return 0 or 1 (§6.5.3.3/5, §6.5.13/3, §6.5.14/3)
defined expands to 0 or 1 (§6.10.1/1)
But all standard library functions e.g. islower just say "nonzero" for truthy (e.g. §7.4.1/1, §7.17.5.1/3, §7.30.2.1/1, §7.30.2.2.1/4).
§6.2.5/2: An object declared as type _Bool is large enough to store the values 0 and 1.
§6.5.5.3/5: The result of the logical negation operator ! is 0 if the value of its operand compares unequal to 0, 1 if the value of its operand compares equal to 0. …
§6.5.8/6: Each of the operators < (less than), > (greater than), <= (less than or equal to), and >= (greater than or equal to) shall yield 1 if the specified relation is true and 0 if it is false.107) …
§6.5.9/3: The == (equal to) and != (not equal to) operators are analogous to the relational operators except for their lower precedence.108) Each of the operators yields 1 if the specified relation is true and 0 if it is false. …
§6.5.13/3: The && operator shall yield 1 if both of its operands compare unequal to 0; …
§6.5.14/3: The || operator shall yield 1 if either of its operands compare unequal to 0; …
§6.10.1/1: … it may contain unary operator expressions of the form — defined identifier — or — defined ( identifier ) — which evaluate to 1 if …
§7.4.1 (Character classification functions)/1: The functions in this subclause return nonzero (true) if and only if …
§7.18/3: The remaining three macros are suitable for use in #if preprocessing directives. They are — true — which expands to the integer constant 1, …
§7.17.5.1/3: The atomic_is_lock_free generic function returns nonzero (true) if and only if the object’s operations are lock-free. …
§7.30.2.1 (Wide character classification functions)/1: The functions in this subclause return nonzero (true) if and only if …
§7.30.2.2.1/4: The iswctype function returns nonzero (true) if and only if …
There are two areas of the standard you need to be aware with when dealing with Boolean values (by which I mean true/false values rather than the specific C bool/_Bool type) in C.
The first has to do with the result of expressions and can be found in various portions of C11 6.5 Expressions (relational and equality operators, for example) . The bottom line is that, whenever a Boolean value is generated by an expression, it ...
... yields 1 if the specified relation is true and 0 if it is false. The result has type int.
So, yes, the result of any Boolean-generating expression will be one for true, or zero for false. This matches what you will find in stdbool.h where the standard macros true and false are defined the same way.
Keep in mind however that, following the robustness principle of "be conservative in what you send, liberal in what you accept", the interpretation of integers in the Boolean context is somewhat more relaxed.
Again, from various parts of 6.5, you'll see language like:
The || operator shall yield 1 if either of its operands compare unequal to 0; otherwise, it yields 0. The result has type int.
From that (and other parts), it's obvious that zero is considered false and any other value is true.
As an aside, the language specifying what value are used for Boolean generation and interpretation also appear back in C99 and C89 so they've been around for quite some time. Even K&R (ANSI-C second edition and the first edition) specified that, with text segments such as:
Relational expressions like i > j and logical expressions connected by && and || are defined to have value 1 if true, and 0 if false.
In the test part of if, while, for, etc, "true" just means "non-zero".
The && operator ... returns 1 if both its operands compare unequal to zero, 0 otherwise.
The || operator ... returns 1 if either its operands compare unequal to zero, and 0 otherwise.
The macros in stdbool.h appear back in C99 as well, but not in C89 or K&R since that header file did not exist at that point.
You are mixing up a lot of different things: control statements, operators and boolean types. Each have their own rules.
Control statements work like for example the if statement, C11 6.4.8.1:
In both forms, the first substatement is executed if the expression
compares unequal to 0.
while, for etc have the same rule. This has nothing to do with "true" or "false".
As for operators that are supposedly yielding a boolean result, they are actually yielding an int with value 1 or 0. For example the equality operators, C11 6.5.9:
Each of the operators yields 1 if the specified relation is true and 0
if it is false
All of the above is because C did not have a boolean type until the year 1999, and even when it did get one, the above rules weren't changed. So unlike most other programming languages where statements and operators yield a boolean type (like C++ and Java), they just yield an int, with a value zero or not zero. For example, sizeof(1==1) will give 4 in C but 1 in C++.
The actual boolean type in C is named _Bool and requires a modern compiler. The header stdbool.h defines macros bool, true and false, that expand to _Bool, 1 and 0 respectively (for compatibility with C++).
It is however considered good programming practice to treat control statements and operators as if they actually required/yielded a boolean type. Certain coding standards like MISRA-C recommend such practice. That is:
if(ptr == NULL) instead of if(ptr).
if((data & mask) != 0) instead of if(data & mask).
The aim of such style is to increase type safety with the aid of static analysis tools, which in turn reduces bugs. Arguably, this style is only meaningful if you do use static analysers. Though in some cases it leads to more readable, self-documenting code, for example
if(c == '\0')
Good, the intent is clear, the code is self-documenting.
versus
if(c)
Bad. Could mean anything, and we have to go look for the type of c to understand the code. Is it an integer, a pointer or a character?
I've programmed in many languages. I've seen true be 1 or -1 depending on the language. The logic behind true being 1 was that a bit was either a 0 or 1. The logic behind true being -1 was that the ! operator was a one's complement. It changed all the 1's to 0's and all the 0's to 1's in an int. So, for an int, !0 = -1 and !(-1) = 0. This has tripped me up enough that I don't compare something to be == true, but instead compare it to be != false. That way, my programming style works in every language. So my answer is to not worry about it, but program so that your code works correctly either way.
This answer needs to be looked at a bit more closely.
The actual definition in C++ is that anything not 0 is treated as true. Why is this relevant? Because C++ doesn't know what an integer is by how we think about it--we create that meaning, all it holds is the shell and rules for what that means. It knows what bits are though, that which make up an integer.
1 as an integer is loosely represented in bits, say an 8-bit signed int as 0000 0001. Many times what we see visually is a bit of a lie, -1 is a much more common way to represent it because of the signed nature of 'integer'. 1 really can't mean true proper, why? Because it's NOT operation is 1111 1110. That's a really major issue for a boolean. When we talk about a boolean, it's just 1 bit--it's really simple, 0 is false and 1 is true. All the logic operations hold as trivial. This is why '-1' should be designated as 'true' for integers (signed). 1111 1111 NOT'ed becomes 0000 0000---the logic holds and we're good. Unsigned ints is a little bit tricky and were a lot more commonly used in the past--where 1 means true because it's easy to imply the logic that 'anything not 0 is true'.
That's the explanation. I say the accepted answer here is wrong--there is no clear definition in the C/C++ definition. A boolean is a boolean, you can treat an integer as a boolean, but the fact the output is an integer says nothing about the operation actually being done is bitwise.
It happened because of the Relational Operators in your printf statement.
Operator == and operator !=
Since (0 == 0) holds true so, it gives a value 1
whereas, (0 != 0) doesn't hold true so, gives a value 0 .
I think I might have found the perfect solution to your problem.
Yes, 0 and any non-zero number are False and True respectively. Though there is no boolean data type in C.
But this is not the problem, the actual problem is how you are dealing with the modification of variable a in the your code :
int a = 16;
while (a--){
printf("%d\n", a);
}
When the compiler comes to the while (condition) statement, first the value of a is read by the compiler for the condition, then the arithmetic operation takes place, in this case,
a = a - 1 / a -= 1. So in the end there will be a case when a = 1 and the condition satisfies and after the arithmetic operation a-- which leads to a = 0, the print statement prints a as 0.
The above scenario depends on whether you use --a or a--. These two statements are read by the compiler in the order they are written.
For --a first the operation is performed on a then its value is read and vice-versa for the other.
So for case --a when a = 1 first the operation is done i.e a = a - 1 / a -= 1 and then a is evaluated for the condition, which then comes out to be Falsy as a = 0. Try the code below :
int a = 16;
while (--a){
printf("%d\n", a); // prints numbers from 15 to 1 as intended
}
OR deal with the modification of a inside the while loop block.
int a = 16;
while(a){
a = a - 1; // or a -= 1
printf("%d\n", a); // also prints numbers from 15 to 1 as intended
}
This question already has answers here:
!! c operator, is a two NOT?
(4 answers)
What is "!!" in C? [duplicate]
(7 answers)
Closed 8 years ago.
doing some exam prep and this is a past question.
Describe what the following pieces of C do and re write in a simple programming style with the same functionality.
(The bad indentation was the intention of the question).
With regards to section A i'm not sure what unsigned cast is doing to a. I have tested it a few times and can't seem to get a result that makes sense.
Similarly In B im not sure what how the while loop is working with the long cast and the !! being another problem
The Code :
//code section A
int f(int a,int b){
return(((unsigned) a)>b);}
//code section B
int h(int *x, int y){
int * z= x-- +y;
w=0;
while( (long) z-- ^(long) x) w += !!(*z);
return w;}
Any help would be appreciated, Thank you.
!! negates a boolean expression twice, essentially converting an expressions value to 0 or 1.
As in C all values other than zero mean true, and zero means false, !! can be used to convert it into 0 or 1, in the case you need to use it later in a function or expression which doesn't accept any value for true, only the number 1.
About the rest: unsigned interprets the internal representation of your int a from your function argument to unsigned int, so for example -1 becomes 4294967295 if your compiler uses two's complement and 4 byte ints.
About the casting to long : I strongly recommend against in similar situations unless you absolutely know what you are doing. In your example, it does some pointer arithmetic, in interpreting your pointer as numeric values, essentially working with the addresses of your variables as if they were just numbers. They have probably chosen long because on their system it had the exact same size as a pointer. This is not guaranteed to be so on all systems.
So, to give a very short answer to your question: The code does undefined behavior with those expressions, except for the !! which just give 0 if the expression was zero, and 1 otherwise.
The operator ! is the logical negation.
!true is false (0), and !false is true (1).
When a value is used as a boolean value, anyhting other than 0 (0 in a large sense) is true; 0 is false.
So, !!(*z) has either the value 0 or 1.
It will be 0 if *z was NULL to begin with
It will be 1 if *z was not NULL.
! is the boolean operator for not.
false is 0, and true is 1 in C.
So when you take any int, which is not 0- and you run ! on it, you'll get 0 (true becomes false), and zero will become one .
So the action of !! is changing each non-zero value to 1 , and leaving each 0 a 0.
I've seen definitions in C
#define TRUE (1==1)
#define FALSE (!TRUE)
Is this necessary? What's the benefit over simply defining TRUE as 1, and FALSE as 0?
This approach will use the actual boolean type (and resolve to true and false) if the compiler supports it. (specifically, C++)
However, it would be better to check whether C++ is in use (via the __cplusplus macro) and actually use true and false.
In a C compiler, this is equivalent to 0 and 1.
(note that removing the parentheses will break that due to order of operations)
The answer is portability. The numeric values of TRUE and FALSE aren't important. What is important is that a statement like if (1 < 2) evaluates to if (TRUE) and a statement like if (1 > 2) evaluates to if (FALSE).
Granted, in C, (1 < 2) evaluates to 1 and (1 > 2) evaluates to 0, so as others have said, there's no practical difference as far as the compiler is concerned. But by letting the compiler define TRUE and FALSE according to its own rules, you're making their meanings explicit to programmers, and you're guaranteeing consistency within your program and any other library (assuming the other library follows C standards ... you'd be amazed).
Some History
Some BASICs defined FALSE as 0 and TRUE as -1. Like many modern languages, they interpreted any non-zero value as TRUE, but they evaluated boolean expressions that were true as -1. Their NOT operation was implemented by adding 1 and flipping the sign, because it was efficient to do it that way. So 'NOT x' became -(x+1). A side effect of this is that a value like 5 evaluates to TRUE, but NOT 5 evaluates to -6, which is also TRUE! Finding this sort of bug is not fun.
Best Practices
Given the de facto rules that zero is interpreted as FALSE and any non-zero value is interpreted as TRUE, you should never compare boolean-looking expressions to TRUE or FALSE. Examples:
if (thisValue == FALSE) // Don't do this!
if (thatValue == TRUE) // Or this!
if (otherValue != TRUE) // Whatever you do, don't do this!
Why? Because many programmers use the shortcut of treating ints as bools. They aren't the same, but compilers generally allow it. So, for example, it's perfectly legal to write
if (strcmp(yourString, myString) == TRUE) // Wrong!!!
That looks legitimate, and the compiler will happily accept it, but it probably doesn't do what you'd want. That's because the return value of strcmp() is
0 if yourString == myString
<0 if yourString < myString
>0 if yourString > myString
So the line above returns TRUE only when yourString > myString.
The right way to do this is either
// Valid, but still treats int as bool.
if (strcmp(yourString, myString))
or
// Better: lingustically clear, compiler will optimize.
if (strcmp(yourString, myString) != 0)
Similarly:
if (someBoolValue == FALSE) // Redundant.
if (!someBoolValue) // Better.
return (x > 0) ? TRUE : FALSE; // You're fired.
return (x > 0); // Simpler, clearer, correct.
if (ptr == NULL) // Perfect: compares pointers.
if (!ptr) // Sleazy, but short and valid.
if (ptr == FALSE) // Whatisthisidonteven.
You'll often find some of these "bad examples" in production code, and many experienced programmers swear by them: they work, some are shorter than their (pedantically?) correct alternatives, and the idioms are almost universally recognized. But consider: the "right" versions are no less efficient, they're guaranteed to be portable, they'll pass even the strictest linters, and even new programmers will understand them.
Isn't that worth it?
The (1 == 1) trick is useful for defining TRUE in a way that is transparent to C, yet provides better typing in C++. The same code can be interpreted as C or C++ if you are writing in a dialect called "Clean C" (which compiles either as C or C++) or if you are writing API header files that can be used by C or C++ programmers.
In C translation units, 1 == 1 has exactly the same meaning as 1; and 1 == 0 has the same meaning as 0. However, in the C++ translation units, 1 == 1 has type bool. So the TRUE macro defined that way integrates better into C++.
An example of how it integrates better is that for instance if function foo has overloads for int and for bool, then foo(TRUE) will choose the bool overload. If TRUE is just defined as 1, then it won't work nicely in the C++. foo(TRUE) will want the int overload.
Of course, C99 introduced bool, true, and false and these can be used in header files that work with C99 and with C.
However:
this practice of defining TRUE and FALSE as (0==0) and (1==0) predates C99.
there are still good reasons to stay away from C99 and work with C90.
If you're working in a mixed C and C++ project, and don't want C99, define the lower-case true, false and bool instead.
#ifndef __cplusplus
typedef int bool;
#define true (0==0)
#define false (!true)
#endif
That being said, the 0==0 trick was (is?) used by some programmers even in code that was never intended to interoperate with C++ in any way. That doesn't buy anything and suggests that the programmer has a misunderstanding of how booleans work in C.
In case the C++ explanation wasn't clear, here is a test program:
#include <cstdio>
void foo(bool x)
{
std::puts("bool");
}
void foo(int x)
{
std::puts("int");
}
int main()
{
foo(1 == 1);
foo(1);
return 0;
}
The output:
bool
int
As to the question from the comments of how are overloaded C++ functions relevant to mixed C and C++ programming. These just illustrate a type difference. A valid reason for wanting a true constant to be bool when compiled as C++ is for clean diagnostics. At its highest warning levels, a C++ compiler might warn us about a conversion if we pass an integer as a bool parameter. One reason for writing in Clean C is not only that our code is more portable (since it is understood by C++ compilers, not only C compilers), but we can benefit from the diagnostic opinions of C++ compilers.
#define TRUE (1==1)
#define FALSE (!TRUE)
is equivalent to
#define TRUE 1
#define FALSE 0
in C.
The result of the relational operators is 0 or 1. 1==1 is guaranteed to be evaluated to 1 and !(1==1) is guaranteed to be evaluated to 0.
There is absolutely no reason to use the first form. Note that the first form is however not less efficient as on nearly all compilers a constant expression is evaluated at compile time rather than at run-time. This is allowed according to this rule:
(C99, 6.6p2) "A constant expression can be evaluated during translation rather than runtime, and accordingly may be used in any place that a constant may be."
PC-Lint will even issue a message (506, constant value boolean) if you don't use a literal for TRUE and FALSE macros:
For C, TRUE should be defined to be 1. However, other languages use quantities other than 1 so some programmers feel that !0 is playing it safe.
Also in C99, the stdbool.h definitions for boolean macros true and false directly use literals:
#define true 1
#define false 0
Aside from C++ (already mentioned), another benefit is for static analysis tools. The compiler will do away with any inefficiencies, but a static analyser can use its own abstract types to distinguish between comparison results and other integer types, so it knows implicitly that TRUE must be the result of a comparison and should not be assumed to be compatible with an integer.
Obviously C says that they are compatible, but you may choose to prohibit deliberate use of that feature to help highlight bugs -- for example, where somebody might have confuse & and &&, or they've bungled their operator precedence.
The pratical difference is none. 0 is evaluated to false and 1 is evaluated to true. The fact that you use a boolean expression (1 == 1) or 1, to define true, doesn't make any difference. They both gets evaluated to int.
Notice that the C standard library provides a specific header for defining booleans: stdbool.h.
We don't know the exact value that TRUE is equal to and the compilers can have their own definitions. So what you privode is to use the compiler's internal one for definition. This is not always necessary if you have good programming habits but can avoid problems for some bad coding style, for example:
if ( (a > b) == TRUE)
This could be a disaster if you mannually define TRUE as 1, while the internal value of TRUE is another one.
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Typically in the C Programming Language, 1 is defined as true and 0 is defined as false. Hence why you see the following quite often:
#define TRUE 1
#define FALSE 0
However, any number not equal to 0 would be evaluated to true as well in a conditional statement. Therefore by using the below:
#define TRUE (1==1)
#define FALSE (!TRUE)
You can just explicitly show that you trying to play it safe by making false equal to whatever isn't true.