I have a string vector of user-input data containing strings. Now I need to make sure program won't execute if strings are different than specified few. Vector contains 4 fields and every has different condition:
vector[0] can only be "1" or "0"
vector[1] can only be "red" or "green
vector[2] can only be "1", "2" or "3"
vector[3] can only be "1" or "0"
I tried writing if for every condition:
if(tokens[0]!="1" || tokens[0]!="0"){
decy = "error";
}
else if(tokens[1]!="red" || tokens[1]!="green"){
decy = "error";
}
else if(tokens[2]!="1" || tokens[2]!="2" || tokens[2]!="3"){
decy = "error";
}
else if(tokens[3]!="1" || tokens[3]!="0"){
decy = "error";
}
else{
switch(){} //working code
}
return decy;
It always enters first if and returns error. I tried with if instead of else if but it doesn't work either. I checked vector[i] contents and it returns correct strings. No " " at the end of it etc. Removing else and releasing switch just makes program check first condition and ignore rest of it.
I'm probably doing something terribly wrong, but I can't find an answer on internet so I decided to ask here.
This line:
if(tokens[0]!="1" || tokens[0]!="0")
should be:
if(tokens[0]!="1" && tokens[0]!="0")
^^
The same goes for the rest of the if statements as well.
The conditions are invalid.
Any distinct value can satisfy your conditions.
You should use && instead of ||.
For example:
if (tokens[0] != "1" || tokens[0] != "0") {
Consider this line. If tokens[0] is "1", which is valid input, it will not satisfy the first condition, but it will satisfy the second. You only want to throw an error when the value is neither of the valid possible inputs.
This means that your condition should be:
if (tokens[0] != "1" && tokens[0] != "0") {
Same goes for all the others.
You should turn those || into &&. If the input can only be X or Y, this means that it is illegal when it is not X and not Y:
if (tokens[0] != "1" && tokens [0] !="0")
// ^^
The first if:
if(tokens[0]!="1" || tokens[0]!="0")
ALWAYS evaluates to true.
Related
I have a variable v in my program, and it may take any value from the set of values
"a", "b", "c", ..., "z"
And my goal is to execute some statement only when v is not "x", "y", or "z".
I have tried,
for C-like languages (where equality operators compare the actual string values; e.g. c#, javascript, php)
if (v != "x" || v != "y" || v != "z")
{
// the statements I want to be executed
// if v is neither "x", nor "y", nor "z"
}
for Pascal-like languages (e.g. plsql)
IF (v != 'x' OR v != 'y' OR v != 'z') THEN
-- the statements I want to be executed
-- if v is neither "x", nor "y", nor "z"
END IF;
The statements inside the if condition always get executed. Am I doing anything wrong?
Use &&/AND/and, not ||/OR/or:
v != "x" && v != "y" && v != "z"
Problem
If an if block is always executed, the condition for the if block always evaluates to true. The logical expression must be wrong.
Let us consider v != "x" || v != "y" || v != "z" for each value of v.
When v = "x",
v != "x" becomes "x" != "x", which is false.
v != "y" becomes "x" != "y", which is true.
v != "z" becomes "x" != "z", which is true.
The expression evaluates to false || true || true, which is true.
When v = "y", the expression becomes
"y" != "x" || "y" != "y" || "y" != "z"
or true || false || true, which is true.
When v = "z", the expression becomes
"z" != "x" || "z" != "y" || "z" != "z"
or true || true || false, which is true.
For any other value for v, the expression evaluates to true || true || true, which is true.
Alternatively, consider the truth-table:
│ A B C │
v │ v != "x" v != "y" v != "z" │ A || B || C
───────┼──────────────────────────────────┼──────────────
"x" │ false true true │ true
"y" │ true false true │ true
"z" │ true true false │ true
other │ true true true │ true
As you can see, your logical expression always evaluates to true.
Solution
What you want to do is, find a logical expression that evaluates to true when
(v is not "x")and(v is not "y")and(v is not "z").
The correct construction is,
for C-like languages (eg. c#, javascript-(may need the strict equality operator !==), php)
if (v != "x" && v != "y" && v != "z")
{
// the statements I want to be executed
// if v is neither "x", nor "y", nor "z"
}
for Pascal-like languages plsql
IF (v != 'x' AND v != 'y' AND v != 'z') THEN
-- the statements I want to be executed
-- if v is neither "x", nor "y", nor "z"
END IF;
De Morgan's law
By De Morgan's law, the expression can also be rewritten as (using C-like syntax)
!(v == "x" || v == "y" || v == "z")
meaning
not((v is "x")or(v is "y")or(v is "z")).
This makes the logic a bit more obvious.
Specific languages
Some languages have specific constructs for testing membership in sets, or you can use array/list operations.
sql: v NOT IN ('x', 'y', 'z')
javascript: ["x", "y", "z"].indexOf(v) == -1
python: v not in {"x", "y", "z"}
java: !Arrays.asList("x", "y", "z").contains(v)
java-9 (and above): !Set.of("x", "y", "z").contains(v)
I figured I'd contribute an answer for Bourne shell script, since the syntax is somewhat peculiar.
In traditional/POSIX sh the string equality test is a feature of the [ command (yes, that is a distinct command name!) which has some pesky requirements on quoting etc.
#### WRONG
if [ "$v" != 'x' ] || [ "$v" != 'y'] || [ "$v" != 'z' ]; then
: some code which should happen when $v is not 'x' or 'y' or 'z'
fi
Modern shells like Ksh, Bash, Zsh etc also have [[ which is somewhat less pesky.
#### STILL WRONG
if [[ $v != 'x' || $v != 'y' || $v != 'z' ]]; then
: some code which should happen when $v is not 'x' or 'y' or 'z'
fi
We should highlight the requirement to have spaces around each token, which is something many beginners overlook (i.e. you can't say if[[$v or $v!='y' without whitespace around the commands and operators), and the apparent optionality of quoting. Failing to quote a value is often not a syntax error, but it will lead to grave undesired semantical troubles if you fail to quote a value which needs to be quoted. (More on this elsewhere.)
The obvious fix here is to use && instead of || but you should also note that [[ typically sports support for regular expressions, so you can say something like
if [[ ! $v =~ ^(x|y|z)$ ]]; then
: yeah
fi
and don't forget the trusty old case statement which is quite natural for this, and portable back into the late 1970s:
case $v in
x | y | z)
;; # don't actually do anything in this switch
*) # anything else, we fall through to this switch
yeah
some more yeah
in fact, lots of yeah;;
esac
The trailing double semicolons cause aneurysms at first, but you quickly recover, and learn to appreciate, even love them. POSIX lets you put an opening parenthesis before the match expression so you don't have unpaired right parentheses, but this usage is rather uncommon.
(This is obviously not a suitable answer for Unix shells which are not from the Bourne family. The C family of shells -- including the still somewhat popular tcsh -- use a syntax which is supposedly "C-like" but that's like being unable to tell apart Alice Cooper from the girl who went to Wonderland; and the Fish shell has its own peculiarities which I'm not even competent to comment on.)
You could use something like this, for PHP:
if(strpos('xyz',$v[0])===false)//example 1
//strpos returns false when the letter isn't in the string
//returns the position (0 based) of the substring
//we must use a strict comparison to see if it isn't in the substring
if(!in_array($v[0],array('x','y','z')))//example 2
//example 3
$out=array('x'=>1,'y'=>1,'z'=>1); //create an array
if(!$out[$v[0]]) //check if it's not 1
if(!preg_match('/^[xyz]$/',$v))//example 4, using regex
if(str_replace(array('x','y','z'),'',$v[0]))//example 5
if(trim($v[0],'xyz'))//example 6
For Javascript:
if(~'xyz'.search(v[0]))//example 1(.indexOf() works too)
if(!(v[0] in {x:0,y:0,z:0}))//example 2
if(~['x','y','z'].indexOf(v[0]))//example 3, incompatible with older browsers.
if(!/^[xyz]$/.match(v))//example 4
if(v.replace(/^[xyz]$/))//example 5
For MySQL:
Select not locate(#v,'xyz'); -- example 1
select #v not in ('x','y','z'); -- example 2
-- repetition of the same pattern for the others
For C:
if(!strstr("xyz",v))//example 1, untested
There are more ways, I'm just too lazy.
Use your imagination and just write the one that you like more!
In the below given code, why the || logical doesn't work, instead the loop terminates specifically when && is used ?
int main() {
char select {};
do {
cout<<"Continue the loop or else quit ? (Y/Q): ";
cin>>select;
} while (select != 'q' && select != 'Q'); // <--- why || (or) doesn't work here ??
return 0;
}
This loop will go on while select is not q and it's not Q:
while (select != 'q' && select != 'Q');
This loop will go on while select is not q or it's not Q.
while (select != 'q' || select != 'Q');
Since one of them must be true, it'll go on forever.
Examples:
The user inputs q
select != 'q' evaluates to false
select != 'Q' evaluates to true
false || true evaluates to true
The user inputs Q
select != 'q' evaluates to true
select != 'Q' evaluates to false
true || false evaluates to true
You want to terminate the loop when select is equal either to 'q' or 'Q'.
The opposite condition can be written like
do {
cout<<"Continue the loop or else quit ? (Y/Q): ";
cin>>select;
} while ( not ( select == 'q' || select == 'Q' ) );
If to open the parentheses then you will get
do {
cout<<"Continue the loop or else quit ? (Y/Q): ";
cin>>select;
} while ( not( select == 'q' ) && not ( select == 'Q' ) );
that in turn is equivalent to
do {
cout<<"Continue the loop or else quit ? (Y/Q): ";
cin>>select;
} while ( select != 'q' && select != 'Q' );
Consider the following diagrams:
The full ellipse are all characters. The white dots is q and Q respectively. The black filled area depicts characters that will make the expression true. First line is select != 'q' && select != 'Q', second line is select != 'q' || select != 'Q'.
&& means both conditions must be true. The resulting black area is the overlap of the two areas on the left.
|| means either of the conditions must be true. The resulting black area is the sum of the two areas on the left.
Trying to test if input is "TF" or "MC". The while condition keeps coming out as true even though argument is written: line != "TF" || line != "MC"
Not understanding how the loop keeps repeating even though I input TF or MC. I have also verified that the tranform method is making the string capital.
do {
cout << "\nEnter the Question type (TF) for True/False or (MC) for Multiple Choice:\n";
getline(cin, line);
transform(line.begin(), line.end(), line.begin(), ::toupper);
} while (line != "TF" || line != "MC");
I expected the loop to only initiate once and exit.
If the person types "MC", the expression
line != "TF"
will be set to true, which will make the OR statement true.
(And thus repeating the while even though the person typed a valid answer)
What you are looking for is to check if the answer is neither one of the options, which can be checked as follows:
(line != "TF" && line != "MC")
The opposite would be reasonable too. That is, to check if the person typed a valid answer, and keep repeating it while it is not the case:
while(!(line == "TF" || line == "MC"))
Both statements are equivalent, as stated in the comments, by the De Morgan's Laws.
This might be overkill, but it does explain the relationships between the OR, AND, and NOT operators: https://en.wikipedia.org/wiki/Boolean_algebra
This question already has answers here:
Why does non-equality check of one variable against many values always return true?
(3 answers)
Closed 6 years ago.
Hello I am doing a very simple while loop in C++ and I can not figure out why I am stuck in it even when the proper input is give.
string itemType = "";
while(!(itemType == "b") || !(itemType == "m") || !(itemType == "d") || !(itemType == "t") || !(itemType == "c")){
cout<<"Enter the item type-b,m,d,t,c:"<<endl;
cin>>itemType;
cout<<itemType<<endl;
}
cout<<itemType;
if someone can point out what I am over looking I'd very much appreciate it. It is suppossed to exit when b,m,d,t or c is entered.
Your problem is in your logic. If you look at your conditions for your while loop, the loop will repeat if the item type is not "b" or not "m" or not "d" etc. That means if your item type is "b", it is obviously not "m", so it will repeat. You want to use && instead of ||.
As other answers and comments wrote correctly your logic is wrong. Using find() would simplify your task:
std::string validCharacters( "bmdtc" );
while ( std::string::npos == validCharacters.find( itemType ) )
{
...
}
This solution is more general and easier to read. See also documentation of std::string::find
The boolean expression to exit the loop is flawed. The way it is, in order to exit the loop the itemType would have to be all those letters at the same time. Try to instead || the letters first, and then negate it:
while(!(itemType == "b" || itemType == "m" || itemType == "d" || itemType == "t" || itemType == "c")
try this
string itemType = "";
while(!(itemType == "b" || itemType == "m" || itemType == "d" || itemType == "t" || itemType == "c")){
cout<<"Enter the item type-b,m,d,t,c:"<<endl;
cin>>itemType;
cout<<itemType<<endl;
}
cout<<itemType;
you condition is always true
The following while loop has two conditions
cin >> user;
while (( user != 'X') || ( user != 'O'))
{
cout << "Please enter either X or O " << endl;
cin >> user;
}
After I enter either X or O, it keeps asking for a new input. I don't understand why? But if I remove on of the conditions it works properly.
Think about the logic of "this thing is not X, or this thing is not Y" — barring overlaps between X and Y, such a condition is always true, even in English!
You've been misled by the colloquial and subtly different "this thing is neither X nor Y", but your code is not "neither X nor Y", but "not X or not Y".
What you meant was "not X and not Y".
while (( user != 'X') && ( user != 'O'))
Use && (and) instead of || (or).
Your condition is always true...
while (( user != 'X') || ( user != 'O'))
If user is 'X' then user is not 'O' thus, even if the first part of the condition is not satisfied, the second part of the condition is satisfied. So the whole condition is true.
Same thing if useris 'O'.
Try this with a "logical and" (a.k.a. &&) instead of a "logical or".
It looks like you want logical AND (&&) instead of logical OR (||).
If you want to understand what happens more intuitively, reverse the logic of your expression:
!= becomes ==, and || (or) becomes && (and).
So what you wrote is:
Quit the loop if user equals 'X' AND user equals 'O'.
As you can see it's impossible for user to have both values at the same time.
What you want is:
while (( user != 'X') && ( user != 'O'))
{
cout << "Please enter either X or O " << endl;
cin >> user;
}
Well, || means or. If you read the condition aloud to yourself, it would sound like "While user is not X or user is not O. If you think about it, it will always be true - when it is X, it is also not O, and when it is O, it is not X. What you probably need is &&:
while (( user != 'X') && ( user != 'O'))
That way, the loop will stop when user is neither X nor O.
Thats because you're using the OR operator.
You're asking if user is not X or user is not O keep asking for input.
change it to && operator