Assembly language program to swap two numbers

How can we swap X and Y without using extra variables?

without using extra variables we know that we can swap x and y by using :
In the same way we can use registers to find solutions for the same. Implementation is provided below :
Load HL register pair as 2050H Which will acts as Memory Pointer. Move content of [2050H] memory location into Accumulator A. Increment memory pointer by 1 .

How do I swap memory with a register?

Two other reasonable (but much worse) options for swapping memory with a register are:
; using scratch space on the stack push [mem] ; [mem] can be any addressing mode
E.g. [bx] mov [mem]
Ax pop ax ; dep chain = load

Is XOR-swap a UOP instruction?

On earlier AMD CPUs
It's still a 2 uop instruction
But with 1c latency each way. xor-swaps or add/sub swaps or any other multi-instruction sequence other than mov are pointless compared to xchg for registers. They all have 2 and 3 cycle latency
And larger code-size. The only thing that's worth considering is mov instructions.

Why do I need separate code blocks for each number to swap?

It requires separate code blocks for each number to swap
Which can be cumbersome for large data sets. The program does not check for errors or boundary conditions
Such as :
When the source or destination address is outside the valid memory range or when the data size is not a multiple of 16 bits.

How to swap two locations?

To swap two locations, you pretty much need the original two and someplace to temporarily store the original values: mov ecx, [eax] ; fetch memory specified by address in eax mov edx, [ebx] ; fetch memory specified by address in ebx mov [eax], edx ; put 2nd value where 1st used to be mov [ebx], ecx ; put 1st value where 2nd used to be

How to swap two numbers in 8085?

In this program we will see how to swap two numbers. Write 8085 Assembly language program to swap two 8-bit number stored at location 8000Hand 8001H. In 8085, there is an instruction XCHG. Using this we can swap the contents of DE and HL values. We are taking the numbers and storing them into H and D, then using XCHG the contents are swapped.

Why do I need separate code blocks for each number to swap?

It requires separate code blocks for each number to swap, which can be cumbersome for large data sets. The program does not check for errors or boundary conditions, such as when the source or destination address is outside the valid memory range or when the data size is not a multiple of 16 bits.

8086 has an instruction for this: xchg ax, bx. If you really need to swap two regs, xchg ax, bx is the most efficient way on all x86 CPUs in mo...15

You can do it using some mathematical operation. I can give you an idea. Hope it helps! I have followed this C code: int i=10; j=20 i=i+j; j=i-j;...Best answer · 1

swapping two integers in assembly is a lot easier than the code you posted. This assumes 32-bit integers on Intel family of 32-bit/64-bit processor...Best answer · 0

I was hoping to do this with the xor commands. Also, at the end of this is the x pointer now pointing to what the y pointer was originally pointin...0

The code I posted does not change the pointers, only the values to which they point. If you want to use xor then do so.0

Swap Two numbers using Stack in assembly language .MODEL SMALL .STACK 100H .CODE MAIN PROC MOV AX,'9' MOV BX,'4' PUSH AX...0

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