Why does this if executes while I have not given any input?

Here is my code snippet:

int a;
if(a=8)
 开发者_JS百科   cout<<"Its right";
else
    cout<<"Not at all";
getch();
return(0);

I'm getting the output Its right while I've not give any input there its just a assignment to the a=8.

**Borland Compiler** gives the following 2 warnings and executes the code.

1: Possible incorrect assignment (a=8)

2: 'a' is assigned a value that is never used.

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When you write:

if(a=8)

You're assigning 8 to the variable a, and returning a, so you're effectively writing:

a = 8;
if(a)

This is non-zero, and hence treated as "true".

I (and the compiler) suspect you intended to write (note == instead of =):

if(a==8)

This is why you get the warnings. (Note that, in this case, a is still uninitialized, so the behavior is undefined. You will still get a warning, and may still return true...)

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a = 8 assign 8 to a and returns the value of the assignment, that is 8. 8 is different from 0 and thus is considered true. the test apsses, and the program outputs "Its right".

you may have written if ( a == 8 ) which correctly tests the value of a without changing its value.

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if(a=8)

implies assign a value of 8 to a and then use it to evaluate for the if statement. Since 8 is nonzero it essentially translates to TRUE (any value that is non-zero is true).

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The first warning is because it's a frequent typo to write = in a condition test where you meant ==. The philosophy behind the warning is that it's a lesser burden for a programmer to rewrite if(a=b) to, say if((a=b)!=0) in the few cases he wants that, than to spend frustrating hours debugging the results if he mistakenly got an assignment instead of a comparison.

The second warning is just what it says: It looks like a mistake to have a local variable that you assign a value to but never use.

Since they are just warnings, you're free to ignore them if you think you know what you're doing. The compiler may even provide a way for you to turn them off.

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Everyone else's answer is correct :)

if (a = 8) // Sets a to be 8 (and returns true so the if statement passes)
if (a == 8) // Check to see if a is equal to 8

I just want to add that I always (force of habit!) write if statements this way :

if (8 == a) {

Now, if I miss out one of the =, the compiler doesn't give me a warning, it gives me an error :)

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You need a == (double equals). Single = is an assignment; when treated as an expression as you have done here, it evaluates to the value assigned, in this case 8. 0 is treated as False and anything else is True, so (a=8) == 8 == True.