Langage machine
quick2
Messages postés
2
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mpmp93 Messages postés 2931 Date d'inscription Statut Membre Dernière intervention -
mpmp93 Messages postés 2931 Date d'inscription Statut Membre Dernière intervention -
bonjour,
j'ai un examen d'informatique mardi; et j'arive pas à trouver les bonne réponse à cet exercice :
10. The following table shows a portion of a machine's memory containing a program
written in the language described in the language description table. Answer the
questions below assuming that the machine is started with its program counter
containing 00. (10%)
address /content address /content
00 25 07 00
01 03 08 C0
02 20 09 00
03 F9 0A C0
04 53 0B 00
05 05 0C C0
06 33 0D 00
(a) What bit pattern will be in register 5 when the machine halts?
(b) What bit pattern will be in register 0 when the machine halts?
(c) What bit pattern will be in register 3 when the machine halts?
(d) What bit pattern will be at memory location 00 when the machine halts?
ANSWER: (a) 03 (b) F9 (c) FC (d) FC
comment arrive -t- on à trouver ces réponses?
j'ai un examen d'informatique mardi; et j'arive pas à trouver les bonne réponse à cet exercice :
10. The following table shows a portion of a machine's memory containing a program
written in the language described in the language description table. Answer the
questions below assuming that the machine is started with its program counter
containing 00. (10%)
address /content address /content
00 25 07 00
01 03 08 C0
02 20 09 00
03 F9 0A C0
04 53 0B 00
05 05 0C C0
06 33 0D 00
(a) What bit pattern will be in register 5 when the machine halts?
(b) What bit pattern will be in register 0 when the machine halts?
(c) What bit pattern will be in register 3 when the machine halts?
(d) What bit pattern will be at memory location 00 when the machine halts?
ANSWER: (a) 03 (b) F9 (c) FC (d) FC
comment arrive -t- on à trouver ces réponses?
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3 réponses
Bonjour,
Cite: a program
written in the language described in the language description table
Sans connaissance du processeur cible, il est rigoureusement impossible de donner une réponse...
A+
Cite: a program
written in the language described in the language description table
Sans connaissance du processeur cible, il est rigoureusement impossible de donner une réponse...
A+
euh j'ai oublié de rajouter l'annexe qui explique un peu comment ça marche le langage machine :
The following table is from Appendix C of the text. It is included here so that it can be incorporated in tests
for student reference. Questions in this test bank refer to this table as the "language description table."
Opcode
Operand Description
1 RXY LOAD the register R with the bit pattern found in the memory cell whose address is XY.
Example: 14A3 would cause the contents of the memory cell located at address A3 to be placed
in register 4.
2 RXY LOAD the register R with the bit pattern XY.
Example: 20A3 would cause the value A3 to be placed in register 0.
3 RXY STORE the bit pattern found in register R in the memory cell whose address is XY.
Example: 35B1 would cause the contents of register 5 to be placed in the memory cell whose
address is B1.
4 0RS MOVE the bit pattern found in register R to register S.
Example: 40A4 would cause the contents of register A to be copied into register 4.
5 RST ADD the bit patterns in registers S and T as though they were two's complement representations
and leave the result in register R.
Example: 5726 would cause the binary values in registers 2 and 6 to be added and the sum placed
in register 7.
6 RST ADD the bit patterns in registers S and T as though they represented values in floating-point
notation and leave the floating-point result in register R.
Example: 634E would cause the values in registers 4 and E to be added as floating-point values
and the result to be placed in register 3.
7 RST OR the bit patterns in registers S and T and place the result in register R.
Example: 7CB4 would cause the result of ORing the contents of registers B and 4 to be placed in
register C.
8 RST AND the bit patterns in register S and T and place the result in register R.
Example: 8045 would cause the result of ANDing the contents of registers 4 and 5 to be placed in
register 0.
9 RST EXCLUSIVE OR the bit patterns in registers S and T and place the result in register R.
Example: 95F3 would cause the result of EXCLUSIVE ORing the contents of registers F and 3 to
be placed in register 5.
A R0X ROTATE the bit pattern in register R one bit to the right X times. Each time place the bit that
started at the low-order end at the high-order end.
Example: A403 would cause the contents of register 4 to be rotated 3 bits to the right in a circular
fashion.
B RXY JUMP to the instruction located in the memory cell at address XY if the bit pattern in register R
is equal to the bit pattern in register number 0. Otherwise, continue with the normal sequence of
execution. (The jump is implemented by copying XY into the program counter during the execute
phase.)
Example: B43C would first compare the contents of register 4 with the contents of register 0. If
the two were equal, the pattern 3C would be placed in the program counter so that the next
instruction executed would be the one located at that memory address. Otherwise, nothing would
be done and program execution would continue in its normal sequence.
C 000 HALT execution.
Example: C000 would cause program execution to stop.
The following table is from Appendix C of the text. It is included here so that it can be incorporated in tests
for student reference. Questions in this test bank refer to this table as the "language description table."
Opcode
Operand Description
1 RXY LOAD the register R with the bit pattern found in the memory cell whose address is XY.
Example: 14A3 would cause the contents of the memory cell located at address A3 to be placed
in register 4.
2 RXY LOAD the register R with the bit pattern XY.
Example: 20A3 would cause the value A3 to be placed in register 0.
3 RXY STORE the bit pattern found in register R in the memory cell whose address is XY.
Example: 35B1 would cause the contents of register 5 to be placed in the memory cell whose
address is B1.
4 0RS MOVE the bit pattern found in register R to register S.
Example: 40A4 would cause the contents of register A to be copied into register 4.
5 RST ADD the bit patterns in registers S and T as though they were two's complement representations
and leave the result in register R.
Example: 5726 would cause the binary values in registers 2 and 6 to be added and the sum placed
in register 7.
6 RST ADD the bit patterns in registers S and T as though they represented values in floating-point
notation and leave the floating-point result in register R.
Example: 634E would cause the values in registers 4 and E to be added as floating-point values
and the result to be placed in register 3.
7 RST OR the bit patterns in registers S and T and place the result in register R.
Example: 7CB4 would cause the result of ORing the contents of registers B and 4 to be placed in
register C.
8 RST AND the bit patterns in register S and T and place the result in register R.
Example: 8045 would cause the result of ANDing the contents of registers 4 and 5 to be placed in
register 0.
9 RST EXCLUSIVE OR the bit patterns in registers S and T and place the result in register R.
Example: 95F3 would cause the result of EXCLUSIVE ORing the contents of registers F and 3 to
be placed in register 5.
A R0X ROTATE the bit pattern in register R one bit to the right X times. Each time place the bit that
started at the low-order end at the high-order end.
Example: A403 would cause the contents of register 4 to be rotated 3 bits to the right in a circular
fashion.
B RXY JUMP to the instruction located in the memory cell at address XY if the bit pattern in register R
is equal to the bit pattern in register number 0. Otherwise, continue with the normal sequence of
execution. (The jump is implemented by copying XY into the program counter during the execute
phase.)
Example: B43C would first compare the contents of register 4 with the contents of register 0. If
the two were equal, the pattern 3C would be placed in the program counter so that the next
instruction executed would be the one located at that memory address. Otherwise, nothing would
be done and program execution would continue in its normal sequence.
C 000 HALT execution.
Example: C000 would cause program execution to stop.
