I have the below code to choose sin or cos to be integrated,
while( x !=1 || y !=(1||0) ){
printf("Sin (1) or Cos (0)?\n");
x = scanf("%d",&y);
_flushall();
if(y==1){
printf("Sin set\n");
}
else if(y==0){
printf("Cos set\n");
}
}
However the
y!= (1||0)
never evaluates to true for y == 0 , can someone explain what's wrong here? Thanks.
You need (y != 1 && y != 0) (or similar, it depends on what you really mean to express there). The || operator is being applied to the operands 1 and 0. Put another way, y != (1 || 0) means "Do (1 || 0) then do y != result".
You are attempting to effectively code directly Boolean algebra, and C doesn't accept it in the manner you've provided.
while( x !=1 || y !=(1||0) )
should be
while( (x!=1) || ( (y!=1) || (y!=0) ) )
Never underestimate the value of using excess parentheses in C. The optimizer will likely optimize the code to be more efficient anyways.
The part of code that generates this error evaluates as follows:
LHS (left hand side), RHS (right hand side)
LHS = y
!= (1||0) [definition given]
!= (1) [b/c (1||0) = (1)]
y != (0||1)
is equivalent to
y != 1
since 0||1 is 1. You'll need two comparisons if you want y != 0 or y != 1.
Related
int num;
scanf("%d", &num);
if (num % 4 == 0 && num%100 != 0 || num % 400 == 0)
printf("%d", 1);
else
printf("%d", 0);
In this logic, I found I do not need to do () in AND condition which is in front of OR condition.
if (*(num % 4 == 0 && num%100 != 0)* || num % 400 == 0)
It's only needed if (num % 4 == 0 && num%100 != 0 || num % 400 == 0) without () in front of OR condition.
so, it seems (A && B || C) works like ((A && B) || C)
but it seemed it could work as (A && (B || C)) condition.
Why is () not needed in this situation? A and B condition are automatically grouped from the beginning?
All operators in C (and in fact all languages) have what's called operator precedence which dictates which operands are grouped first.
The logical AND operator && has a higher precedence than the logical OR operator ||, which means that this:
A && B || C
Is the same as this:
(A && B) || C
So if you want B || C to be grouped together, you need to explicitly add parenthesis, i.e.:
A && (B || C)
Parentheses decide order of operations, if you move the parentheses around you may change what the output is. In the same way that (A + B) / C is different from A + (B / C) but still a valid equation.
See Order of Operations in C
Logical AND has higher priority than OR:
https://en.cppreference.com/w/c/language/operator_precedence
dg
I was solving some problems related to "or" operators in C.
The body of the program was like what mentioned below:
#include<stdio.h>
main() {
int i = 4, j = -1, k = 0, w, x, y, z;
w = i || j || k ;
x = i && j && k ;
z = i && j || k ;
printf("\n w=%d , x = %d", w, x);
printf("\n y=%d , z = %d", y, z);
}
Can someone please tell me the mechanism of these statements?
These are binary operators, they take two arguments. When confronted with a more complicated expression you can break it down into sets of binary operations.
For example, z = i && j || k; could be written as z = ( (i && j) || k ). Or in pictorial form:
z
=
||
/ \
&& k
/ \
i j
To know that the tree has this layout we look up operator precedence (or language grammar rules), the rule is that to choose between && and ||, the || is the 'outer' operation (i.e. lower precedence).
These operators also do short-circuiting, although that is not relevant in this particular example.
So, looking at the above table, i && j gives 1 because both i and j are non-zero. The 1 || k gives 1 because at least one of the operands is non-zero. Finaly 1 is assigned to z.
You can find the right values for w and x in a similar way, using the precedence rule that the left-most && is the inner one, when the situation is a && b && c, and similarly for ||.
the || and && operators are logical operators, which means they evaluate the numbers as either 'true' or 'false'
In that sense, i and j are 'true' and k is 'false'. So, calculating the outcome:
w = 'true' or 'true' or 'false' = true = 1
x = 'true' and 'true' and 'false' = false = 0
z = 'true' and 'true' or 'false' = true = 1
y isn't defined and will contain 0 or garbage
--edit--
note that && and || are left to right operators, therefore the first operator is always evaluated first.
Why does this following program print "Yes" instead of "No"?
None of the variables is initialized to 2.
bool hello = 0;
int a = 1;
int b = 3;
int c = 4;
int d = 5;
if (a || b || c || d == 2) {
hello = 1;
}
if (hello == 1) {
printf("Yes");
}
if (hello == 0) {
printf("No");
}
return 0;
}
The statement
if (a || b || c || d == 2)
is equivalent to:
if (a != 0 || b != 0 || c != 0 || d == 2)
The equality comparison does not automatically distribute across all the variables. If you want to do that, you need to perform all the comparisons explicitly:
if (a == 2 || b == 2 || c ==2 || d == 2)
The expression (a || b || c || d == 2) evalutates to true because it treats a, b, c as booleans, and any non-zero integer is true.
You have given logical operator in the expression It means that if non zero value came then the expression is true. Then hello=1 is set and in next f statement it prints YES
You just meet the short circuit behavior of logical expressions OR.
The order of evaluation of logical OR || is left to right.
So in the following expression:
left || right
if left = true then right will never going to be executed (short circuit). In your code exactly same happened.
As you know, any non zero value treated as true in C, hence, a which is 1 is true. So, take a look:
if (a || b || c || d == 2)
if (true || bla bla bla) //rights are not even checked!
if (true)
hello = 1;
Tada! So the program print "Yes"!
None of the variables is initialized to 2.
Yes of course! But your if condition is not going to check that. To do so, try this:
if (a == 2 || b == 2 || c ==2 || d == 2) {
//...
because if judge num is not zero , if think this is true. so your code
if (a || b || c || d == 2)
like
if ( true || true || true || false)
the result is true, programe print "YES"
I should write a C while statement that continues while x is both larger than 10 and a multiple of either 6 or 7. I only need to include the actual while statement, not the entire loop!
I know it may look funny to you, but this is what I have so far.
#include<stdio.h>
int main(){
int x;
while (x > 10, x % 7 = 0 || x % 6 = 0) {
printf("%d\n", x);
}
}
Logical and is &&. As an operator, , is "ignore the result of the thing to the left".
Your Boolean logic and comparison both use the wrong syntax. It should be like this:
while (x > 10 && (x % 7 == 0 || x % 6 == 0)) {
The comma (,) is not a valid Boolean operator. What you wanted is && (logical AND). You need parentheses around the other two expressions because AND (&&) has higher precedence than OR (||).
And = is for setting values; == is for comparing them.
Finally, as others pointed out, you don't set x anywhere in the code that you have posted. So, your loop will not run. And you don't modify x in the loop, so, if you ever get into the loop, you will never get out.
The program can look like
#include <stdio.h>
int main( void )
{
int x;
printf( "Enter an integer number: " );
scanf( "%d", &x );
while ( x > 10 && ( x % 7 == 0 || x % 6 == 0 ) )
{
printf( "%d\n", x );
printf( "Enter next integer number: " );
scanf( "%d", &x );
}
}
Or you could enlarge the condition in the while loop the following way (to check that the input is correct)
#include <stdio.h>
int main( void )
{
int x;
printf( "Enter an integer number: " );
while ( scanf( "%d", &x ) == 1 && x > 10 && ( x % 7 == 0 || x % 6 == 0 ) )
{
printf( "%d\n", x );
printf( "Enter next integer number: " );
}
}
As for your condition in the while loop then it has two errors
while (x > 10, x % 7 = 0 || x % 6 = 0)
It uses the comma operator instead of logical operator && and it uses the assignment operator as for example x % 7 = 0 instead of the comparison operator x % 7 == 0. Also variable x was not initialized.
Your while statement should read
while (x > 10 && (x % 7 == 0 || x % 6 == 0))
Note the use of && for "boolean and", and == for "equality comparison".
I should Write a C while statement that continues while x is both
larger than 10 and a multiple of either 6 or 7. I Only need to include
the actual while statement, not the entire loop!
Let's analyze the condition that will go in the while() statement:
x is both larger than 10 and a multiple of either 6 or 7
Can be expressed as
x is greater than 10 AND a multiple of 6 OR multiple of 7
So let's code that in C:
x is greater than 10
x>10
AND
&&
x multiple of 6 OR x multiple of 7
(x%6)==0 || (x%7)==0
Then, all together
x>10 && ( (x%6)==0 || (x%7)==0 )
So your while sentence is like this:
while (x>10 && ( (x%6)==0 || (x%7)==0 ))
{
stuff...
}
Here can be the Solution Ava
while (x > 10 && (x % 7 == 0 || x % 6 == 0)) {
Can someone explain why the output of this program is false??
x && y gives 1. Still the output is false.
#include <stdio.h>
int main()
{
int x = 1, y = 2;
if(x && y == 1)
{
printf("true.");
}
else
{
printf("false.");
}
return 0;
}
Because == has a higher precedence than && So first this get's evaluated:
x && (y == 1)
y == 1 // 2 == 1
//Result: false
Which is false and then second:
x && false //1 && false
//Result: false
So the if statement will be false
For more information about operator precedence see here: http://en.cppreference.com/w/cpp/language/operator_precedence
if(x && y == 1)
Is the same as
if( ( x != 0 ) && ( y == 1 ) )
Here,x != 0 is true, but y == 1 is false. And since at least one of the operands of && is false, the condition evaluates to false and the else part executes.
It clearly stated X = 1 & Y = 2;
Now with your expression
X && Y == 1
The expression is evaluated as
Y == 1 (Precedence Rule, Also output is False)
X != 0 (Its True)
Now && is Logical And Operator, so it evaluates to True only if both the parts in expression evaluates to True!!!
It's okay to false, then 2 and 2 and it is different from one.
What you're asking is whether both x and y both are worth 1. If this happens say true but false