Return true/false function function always returns true

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1.00/5 (1 vote)

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Im a beginner and I am trying to write a function that checks if the number is in a given interval. If it is, the program is supposed to return true, else return false. For some reason the output is always "Process returned 1", no matter which values I assign to the variables. Does return do something else than I think or what is the problem?

What I have tried:

C++

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

int IsInRange(int number, int high, int low){
    if (low <= number && number <= high){
        return true;
    }
    else{
        return false;
    }
}

int main()
{
    int x, low = 15, high = 25, number = 2;
    return IsInRange(low, high, number);

    return 0;
}

Posted 4-Nov-20 10:12am

Member 14955513

Updated 4-Nov-20 20:31pm

KarstenK

v4

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3 solutions

Solution 2

C++

int IsInRange(int number, int high, int low){
    if (low <= number <= high){ // this is not c
        return true;
    }
    else{
        return false;
    }
}

In most programming languages, a double condition is written as 2 simple conditions with logical operation in between.

C++

int IsInRange(int number, int high, int low){
    if (low <= number && number <= high){
        return true;
    }
    else{
        return false;
    }
}

which means: (low <= number) and (number <= high)
[Update]
By the way, position of arguments matters !

C++

int IsInRange(int number, int high, int low)
...
return IsInRange(low, high, number);

Posted 4-Nov-20 10:29am

Patrice T

Updated 4-Nov-20 11:00am

v2

Comments

Member 14955513 4-Nov-20 16:43pm

Changed it this way but now the output is only true when number == low

Patrice T 4-Nov-20 16:55pm

Show your new code.

Member 14955513 4-Nov-20 16:59pm

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

int IsInRange(int number, int high, int low){
if ((low <= number) && (number <= high)){
return true;
}
else{
return false;
}

}

int main()
{
int x, low = 20, high = 30, number = 21;
return IsInRange(low, high, number);

return 0;
}

Patrice T 4-Nov-20 17:00pm

Use Improve question to update your question.
So that everyone can pay attention to this information.

Solution 3

I prefer to write code in very simple terms. I would do it like this :

C++

int IsInRange( int number, int high, int low )
{
    if( number < low )
       return false;
    else if( number > high )
       return false;
    else
       return true;
}

and you can test it :

C++

int main()
{
    const char * str;
    int result;
    int low = 15;
    int high = 25;
    int n;
    for( int n = 10; n <= 30; ++n )
    {
        result = IsInRange( n, low, high );
        if( result )
           str = "true";
        else
           str = "false";
        printf( "when n is %d  result is %s\n", n, str );
    }
}

Posted 4-Nov-20 20:32pm

Rick York

Updated 4-Nov-20 20:39pm

v3

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OriginalGriff · Accepted Answer · 2020-11-04T10:23:00

Solution 1

The condition you use does not produce the results you expect:

low <= number <= high

Actually evaluates as either

(low <= number) <= high

Or

low <= (number <= high)

Since the result of a <= b will always be zero or nonzero only (and most likely 0 or 1) the result isn't what you expected.
Try this instead:

if (low <= number && number <= high)

Posted 4-Nov-20 10:23am

OriginalGriff

Comments

Member 14955513 4-Nov-20 16:36pm

if ((low <= number) && (number <= high))

Modified it like this but now it only return true when number == low.

"Process returned 1 (0*1)"

OriginalGriff 4-Nov-20 17:09pm

That's because you are passing the values in the wrong order ... check the names in the call list against the parameters in the function.
When you call a function, it's parameters are passed by order into the variables named in the function declaration regardless of the names in the "outside world".

int IsInRange(int number, int high, int low){
...
return IsInRange(low, high, number);

You need:

int IsInRange(int number, int high, int low){
...
return IsInRange(number, high, low);