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LABEL :
The Label directive which is used to assign a name to the current content of the location counter. At the beginning of the assembly process, the assembler start a location counter to keep track of memory locations assigned to the program. As the program assembly proceeds, the specific contents of location counter are updated. At the time the assembly process, whenever the assembler across the LABEL directive, itassigns the declared label with the current contents of the location counter. The type of the label ought to be specified, for example whether it is a NEAR or a FAR label, WORD or BYTE label, etc.
A LABEL directive can be used to make a FAR jump as shown. A FAR jump can't be made at a normal label with a colon. The label CONTINUES is used for a FAR jump, if the program contains the below describe statement.
CONTINUE LABEL FAR
The LABEL directive is used to refer to the data segment along with the data type, word or byte as showing.
DATA SEGMENT DATAS DB 5 OH DUP (?) DATA-LAST LABEL BYTE FAR DATA ENDS
After reserving the 50H locations for DATAS, the next location will be assigned a label DATALAST and its type will be far and byte.
what is double hashing
Memory Segmentation : The memory in an 8086/8088 based system is organized as segmented memory. In this scheme, the whole physically available memory can be divided into a n
The modes are determined by the contents of the control register, whose format is given in Figure These modes are: Mode 0: If a group is in mode 0, it is divided into 2 sets.
Comparison between 8086 and 8088 All the changes in 8088 above 8086 are indirectly or directly related to the 8-bit, 8085 compatible data and control bus interface. 1) The p
SHORT : The SHORT operator denoted to the assembler that only one byte is needed to code the displacement for a jump (for example displacement is within -128 to +127 bytes fr
Prime Finder - assembly program: Problem: Prime Finder In this problem you will write a small program that tests whether a given integer is a prime number or not. Let's
move a byte string ,16 bytes long from the offset 0200H to 0300H in the segment 7000H..
from pin description it seems that 8086 has 16 address/data lines i.e.AD0_AD15.The physical address is however is larger than 2^16.How this condition can be handled
I am assigned to implement dijkstra algorithm in assembly language. I am not a novice in assembly. I need help implementing it.Kindly if anyone then please.
INTO : Interrupt on Overflow:- It is executed, when the overflow flag OF is set. The new contents of IP and CS register are taken from the address 0000:0000 as described in INT
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