Why does the code below return true only for a = 1?

Question

Why does the code below return true only for a = 1?

main(){
int a = 10;
if (true == a)
     cout<<"Why am I not getting executed";
}

Answer 1

When a Bool true is converted to an int, it's always converted to 1. Your code is thus, equivalent to:

main(){
   int a = 10;
   if (1 == a)
      cout<<"y i am not getting executed";
   }

This is part of the C++ standard, so it's something you would expect to happen with every C++ standards compliant compiler.

Answer 2

The reason your print statement is not getting executed is because your boolean is getting implicitly converted to a number instead of the other way around. I.e. your if statement is equivalent to this: if (1 == a)

You could get around this by first explicitly converting it to a boolean:

main(){
int a = 10;
if (((bool)a) == true)
     cout<<"I am definitely getting executed";
}

In C/C++ false is represented as 0.

Everything else is represented as non zero. That is sometimes 1, sometimes anything else. So you should never test for equality (==) to something that is true.

Instead you should test for equality to something that is false. Since false has only 1 valid value.

Here we are testing for all non false values, any of them is fine:

main(){
int a = 10;
if (a)
     cout<<"I am definitely getting executed";
}

And one third example just to prove that it is safe to compare any integer that is considered false to a false (which is only 0):

main(){
int a = 0;
if (0 == false)
     cout<<"I am definitely getting executed";
}

Answer 3

Your boolean is promoted to an integer, and becomes 1.

Answer 4

in C and C++, 0 is false and anything but zero is true:

if ( 0 )
{
// never run
}

if ( 1 )
{
// always run
}

if ( var1 == 1 )
{
// run when var1 is "1"
}

When compiler calculates a boolean expression it is obliged to produce 0 or 1. Also, there's a couple handy typedefs and defines, which allow you to use "true" and "false" instead of 1 and 0 in your expressions.

So your code actually looks like this:

main(){
int a = 10;
if (1 == a)
     cout<<"y i am not getting executed";
}

You probably want:

main(){
int a = 10;
if (true == (bool)a)
     cout<<"if you want to explicitly use true/false";
}

or really just:

main(){
int a = 10;
if ( a )
     cout<<"usual C++ style";
}

Answer 5

Because true is 1. If you want to test a for a non-zero value, just write if(a).

Answer 6

I suggest you switch to a compiler that warns you about this... (VC++ yields this: warning C4806: '==' : unsafe operation: no value of type 'bool' promoted to type 'int' can equal the given constant; I don't have another compiler at hand.)

I agree with Lou Franco - you want to know if a variable is bigger than zero (or unequal to it), test for that.

Everything that's done implicitly by the compiler is hazardous if you don't know the last detail.

Answer 7

Here is the way most people write that kind of code:

main(){
int a = 10;
if (a) // all non-zero satisfy 'truth'
     cout<<"y i am not getting executed";
}

I have also seen:

main(){
int a = 10;
if (!!a == true) // ! result is guaranteed to be == true or == false
     cout<<"y i am not getting executed";
}

Answer 8

I wouldn't expect that code to be defined and you shouldn't depend on whatever behavior your compiler is giving you. Probably the true is being converted to an int (1), and a is not being converted to a bool (true) as you expect. Better to write what you mean (a != 0) then to depend on this (even if it turns out to be defined).

Answer 9

something different from 0 (that is false) is not necessary true (that is 1)

Answer 10

Because true is equal to 1. It is defined in a pre-proccesor directive, so all code with true in it is turnbed into 1 before compile time.

Answer 11

Because a boolean is a bit in C/C++ and true is represented by 1, false by 0.

Update: as said in the comment my original Answer is false. So bypass it.