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Logical Operators, Expressions & Condition Statements (boolean values)
Improve INSERT-per-second performance of SQLite?Using boolean values in CPython's equivalent of && (logical-and) in an if-statementWhy does C not have a logical assignment operator?Performance efficiency… conditional statements vs logical expressions in JavascriptRetrieve the result of an OR logic operation in JSLogical Boolean Negation Operator Precedence and AssociationAre || and ! operators sufficient to make every possible logical expression?Compare multiple boolean valuesLogical equality expression in Javascript
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So I am going over a few things in C programming, which I've gone over before.
However, I'm having a hard time recalling a couple of things that are becoming relevant again. I'm starting to write more complex programs, which are using more and more conditional statements, and I can't quite get them down right.
To recap, I know that in C logical operators determine a value of a condition in one of two ways, true or false. Which really equates to either 1 or 0.
Take the following expression as an example:
(if x is 4, y is 5, z is 3)
x > z && y > z
Plug in the values...
4 > 3 && 5 > 3
So in terms of Boolean logic...
I know that the value of this statement is actually:1 && 1 which is 1 (true)
or...
(with same variable values as declared above)
z >= x && x <= y
The value of this statement is actually:0 && 1 which is 0 (false because of the logical &&)
So here's where I need help...
I cant remember how to equate things in a few different forms of expressions like this one:
new values: x = 3, y = 0, z = -3)
x && y || z
what is the Boolean value of something like this? Without any operators like <, >, ==, and !=?
would it be...
x && y || z
1 0 1
which is 1 for true? because the x && y becomes false (because any false with && is a false) but then its followed with the || (OR) which if there is true it is true?
Is my question making sense?
or what about an expression that looks like this...
(if x = 5, y = 1)
!x + !!y
what would be the Boolean value here? Is it
!(5) + !(!)(1)
0 1 which would be 1 true?
Do I add the zero and one? Probably not.
I'm probably overthinking this.
c conditional logical-operators boolean-logic
edited Mar 8 at 0:41
Jonathan Leffler
574k956881041
asked Mar 8 at 0:09
Jedi_Sk8TricksJedi_Sk8Tricks
215
add a comment |
So I am going over a few things in C programming, which I've gone over before.
However, I'm having a hard time recalling a couple of things that are becoming relevant again. I'm starting to write more complex programs, which are using more and more conditional statements, and I can't quite get them down right.
To recap, I know that in C logical operators determine a value of a condition in one of two ways, true or false. Which really equates to either 1 or 0.
Take the following expression as an example:
(if x is 4, y is 5, z is 3)
x > z && y > z
Plug in the values...
4 > 3 && 5 > 3
So in terms of Boolean logic...
I know that the value of this statement is actually:1 && 1 which is 1 (true)
or...
(with same variable values as declared above)
z >= x && x <= y
The value of this statement is actually:0 && 1 which is 0 (false because of the logical &&)
So here's where I need help...
I cant remember how to equate things in a few different forms of expressions like this one:
new values: x = 3, y = 0, z = -3)
x && y || z
what is the Boolean value of something like this? Without any operators like <, >, ==, and !=?
would it be...
x && y || z
1 0 1
which is 1 for true? because the x && y becomes false (because any false with && is a false) but then its followed with the || (OR) which if there is true it is true?
Is my question making sense?
or what about an expression that looks like this...
(if x = 5, y = 1)
!x + !!y
what would be the Boolean value here? Is it
!(5) + !(!)(1)
0 1 which would be 1 true?
Do I add the zero and one? Probably not.
I'm probably overthinking this.
c conditional logical-operators boolean-logic
edited Mar 8 at 0:41
Jonathan Leffler
574k956881041
asked Mar 8 at 0:09
Jedi_Sk8TricksJedi_Sk8Tricks
215
1
What is the value ofx && y || z? I've been programming professionally in C for over thirty years, and I haven't a clue. Sure, I could go to the standard docs and look up operator precedence and find out, or I could just not write such monstrosities in the first place--put in parentheses to make it obvious and get on with my work.
– Lee Daniel Crocker
Mar 8 at 0:37
add a comment |
So I am going over a few things in C programming, which I've gone over before.
However, I'm having a hard time recalling a couple of things that are becoming relevant again. I'm starting to write more complex programs, which are using more and more conditional statements, and I can't quite get them down right.
To recap, I know that in C logical operators determine a value of a condition in one of two ways, true or false. Which really equates to either 1 or 0.
Take the following expression as an example:
(if x is 4, y is 5, z is 3)
x > z && y > z
Plug in the values...
4 > 3 && 5 > 3
So in terms of Boolean logic...
I know that the value of this statement is actually:1 && 1 which is 1 (true)
or...
(with same variable values as declared above)
z >= x && x <= y
The value of this statement is actually:0 && 1 which is 0 (false because of the logical &&)
So here's where I need help...
I cant remember how to equate things in a few different forms of expressions like this one:
new values: x = 3, y = 0, z = -3)
x && y || z
what is the Boolean value of something like this? Without any operators like <, >, ==, and !=?
would it be...
x && y || z
1 0 1
which is 1 for true? because the x && y becomes false (because any false with && is a false) but then its followed with the || (OR) which if there is true it is true?
Is my question making sense?
or what about an expression that looks like this...
(if x = 5, y = 1)
!x + !!y
what would be the Boolean value here? Is it
!(5) + !(!)(1)
0 1 which would be 1 true?
Do I add the zero and one? Probably not.
I'm probably overthinking this.
c conditional logical-operators boolean-logic
edited Mar 8 at 0:41
Jonathan Leffler
574k956881041
asked Mar 8 at 0:09
Jedi_Sk8TricksJedi_Sk8Tricks
215
So I am going over a few things in C programming, which I've gone over before.
However, I'm having a hard time recalling a couple of things that are becoming relevant again. I'm starting to write more complex programs, which are using more and more conditional statements, and I can't quite get them down right.
To recap, I know that in C logical operators determine a value of a condition in one of two ways, true or false. Which really equates to either 1 or 0.
Take the following expression as an example:
(if x is 4, y is 5, z is 3)
x > z && y > z
Plug in the values...
4 > 3 && 5 > 3
So in terms of Boolean logic...
I know that the value of this statement is actually:1 && 1 which is 1 (true)
or...
(with same variable values as declared above)
z >= x && x <= y
The value of this statement is actually:0 && 1 which is 0 (false because of the logical &&)
So here's where I need help...
I cant remember how to equate things in a few different forms of expressions like this one:
new values: x = 3, y = 0, z = -3)
x && y || z
what is the Boolean value of something like this? Without any operators like <, >, ==, and !=?
would it be...
x && y || z
1 0 1
which is 1 for true? because the x && y becomes false (because any false with && is a false) but then its followed with the || (OR) which if there is true it is true?
Is my question making sense?
or what about an expression that looks like this...
(if x = 5, y = 1)
!x + !!y
what would be the Boolean value here? Is it
!(5) + !(!)(1)
0 1 which would be 1 true?
Do I add the zero and one? Probably not.
I'm probably overthinking this.
c conditional logical-operators boolean-logic
c conditional logical-operators boolean-logic
edited Mar 8 at 0:41
Jonathan Leffler
574k956881041
asked Mar 8 at 0:09
Jedi_Sk8TricksJedi_Sk8Tricks
215
edited Mar 8 at 0:41
Jonathan Leffler
574k956881041
asked Mar 8 at 0:09
Jedi_Sk8TricksJedi_Sk8Tricks
215
edited Mar 8 at 0:41
Jonathan Leffler
574k956881041
edited Mar 8 at 0:41
Jonathan Leffler
574k956881041
edited Mar 8 at 0:41
Jonathan Leffler
574k956881041
574k956881041
asked Mar 8 at 0:09
Jedi_Sk8TricksJedi_Sk8Tricks
215
asked Mar 8 at 0:09
Jedi_Sk8TricksJedi_Sk8Tricks
215
asked Mar 8 at 0:09
Jedi_Sk8TricksJedi_Sk8Tricks
215
215
1
What is the value ofx && y || z? I've been programming professionally in C for over thirty years, and I haven't a clue. Sure, I could go to the standard docs and look up operator precedence and find out, or I could just not write such monstrosities in the first place--put in parentheses to make it obvious and get on with my work.
– Lee Daniel Crocker
Mar 8 at 0:37
add a comment |
1
What is the value ofx && y || z? I've been programming professionally in C for over thirty years, and I haven't a clue. Sure, I could go to the standard docs and look up operator precedence and find out, or I could just not write such monstrosities in the first place--put in parentheses to make it obvious and get on with my work.
– Lee Daniel Crocker
Mar 8 at 0:37
1
1
What is the value of
x && y || z? I've been programming professionally in C for over thirty years, and I haven't a clue. Sure, I could go to the standard docs and look up operator precedence and find out, or I could just not write such monstrosities in the first place--put in parentheses to make it obvious and get on with my work.– Lee Daniel Crocker
Mar 8 at 0:37
What is the value of
x && y || z? I've been programming professionally in C for over thirty years, and I haven't a clue. Sure, I could go to the standard docs and look up operator precedence and find out, or I could just not write such monstrosities in the first place--put in parentheses to make it obvious and get on with my work.– Lee Daniel Crocker
Mar 8 at 0:37
add a comment |
1 Answer
1
active
oldest
votes
We know that (cppreference conversion):
- zero value evaluates to false.
- nonzero value evaluates to true. Any nonzero value.
-3is true.1is true.INT_MAXis true.
We also know about operator precedence:
- && has higher precedence than ||
- which means, first && is evaluated, then ||
We also know that true and false in C are just macros for 1 and 0 defined in stdbool.h, so they have the int type. C does not has "real" boolean values, only boolean _Bool type. The logical operators && || and ! evaluate to int(!) type values 1 or 0 only, see cppreference.
So:
3 && 0 || -3
is equal to:
(true && false) || true
which evaluates to 1.
!5 + !!1
The ! has higher precedence.
The value of ! operator is 1 (true) in case of zero. The value of ! operator is 0 (false) in case of nonzero expression.
So it's:
(! true) + (! (! true) )
false + true
0 + 1
1
answered Mar 8 at 0:22
Kamil CukKamil Cuk
13.4k1529
Thank you @KamilCuk . This makes sense. So to be sure I'm understanding you correctly, here's another example: (c = 2, d = -3, a = 0) with the expression: 2 < !d + d Which would be: 1 < 0 + 1 and ultimately be false: 1 < 1 (false) am I on the right track here? Thanks so much!
– Jedi_Sk8Tricks
Mar 8 at 19:04
I think yes, you are correct.
– Kamil Cuk
Mar 9 at 6:34
add a comment |
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votes
We know that (cppreference conversion):
- zero value evaluates to false.
- nonzero value evaluates to true. Any nonzero value.
-3is true.1is true.INT_MAXis true.
We also know about operator precedence:
- && has higher precedence than ||
- which means, first && is evaluated, then ||
We also know that true and false in C are just macros for 1 and 0 defined in stdbool.h, so they have the int type. C does not has "real" boolean values, only boolean _Bool type. The logical operators && || and ! evaluate to int(!) type values 1 or 0 only, see cppreference.
So:
3 && 0 || -3
is equal to:
(true && false) || true
which evaluates to 1.
!5 + !!1
The ! has higher precedence.
The value of ! operator is 1 (true) in case of zero. The value of ! operator is 0 (false) in case of nonzero expression.
So it's:
(! true) + (! (! true) )
false + true
0 + 1
1
answered Mar 8 at 0:22
Kamil CukKamil Cuk
13.4k1529
Thank you @KamilCuk . This makes sense. So to be sure I'm understanding you correctly, here's another example: (c = 2, d = -3, a = 0) with the expression: 2 < !d + d Which would be: 1 < 0 + 1 and ultimately be false: 1 < 1 (false) am I on the right track here? Thanks so much!
– Jedi_Sk8Tricks
Mar 8 at 19:04
I think yes, you are correct.
– Kamil Cuk
Mar 9 at 6:34
add a comment |
We know that (cppreference conversion):
- zero value evaluates to false.
- nonzero value evaluates to true. Any nonzero value.
-3is true.1is true.INT_MAXis true.
We also know about operator precedence:
- && has higher precedence than ||
- which means, first && is evaluated, then ||
We also know that true and false in C are just macros for 1 and 0 defined in stdbool.h, so they have the int type. C does not has "real" boolean values, only boolean _Bool type. The logical operators && || and ! evaluate to int(!) type values 1 or 0 only, see cppreference.
So:
3 && 0 || -3
is equal to:
(true && false) || true
which evaluates to 1.
!5 + !!1
The ! has higher precedence.
The value of ! operator is 1 (true) in case of zero. The value of ! operator is 0 (false) in case of nonzero expression.
So it's:
(! true) + (! (! true) )
false + true
0 + 1
1
answered Mar 8 at 0:22
Kamil CukKamil Cuk
13.4k1529
Thank you @KamilCuk . This makes sense. So to be sure I'm understanding you correctly, here's another example: (c = 2, d = -3, a = 0) with the expression: 2 < !d + d Which would be: 1 < 0 + 1 and ultimately be false: 1 < 1 (false) am I on the right track here? Thanks so much!
– Jedi_Sk8Tricks
Mar 8 at 19:04
I think yes, you are correct.
– Kamil Cuk
Mar 9 at 6:34
add a comment |
We know that (cppreference conversion):
- zero value evaluates to false.
- nonzero value evaluates to true. Any nonzero value.
-3is true.1is true.INT_MAXis true.
We also know about operator precedence:
- && has higher precedence than ||
- which means, first && is evaluated, then ||
We also know that true and false in C are just macros for 1 and 0 defined in stdbool.h, so they have the int type. C does not has "real" boolean values, only boolean _Bool type. The logical operators && || and ! evaluate to int(!) type values 1 or 0 only, see cppreference.
So:
3 && 0 || -3
is equal to:
(true && false) || true
which evaluates to 1.
!5 + !!1
The ! has higher precedence.
The value of ! operator is 1 (true) in case of zero. The value of ! operator is 0 (false) in case of nonzero expression.
So it's:
(! true) + (! (! true) )
false + true
0 + 1
1
answered Mar 8 at 0:22
Kamil CukKamil Cuk
13.4k1529
We know that (cppreference conversion):
- zero value evaluates to false.
- nonzero value evaluates to true. Any nonzero value.
-3is true.1is true.INT_MAXis true.
We also know about operator precedence:
- && has higher precedence than ||
- which means, first && is evaluated, then ||
We also know that true and false in C are just macros for 1 and 0 defined in stdbool.h, so they have the int type. C does not has "real" boolean values, only boolean _Bool type. The logical operators && || and ! evaluate to int(!) type values 1 or 0 only, see cppreference.
So:
3 && 0 || -3
is equal to:
(true && false) || true
which evaluates to 1.
!5 + !!1
The ! has higher precedence.
The value of ! operator is 1 (true) in case of zero. The value of ! operator is 0 (false) in case of nonzero expression.
So it's:
(! true) + (! (! true) )
false + true
0 + 1
1
answered Mar 8 at 0:22
Kamil CukKamil Cuk
13.4k1529
edited Mar 8 at 0:29
answered Mar 8 at 0:22
Kamil CukKamil Cuk
13.4k1529
answered Mar 8 at 0:22
Kamil CukKamil Cuk
13.4k1529
answered Mar 8 at 0:22
Kamil CukKamil Cuk
13.4k1529
13.4k1529
Thank you @KamilCuk . This makes sense. So to be sure I'm understanding you correctly, here's another example: (c = 2, d = -3, a = 0) with the expression: 2 < !d + d Which would be: 1 < 0 + 1 and ultimately be false: 1 < 1 (false) am I on the right track here? Thanks so much!
– Jedi_Sk8Tricks
Mar 8 at 19:04
I think yes, you are correct.
– Kamil Cuk
Mar 9 at 6:34
add a comment |
Thank you @KamilCuk . This makes sense. So to be sure I'm understanding you correctly, here's another example: (c = 2, d = -3, a = 0) with the expression: 2 < !d + d Which would be: 1 < 0 + 1 and ultimately be false: 1 < 1 (false) am I on the right track here? Thanks so much!
– Jedi_Sk8Tricks
Mar 8 at 19:04
I think yes, you are correct.
– Kamil Cuk
Mar 9 at 6:34
Thank you @KamilCuk . This makes sense. So to be sure I'm understanding you correctly, here's another example: (c = 2, d = -3, a = 0) with the expression: 2 < !d + d Which would be: 1 < 0 + 1 and ultimately be false: 1 < 1 (false) am I on the right track here? Thanks so much!
– Jedi_Sk8Tricks
Mar 8 at 19:04
Thank you @KamilCuk . This makes sense. So to be sure I'm understanding you correctly, here's another example: (c = 2, d = -3, a = 0) with the expression: 2 < !d + d Which would be: 1 < 0 + 1 and ultimately be false: 1 < 1 (false) am I on the right track here? Thanks so much!
– Jedi_Sk8Tricks
Mar 8 at 19:04
I think yes, you are correct.
– Kamil Cuk
Mar 9 at 6:34
I think yes, you are correct.
– Kamil Cuk
Mar 9 at 6:34
add a comment |
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1
What is the value of
x && y || z? I've been programming professionally in C for over thirty years, and I haven't a clue. Sure, I could go to the standard docs and look up operator precedence and find out, or I could just not write such monstrosities in the first place--put in parentheses to make it obvious and get on with my work.– Lee Daniel Crocker
Mar 8 at 0:37