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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 from the address of the byte next to the jump opcode). This process of specifying the jump address saves the memory. Or else, the assembler may reserve two bytes for the displacement. The sentence structure of the statement is as given below.
JMP SHORT LABEL
TYPE :
The TYPE operator directs the assembler to decide the data type of the mention label and replaces the 'TYPE label' by the decided data type. For the particular word type variable, the data type is 2, for double word type, it is four, and for byte type, it is one. Imagine, the STRING is a word array. The instruction TYPE STRING, MOV AX, moves the value 0002H in AX.
GLOBAL :
The, variables, labels, constants or procedures declared GLOBAL can be used by other modules of the program. Once a variable is declared GLOBAL, it may be used by any module in the program. The following statement declares the process ROUTINE as a global label.
Data copy/transfer Instructions MOV: This data transfer instruction transfers data from one register or memory location to another register or memory location. The source can
The processor 8088 The launching of the processor 8086 is consider as a remarkable step in the development of high speed computing machines. Before the introduction of 8086 mo
ASSUME: Assume Logical Segment Name:- The ASSUME directive which is used to inform the assembler, the specified names of the logical segments to be consider for different segme
code to add two matrices
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General Data Registers Given figure indicate the register organization of 8086. The registers DX, CX, BX and AX are the general purpose 16-bit registers. AX is behaved as 16-bi
There are two parts to this assignment. The first part has you reading 4 integers representing; #QUARTERS, #DIMES, #NICKELS & #PENNIES, respectively. Your program should compute t
Read Architecture : Look Aside Cache In "look aside" cache architecture the main memory is located conflictingthe system interface. Both the cache main memory sees a bus cycle
Assume that the registers are initialized to EAX=12345h,EBX =9528h ECX=1275h,EDX=3001h sub AH,AH sub DH,DH mov DL,AL mov CL,3 shl DX,CL shl AX,1 add DX,AX
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