Defijne meaning of statement in C++

Question

I am working with some C++ code and I am just a novice and I do not understand what this conditional statement means for a true or false result.

This is what I have:

Font contains values related to bitmap font

for(j = 0; j < COUNT; j++)   {
    for(i = 0; i < 8; i++)  {
        Font[j][i] <<= 1;
        if((j != COUNT -1) && (Font[j + 1][i] & 0x80))
            Font[j][i] |= 0x01;
    }
}

I understand most of this, the Boolean && , then the & is confusing me for relevancy in this use and then the lone 0x80 , I just don't understand how that relates to the (Font[j + 1][i] & 0x80) 0x80 is 128 decimal... The font is a 128 x 8 font , but is that a relationship? Can someone help me put this together so that I can understand how it is providing the condition? I also need to know how |= 0x01 affects Font[j][i]. Is that a form of piping?

Answer 1

The if statement generic format is

if (conditional_expression) ...

The conditional_expression is any expression which yields a result. The result zero ( 0 ) is false, and anything non-zero is true.

If the conditional_expression is simple and without any kind of comparison, then it's implicitly compared to zero. For example

int my_variable = ...;  // Actual value irrelevant for example

if (my_variable) { ... }

The if statement above is implicitly equal to

if (my_variable != 0) { ... }

This implicit comparison to zero is also done for compound conditions, for example

if (some_condition && my_variable) { ... }

is equal to

if (some_condition && my_variable != 0) { ... }

---

Now we get back to your code and the condition:

if((j != COUNT -1) && (Font[j + 1][i] & 0x80))

With the implicit comparison against zero, the above is equal to

if((j != COUNT -1) && (Font[j + 1][i] & 0x80) != 0)

That is, the right-hand side of the && check is Font[j + 1][i] & 0x80 is zero or not.

As for the & operator itself, it's the bitwise AND , and in essence can be used to check if a bit is set or not. For your code it checks if the bit corresponding to the value 0x80 (the eight bit) is set.