This question already has answers here:
Why can't we use bitwise operators on float & double data types
(2 answers)
Closed 7 years ago.
These are two simple samples in C++ written on Dev-cpp C++ 5.4.2:
float a, b, c;
if (a | b & a | c)
printf("x = %.2f\tF = %.0f\n", x, F);
else
printf("x = %.2f\tF = %.2f\n", x, F);
and this code :
float a, b, c;
if (a || b && a || c)
printf("x = %.2f\tF = %.0f\n", x, F);
else
printf("x = %.2f\tF = %.2f\n", x, F);
Can somebody tell my difference between || > | and & > &&. The second code works , but first does not.
And compiler gives an error message :
[Error] invalid operands of types 'float' and 'float' to binary 'operator&'.
The operators |, &, and ~ act on individual bits in parallel. They can be used only on integer types. a | b does an independent OR operation of each bit of a with the corresponding bit of b to generate that bit of the result.
The operators ||, &&, and ! act on each entire operand as a single true/false value. Any data type can be used that implicitly converts to bool. Many data types, including float implicitly convert to bool with an implied !=0 operation.
|| and && also "short circuit". That means whenever the value of the result can be determined by just the first operand, the second is not evaluated. Example:
ptr && (*ptr==7) If ptr is zero, the result is false without any risk of seg faulting by dereferencing zero.
You could contrast that with (int)ptr & (*ptr). Ignoring the fact that this would be a bizarre operation to even want, if (int)ptr were zero, the entire result would be zero, so a human might think you don't need the second operand in that case. But the program will likely compute both anyway.
You seems be confused with the symbols of the operators. Theses symbols are actually split in two different categories, which are bit-wise operators and logical operators. Although they use the same symbols, you should regard them as different operators. The truth tables for both categories are similar, but the meanings are different. Maybe that's why people use the similar symbols for the operators.
bit-wise operators
~ // NOT
& // AND
| // OR
^ // XOR
The bit-wise operators will regard all its operands as binary numerals and act according to the bit-wise truth tables on every bit of the operands.
Bit-wise Truth Table
x y x&y x|y x^y
0 0 0 0 0
1 0 0 1 1
0 1 0 1 1
1 1 1 1 0
x ~x
0 1
1 0
logical operators
! // Logical NOT (negation)
&& // Logical AND (conjunction)
|| // Logical OR (disjunction)
The logical operator will regard all its operands as bools and act according the operator truth tables. Any number that is not equal to 0 will be true, else will be false.
Logical Truth Table
x y x&&y x||y
F F F F
T F F T
F T F T
T T T T
x !x
F T
T F
For example:
int a = 10; // a = 0000 .... 0000 1010 <-- a 32 bits integer
// a is not zero -> true
int b = 7; // b = 0000 .... 0000 0111 <-- a 32 bits integer
// b is not zero -> true
Then for bit-wise operator:
assert(a & b == 2); // 2 = 0000 .... 0000 0010 <-- every bit will & separately
For logic operator:
assert(a && b == true); // true && true -> true
The bitwise operators, which are | (OR), & (AND), ^ (XOR), and ~ (complement) do what you expect them to do: they perform the aforementioned operations on bits.
And regarding your compilation issue, there are no bitwise operations for floating point numbers.
The logical operators, which are || (OR), && (AND), and ! (NOT) only know the values true and false.
An expression is true if its value is not 0. It is false if its value equals 0.
The logical operators do this operation first. Then they perform their corresponding operation:
||: true if at least one the operands is true
&&: true if both operands are true
!: true if the operand is false
Note that all logical operators are short-circuit operators.
Bitwise operation is not supported for floating points
Alternatively if you really need to check, you can cast before you use them (highly discouraged),
Check here how to convert a float into integrals, https://www.cs.tut.fi/~jkorpela/round.html
Related
I have a problem relative to some C++ code about the !! operator. It gives me an unexpected result and I don't understand why:
int x=-12;
x=!!x;
print("value=",x);
The output of this is 1. But i do not know how. Can anyone explain this ambiguous result?
!!x is grouped as !(!x).
!x is 0 if x is non-zero, and 1 if x is zero.
Applying ! to that reverses the result.
So, !!x can be viewed as a way of setting x to 1 if it's not zero, and remaining at 0 if it's zero. In other words x = !!x is the same as x = x ? 1 : 0.
... !(-12) kindly explain this expression.
It's "logical not of -12". In C++ numeric value 0 is false in logical way, and any non-zero value is true in logical way. This is how C and C++ evaluates numerical values in boolean context, like if (expression) ..., i.e. if (-12) exit(1); will exit your application, because -12 is "true".
When you typecast numeric value to bool type and then back to int, the true will become value 1, but most of the time you can avoid these conversions and use intermediate results of numerical calculations directly, where any non-zero value is "true".
So "not of true" is value false. I.e. !(-12) == false. And !false == true. And you convert the logical value "true" back to int, which will make x == 1.
The !!(numeric_value) is idiomatic way in C++ (and in C too) to "normalize" any numeric value into precisely 0 or 1 (and you may encounter it in source code of many applications).
The "logical" distinction is important, because C++ has also binary operators, where the calculation is working per individual bits, not with value as whole. In "binary" way the "!" operator sibling is ~. Similar example to yours with bit-complement operator x=~~x; would then result into original value -12, because every bit of x is flipped twice, ending with same value as it did start. So ~~x is not something to encounter in regular source, as it's basically "no operation" (contrary to !! which is "no operation" only if the original value was already 0 or 1).
Also this is sometimes source of bugs for people learning the language, as they forget about the distinction and write something like
if (4 & 2) { /* this is "and" so it should be true??? */ }
// it will be false, because in binary way 4&2 == 0
if (4 && 2) { /* this is logical "and" and will be "true" */ }
// because 4 is non-zero, 2 is non-zero, and "true and true" is "true"
I.e. binary operators are & | ~ ^ (and, or, not, xor), and their logical siblings are && || ! for "and, or, not", the logical "xor" doesn't have operator in C++.
This question already has answers here:
What is the "-->" operator in C++?
(29 answers)
Closed 8 years ago.
In the question What is the "-->" operator in C++? it asks what --> does and gives a link to a comp.lang.c++.moderated thread. scrolling down the thread a bit further found me this:
> There is no such operator in C++.
> It's just a combination of two operators: postfix decrement "--" and
> greater ">".
> That's why this example works.
> Try ( x --> 20 ) and you'll get no output in this case;)
Of course there is. It is described together with "runs to" operator:
#include <stdio.h>
int main()
{
int x = 10;
while( x -->> 0 ) // x runs to 0
printf("%d ", x);
}
What does the "runs to" operator actually do?
while( x -->> 0 ) // x runs to 0
This is actually a hybrid of the -- (post-decrement) and >> (bitshift right) operators, better formatted as:
while (x-- >> 0) ...
For this specific usage, with 0 on the right hand side, x is decremented with each loop iteration due to the postfix --, and the prior (pre-decrement) value is shifted right 0 bits by >> 0 which does nothing at all when x is non-negative, so the statement could be simplified to:
while (x--) ...
When x is 1 that's non-zero so found true for the purposes of the while test, then post-decrement reduces it to 0 and the loop executes for the last time (with x being 0 during that iteration); the next time while (x--) is checked with x already 0, the while loop terminates, with x left wrapping to the highest representable value for the unsigned type.
More generally, if you try to use >> on a negative value (e.g. x starts at 0 or a negative value great than INT_MIN, so x-- yields a negative value) the result is implementation defined, which means you have to consult your compiler documentation. You could use your compiler documentation to reason about how it would behave in the loop....
Relevant part of the Standard: 5.8/3:
The value of E1 >> E2 is E1 right-shifted E2 bit positions. If E1 has an unsigned type or if E1 has a signed type and a non-negative value, the value of the result is the integral part of the quotient of E1/2^E2. If E1 has a signed type and a negative value, the resulting value is implementation-defined.
BTW /- for Visual Studio, per http://msdn.microsoft.com/en-us/library/336xbhcz.aspx, the implementation defined behaviour is "No shift operation is performed if additive-expression is 0.". I can't find anything in the GCC manual about this (would have expected it here).
while( x -->> 0 ) // x runs to 0
No, the "goes to operator" is --> with only one > sign. It decreases x by one and then compares the result to zero.
The -- >> 0 "runs to operator" decreases x and then bitshifts the result rightward by zero. Bitshifting by zero does nothing for nonnegative x, otherwise it's implementation-defined (usually does nothing, but could be random). Zero bitshifted by zero is zero, which is interpreted as false, at which point the loop will terminate.
So it "works" but it's a terrible way of expressing a loop.
-- decrements but returns the value of the variable before it was decremented, >> shifts to the right by the right operand, which is 0 (a.k.a. a no-op), then it implicitly compares the result against 0.
I have written this C++ program, and I am not able to understand why it is printing 1 in the third cout statement.
#include<iostream>
using namespace std;
int main()
{
bool b = false;
cout << b << "\n"; // Print 0
b = ~b;
cout << b << "\n"; // Print 1
b = ~b;
cout << b << "\n"; // Print 1 **Why?**
return 0;
}
Output:
0
1
1
Why is it not printing the following?
0
1
0
This is due to C legacy operator mechanization (also recalling that ~ is bitwise complement). Integral operands to ~ are promoted to int before doing the operation, then converted back to bool. So effectively what you're getting is (using unsigned 32 bit representation) false -> 0 -> 0xFFFFFFFF -> true. Then true -> 1 -> 0xFFFFFFFE -> 1 -> true.
You're looking for the ! operator to invert a boolean value.
You probably want to do this:
b = !b;
which is logical negation. What you did is bitwise negation of a bool cast to an integer. The second time the statement b = ~b; is executed, the prior value of b is true. Cast to an integer this gives 1 whose bitwise complement is -2 and hence cast back to bool true. Therefore, true values of b will remain true while false values will be assigned true. This is due to the C legacy.
As pretty much everyone else has said, the bool is getting promoted to an integer before the complement operator is getting its work done. ~ is a bitwise operator and thus inverts each individual bit of the integer; if you apply ~ to 00000001, the result is 11111110. When you apply this to a 32-bit signed integer, ~1 gives you -2. If you're confused why, just take a look at a binary converter. For example: http://www.binaryconvert.com/result_signed_int.html?decimal=045050
To your revised question:
False to true works for the same reason as above. If you flip 00000000 (out to 32 bits), you get 11111111... which I believe is -1 in integer. When comparing boolean values, anything that is -not- 0 is considered to be true, while 0 alone is false.
You should use logical operators, not binary operators. Use ! instead of ~.
I've been struggling for a while with a part of my code and I finally found that the problem lies with a simple test that don't give me the result I expect.
if (2) //=> true
if (2 & true) //=> false
if (bool(2) & true) //=> true
What I don't understand is why the second line results in false.
My understanding was that every non-zero integer was considered as true in a test.
Because the bitwise and between 2 and true is false.
& (bitwise operator) is different than && (logical operator).
true cast to int is 1.
So 2 & true is 2 & 1 which is false - because 0000000000000010 & 0000000000000001 == 0. (bits may vary)
Whereas bool(2) == 1, and 1 & 1 is true.
if (2) //=> true
So far, so good.
if (2 & true) //=> false
The condition here evaluates to 2 & 1 == 0, because & is a bitwise operator and 2 and 1 are respectively 00000010 and 00000001 in binary.
if (bool(2) & true) //=> true
Interestingly enough, on my compiler I seem to recall erratic behavior in some cases like this; and, if sect. 4.12 of the C++11 standard addresses the matter, it does so in a manner I do not understand. I seem to recall seeing my compiler let bool(2) == 2, which one would not expect. Whether this represents a bug in my compiler or a fault in my recollection, I do not know.
I suspect however that you want the logical operator && rather than the bitwise operator &.
QUIZ
To check your understanding, try
if (3 & true) //=> true
Do you understand why? (Hint: the binary representation of 3 is 00000011.)
You need && instead of &.
&& is the boolean and operator, whereas & is the binary 'and' so 2 & true is the same as 0010 & 0001 = 0000 -> false whereas 2 && true = true.
& does an AND between all the bits (call bitwise AND) , what you need is the && operator (boolean AND).
2 in binary is '10' and true is 1 (01) in binary, the result 10 & 01 is therefore 0 .
bool(2) convert 2 to true , is 01 in binary, and 01 & 01 is 01.
I have an assignment that I'm supposed to implement the MIPS processor in C++ and one of the MIPS instructions is "AND" and "OR" the MIPS instruction is represented as and $s1,$s2,$s3 which means that $s1=$s2(and)$s3 the $s2 and $s3 registers are represented into bits ,,, how can I perform the "AND" and "OR" operations using C++?
There are both binary and logical and and or operators in C++.
int a, b = 1;
int x = a | b; // binary OR
int x = a & b; // binary AND
bool x = a || b; // boolean OR
bool x = a && b; // boolean AND
A boolean comparison will return true or false depending on the value of the two operands. Logical "and" will return true if both operands are non-zero. Logical "or" will return false only if both operands are false.
Bitwise operators are different and operate on the bits of the operands. A bit wise "and" will set a bit to true only if both corresponding bits are true:
101 & 110 = 100
Bitwise "or" sets a bit to zero only if both corresponding bits are zero:
010 | 001 = 011
The two bitwise comparison operators are more closely related to the shift operators (<< and >>) and the one's complement operator (~) in that they are low level operations.