C Equivalent of assembler instruction ANL

"Can some one shed a light on the correct interpretation"

What you need is a 'C' textbook!

Always a good idea to have a language reference handy. I imagine you can find a tutorial on the web with a bit of Googling.

Here's an online reference:

http://www-ccs.ucsd.edu/c/

but it's just that -- a reference, and so may be short on explanation.

C has both "bitwise" and "logical" AND and OR operators. The bitwise operators act on each bit in a type independently. The logical operators interpret the operands as Boolean values (0 is false, non-zero is true, which is 1) and then produce a result. To set and clear bits, you generally want the bitwise operators, which are the single-character version ('&' instead of '&&').

ccapm1 = ccapm1 & 0x0d;

is the match for

ANL CCAPM1, #0dH

It performs a bitwise AND of the value in ccamp1 with 0x0d, and then stores the result back into ccapm1.

ccapm1 = ccapm1 && 0x0d;

will probably not do what you want. If ccapm1 is non-zero, then the result of the logical expression is 1, which will be assigned to ccapm1. If ccapm1 is zero, it will remain zero.

~ is the bitwise NOT operator. It inverts all the bits in the operand. So

ccapm1 = ccapm1 & ~0x0d;

is the same as

ccapm1 = ccapm1 & 0xf2;

Since using a variable in this sort of expression and then reassigning the result is such a common thing to do, C provides a shorthand notation.

ccapm1 &= 0x0d;

means exactly the same thing as

ccapm1 = ccapm1 & 0x0d;

The usual idioms for setting and clearing bits are thus:

var |= mask; // set 1 bits of mask in var
var &= ~mask; // clr 1 bits in mask from var