subtraction with borrow

"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!

The whole point of high-level languages is that they allow you to concentrate on what you want to do, rather than how to achieve it with the available processor instructions!

"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!

The whole point of high-level languages is that they allow you to concentrate on what you want to do, rather than how to achieve it with the available processor instructions!