subtraction with borrow

I am trying a C code, the "xbuffer=(EX-X)+1;" is correct if original code as following:
.
.
MOV A,EX
SUBB A,X
INC A
MOV xbuffer,A
.
.

I am trying a C code, the "xbuffer=(EX-X)+1;" is correct if original code as following:
.
.
MOV A,EX
SUBB A,X
INC A
MOV xbuffer,A
.
.

"I am trying a C code, the "xbuffer=(EX-X)+1;" is correct if original code as following:
.
.
MOV A,EX
SUBB A,X
INC A
MOV xbuffer,A"

The assembler code behaves differently depending on the value of the carry flag, so I guess the answer is no.

Are you trying to translate an assembler program into 'C'?

Yes, if it isn't correct. How to write a C code?

"Yes, if it isn't correct. How to write a C code?"

You can't, or at least not meaningfully. While it is safe enough to read the carry flag in 'C' its value doesn't mean anything useful.

Look at the bigger picture - figure out what the assembler program does and reproduce that in 'C'. Don't try to translate it one instruction at a time. Translate the algorithm instead.

Translate the algorithm instead.

So anyone can for explain a simple program.

"So anyone can for explain a simple program."

The subject of your question is a perfect example.
The 8051 is an 8-bit processor, so all its instructions deal with only 8 bits at a time. Therefore, if you want to handle data larger than 8 bits in Assembler, you have to "manually" code it 8 bits at a time.

The Subtract-with-Borrow (and, similarly, Add-with-Carry) instruction allows you to do subtraction on data larger than 8 bits by handling 8 bits at a time, and including a "borrow" from the previous 8-bit operation.

In 'C', you don't mess about with all this detail - you just use the 16-bit data type (int) or 32-bit data type (long) built into the language!

long a, b, c;

a = b - c; // 32-bit subtraction in 'C' is just a simple use of the '-' operator!