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CMPS : Compare String Byte or String Word:-The CMPS instruction may be utilized to compare two strings of Words or byte. The length of the string ought to be stored in the CX. If both the word or byte strings are equal, zero flag is set. Flags are affected in the similar way as CMP instruction. The DS: SI and ES: DI point to the 2 strings. The REP instruction prefix is utilized to repeat the operation till register CX (counter) becomes zero or the condition specified by the REP prefix is false.
Following string of instructions describe the instruction. Comparison of the string begin from initial byte / word of the string, after each comparison the index registers are updated depending on the direction flag and the counter is decremented. This byte by byte / word by word comparison continues until a mismatch is found. When, a mismatch is found, the carry flag and zero flags are modified properly and the execution proceeds further.
Example :
If both strings are fully equal, for example the register CX becomes zero, the ZF is set, or else, ZF is reset.
how to store a bulk data in a external eeprom
Control Transfer or Branching Instruction Control transfer instructions transfer the flow of execution of the program to a new address specified in the instruction indirectly o
Write an application that does the following:(1) fill an array with 50 random integers; (2) loop through the array, displaying each value, and count the number of negative values;
Conditional branch Instruction When these type of instructions are executed, they transfer control of execution to the address mention relatively in the instruction, provided t
DIV: Unsigned Division:- This instruction performs unsigned division operation. It divides an unsigned word or double word by a 16-bit or 8-bit operand. The dividend might be in t
init_lcd ;(this initialises a 2 row lcd) bcf TRISA,0 ;PORTA bit 0 as an output (lcd RS pin) bcf TRISA,1 ;PORTA bit 1
* * * * **** * * * * * How can i print this help me pls
Assembly Language Example Programs We studied the entire instruction set of 8086/88, pseudo-ops and assembler directives. We have explained the process of entering an assembly
ORG : Origin:- The ORG directive directs the assembler to begin the memory allotment for the specific segment, code or block from the declared address in the ORG statement. W
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
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