How to fix "CONSTANT EXPRESSION RESULT" type coverity warnings in c code

Constant Expression Result - Coverity Warning:
    In this post we can discuss about a common logical error in the "if" condition evaluation statement which may lead to the "Constant Expression Result" type coverity warning in C code. Because of this logical error, the expression in the "if" condition evaluates to be either true or false always.
   Coverity tool will report "constant expression result" error when an expression is evaluated to be either true or false always irrespective of the value of the operands in that expression. That is, the expression's value does not depend on its operands. Consider the below C program.

Example 1: Expression always evaluates to be "false"

    In this program, we wanted to print the statement "Condition Met" only for the inputs red and green.

 #include <stdio.h>
typedef enum
{
    violet = 1,
    indigo,
    blue,
    green,
    yellow,
    orange,
    red
}tVIBGYOR;

void VerifyColour (tVIBGYOR input)
{
    if ((input == red) && (input == green))
    {
        if (input == red)
            printf ("Condition Met for RED\n");
        if (input == green)
            printf ("Condition Met for GREEN\n");
    }
}
int main ()
{
    VerifyColour (violet);
    VerifyColour (green);
    VerifyColour (red);
    VerifyColour (orange);
  
    return 0;
}

Output:

root@ubuntu:~/Programs# ./a.out
root@ubuntu:~/Programs#

    When we execute the program, the "Condition Met" statement is not at all being printed for any input value passed. The  reason is that, the below condition check in the "if" statement in the function "VerifyColour" always evaluated to be false.

     if ((input == red) && (input == green))

    In this statement we expect the variable "input" to have two different values at the same time whic is not possible. Even though when we pass the input as "red", the other condition of "input == green" will not be met. Similarly for the input "violet", none of the condition would be met. Eventually, the statement evaluated to be false irrespective of input arguments. So coverity reports this as "CONSTANT_EXPRESSION_RESULT" warning for this condition check.

How to Solve:

    If we change the "if" statement as below, the issue will get resolved.

    if ((input == red) || (input == green))

   With this change, the program will print the "Condition Met" statement when the input is either "red" or "green".

root@ubuntu:~/Programs# ./a.out
Condition Met for GREEN
Condition Met for RED
root@ubuntu:~/Programs#


Example 2: Expression always evaluates to be "true"

    In this example, we want to print the input value along with "Condition Met" statement except for the input values "red" and "green".

 #include <stdio.h>
typedef enum
{
    violet = 1,
    indigo,
    blue,
    green,
    yellow,
    orange,
    red
}tVIBGYOR;

void VerifyColour (tVIBGYOR input)
{
    if ((input != red) || (input != green))
    {
        printf ("Condition Met. Value=%d\n",(int)input);
    }
}
int main ()
{
    VerifyColour (violet);
    VerifyColour (green);
    VerifyColour (red);
    VerifyColour (orange);
  
    return 0;
}

Output:
root@ubuntu:~/Programs# ./a.out
Condition Met. Value=1
Condition Met. Value=4
Condition Met. Value=7
Condition Met. Value=6
root@ubuntu:~/Programs#

    In this program, the below "if" statement in the function "VerifyColour" is evaluated to be "true"  always.

    if ((input != red) || (input != green))

   When we pass the input as "red",  the second condition "input != green" gets satisfied and the condition passes. Similarly for input "violet", the expression is again is evaluated to be "true".

How to Solve:
    If we modify the expression as below, the program works as expected.

    if ((input != red) && (input != green))

    With this change, the program will print the input value except for the inputs "red" and "green".

root@ubuntu:~/Programs# ./a.out
Condition Met. Value=1
Condition Met. Value=6
root@ubuntu:~/Programs#