Interrupt 20 ( AKA INT 20, int 20h, INT 20H) is the 33rd entry in the standard PC interrupt table. The '20' is hexadecimal notation (hence the use of the H or h, which may actually be required by some assemblers whereas others assume entries are hexadecimal) for decimal thirty-two (32) but counting starts from zero, as is usually the case in Computer Science.
Int 20 is known as "DOS Terminate" and will end execution of the code at that point and return control to DOS (or the DOS-like environment which called it). It looks at the CS register (a standard 80x86 register which was intended to 'point' to the segment where the code was loaded - Code Segment) and returns nothing by way of register or stack.
Simple Answer: this is an instruction used in very small and typically very old DOS programs to terminate themselves quickly and with a minimum of effort, it probably ought not to be used today, research Int 21 instead (the so-called 'DOS Interrupt').
N.B.: This instruction could also refer to other architectures and thus be very different, e.g. AIX, VMS, etc. My answer is PC & DOS specific as this is the most likely context for this question.
EPIC, which stands for Explicitly Parallel Instruction Computing.
It is employed to hold temporarily the right hand instruction from a word in memory.. For example, The IAS machine's basic unit of information was a 40-bit, so that two instructions could be stored in each 40-bit memory location. Each instruction consisted of an 8-bit {operation code} and a 12-bit address. Hence the IBR (Instruction Buffer Register) is used to temporarily hold Right hand instruction for the next use.
• The processor fetches the instruction from memory • Program counter (PC) holds address of the instruction to be fetched next • PC is incremented after each fetch • Fetched instruction loaded into instruction register
test and branch instruction
Because it could not be used without training beyond a instruction booklet, and it wasn't mostly electrical
20H
The instruction 20H is RIM (Read Interrupt Mask). The instruction 57H is MOV D,A. The contents of PC after these two instruction will be 2 greater than it was before starting.
The INT 21H instruction in the 8086 is a software interrupt to vector 21H. In order for it to be used for input/output, the programming that responds to INT 21H must be present. This is part of the Operating System.
-17h + 27 = 20h so 27 = 20h + 17h = 37h Then h = 27/37 = 0.7297 approx.
12g + 20h You could write this as 4(3g + 5h)
How do you solve this equation 20h-40=8h+152 ?
STA[20H] and STA20H
instruction register is used to store the next instruction to be executed. instruction pointer is used to store the address of the next instruction to be executed.
8-20h
The INT 03 instruction on the 8086/8088 and higher class processors is a program generated interrupt that only requires one byte in the opcode. Often, this is used by a debugger, to plant breakpoints at certain points in the code. During the interrupt servicing routine, the original opcode would be restored so that it could be executed if desired. Contrast this with the INT 01 instruction, which is actually a single step type of interrupt. In this case, the debugger sets the single step flag in the return PSW, and then simply returns to the program. No opcode needed to be planted, as there will be an automatic execution of just one instruction, and then the interrupt will occur. The difference is in performance. INT 03 can allow the program to run at full speed until it hits the breakpoint. The downside is that, if the program does not make it to the breakpoint, the debugger will not be able to regain control without forcing an interrupt. INT 01 allows the debugger to examine the state of the program at every single instruction, allowing the implementation of complex rule based breakpoints. The downside is that program execution will be very slow.
int[] marks = new int[3]; int marks[3];
Instruction is simmilar to the MUL except that operands are assumed to be signed numbers. The source operand specified in the instuction is multiplied with accumulator. Example: IMUL BL If AL=80H, BL=20H content of AL is treated as signed no. Which is 2'compliment of 128 -128*32=-4096 2'compliment of it is stored in AX i.e. F000H