I am trying to form an if statement like so:
int myVariable = -1;
if (0 <= myVariable <= 99)
{
// Do something
}
However, if I assigned -1 to myVariable, which is an int, the if-statement still evaluations to true.
What am I doing wrong ?
That is not the correct way to say what you want in C++ syntax. You want:
int myVariable = -1;
if (0 <= myVariable && myVariable <= 99)
{
// Do something
}
What you wrote does the following:
1) Evaluate 0 <= myVariable, which in your example will be false (converted to 0 in C++ in this context)
2) Then this result (0) is compared against <= 99, so C++ reads "0 <= 99", which is true, so the if statement is true.
Related
Any idea why the else if statment will be never executed ? The value of difference is constantly changing when the program runs.
double difference = abs(reale_x[0] - reale_x[1]);
if (0 <= difference < 45) {
timer_counter += 1;
if (timer_counter == 30) {
cout << "CLICK" << '\n';
}
}
else if (difference > 50) {
timer_counter = 0;
}
That is not how comparation works in c++.
What this code
if (0 <= difference < 45) {
does is it first compares if 0 is smaller or equal to difference. It is then "replaced" by a bool value either true or false. And then a bool value (so either 1 or 0) is compared to 45. And it will always be smaller than 45. What you have there is an always true statement.
So the way you would write this if statement is
if (difference >= 0 && difference < 45){
Note that because of your else if statement it will not execute if the difference is >44 and <51
if (0 <= difference < 45) will be executed as if ((0 <= difference) < 45), which will be either 0<45 or 1<45 and will always be true. That's why the else part is not getting executed.
in mathematics, we see and write 0 <= x < 45 or something like that to define the range of the variable x. But in order to tell the computer the same thing, you have to tell more clearly. Saying, to have to tell the compiler, that the value of x is greater than or equal to zero and at the same time, that value will be less than 45, and you can tell the compiler by this statement: difference >= && difference < 45 . the && is an 'AND' operator in most of the languages.
I have a loop going through an array trying to find which index is a string. It should solve for what that value should be.
I can't figure out why, but as soon as the if statements start i becomes 1 which gives my code an error.
I'm not very fluent in C++.
for(int i = 0; i < 4; i++) {
if(auto value = std::get_if<std::string>(&varArr[i])) {
solvedIndex = i;
auto value0 = std::get_if<float>(&varArr[0]);
auto value1 = std::get_if<float>(&varArr[1]);
auto value2 = std::get_if<float>(&varArr[2]);
auto value3 = std::get_if<float>(&varArr[3]);
//i changes to 1 when this if starts??
if(i = 0) {
solvedVar = (*value3 / *value1) * *value2;
} else if (i = 1) {
solvedVar = *value3 / (*value0 / *value2);
} else if (i = 2) {
solvedVar = *value0 / (*value3 / *value1);
} else {
solvedVar = *value1 * (*value0 / *value2);
}
break;
}
}
Note that these variables are declared above. Also, varArr is filled with values:
std::variant<std::string, float> varArr[4];
int solvedIndex;
float solvedVar;
As has been noted, in your if statements, you are using the assignment operator (=) but want the equality comparison operator (==). For your variable i the first if statement sets i equal to 0 and if(0) is the same as if(false). So your program goes to the first else-if which sets i equal to 1 and if(1) evaluates to true. Your code then finishes the block within else if (i = 1) {...} and then ends.
That's because operator= is the assignment operator in C++ (and most languages, actually). That changes the value of the variable to the value on the other side. So, for instance:
x = 0
will change the value of x to 0. Doesn't matter if it's in an if statement. It will always change the value to 0 (or whatever the right hand side value is).
What you are looking for is operator==, which is the comparison (aka relational) operator in C++/ That asks the question "Are these two things equal?" So, for instance:
x == 0
asks is x is equal to 0.
This question already has answers here:
Is (4 > y > 1) a valid statement in C++? How do you evaluate it if so?
(5 answers)
Closed 4 years ago.
We usually use logical operators if need to combine boolean expressions. I was wondering about the expressions if don't use logical operators.
int x=101;
if(90<=x<=100)
cout<<'A';
This code still prints 'A' on console. Can you please help me to understand that how and in which order this boolean expression would be evaluated.
Since the operators have equal precedence, the expression is evaluated left-to-right :
if( (90 <= x) <= 100 )
if( (90 <= 101) <= 100 ) // substitute x's value
if( true <= 100 ) // evaluate the first <= operator
if( 1 <= 100 ) // implicit integer promotion for true to int
if( true ) // evaluate the second <= operator
To achieve the comparison you want, you would use the condition :
if( 90 <= x && x <= 100)
This is a common source of errors, because it looks right, and sytactically it is correct.
int x=101;
if(90<=x<=100)
This is equivalent to
if ( (90 <= x) <= 100)
which is
if ( true <= 100 )
and as true can convert to 1 this is
if ( true )
This expression is roughly equals to
int x=101;
bool b1 = 90 <= x; // true
bool b2 = int(b1) <= 100; // 1 <= 100 // true
if(b2)
cout<<'A';
So here is true result.
So I'm just curious if there is a short hand statement to this:
if(number < 0 )
bigInt.sign = 0;
else
bigInt.sign = 1;
I see all these short hand statements for if a < b and such.
I'm not sure on how to do it properly and would like some input on this.
Thanks!
I actually just figured it out right before you guys had answered.
The shortest solution is bigInt.sign = (number < 0) ? 0 : 1
The basic syntax for using ternary operator is like this:
(condition) ? (if_true) : (if_false)
For you case it is like this:
number < 0 ? bigInt.sign = 0 : bigInt.sign = 1;
try this:
bigInt.sign = number < 0 ? 0 : 1
Yes:
bigInt.sign = !(number < 0);
The ! operator always evaluates to true or false. When converted to int, these become 1 and 0 respectively.
Of course this is equivalent to:
bigInt.sign = (number >= 0);
Here the parentheses are redundant but I add them for clarity. All of the comparison and relational operator evaluate to true or false.
Depending on how often you use this in your code you could consider the following:
macro
#define SIGN(x) ( (x) >= 0 )
Inline function
inline int sign(int x)
{
return x >= 0;
}
Then you would just go:
bigInt.sign = sign(number);
you can also try this :
bigInt.sign = (number<0)*0 + (number>=0)*1;
If we need to assign another value other than 0 and 1 then this code can be used like :
bigInt.sign = (number<0)*(replacement_of_0) + (number>=0)*(replacement_of_1);
I have the following problem
in my app i have severeal if-statements
if ( (number >= 1 && number <= 18) && !strcmp("half1-18", _myBetCh) ) {
}
Now I realized that I have to split this condition because I need a boolean variable after one condition
bool success = false,
if(!strcmp("half1-18", _myBetCh) {
success = true;
if (number >= 1 && number <= 18) {
}
}
Is there a workaround to this? Is it possible, for instance, to make an assignment withing the if-statement?
It's possible, like this:
if ((success = !strcmp("half1-18", _myBatCh)) && number > 1 && number < 18)
but I personally think assignments in conditions are messy and hard to read, and prefer this variation:
bool success = strcmp("half1-18", _myBetCh) == 0;
if (success && number >= 1 && number <= 18) {
// ...
}
Well, there is:
if ( !strcmp("half1-18", _myBatCh) && (success = true, number > 1 && number < 18) )
or, obviating the need for the success = false earlier
if ( (success = !strcmp("half1-18", _myBatCh)) && number > 1 && number < 18 )
Your way is easier to read though, so I would consider sticking with what you've got.