GATE EC 2016 Set 1 — Question 44
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Digital Circuits → Computer Organization → Machine Instructions & Addressing
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Question
In an 8085 system, a PUSH operation requires more clock cycles than a POP operation. Which one of the following options is the correct reason for this?
Correct answer
0.78 to 0.82
Solution
In an 8085 microprocessor, both PUSH and POP operations involve memory access.A PUSH operation involves:
(A) Data transceivers direction: While the direction of data flow changes, this alone does not fully explain the difference in clock cycles for the instruction itself.
(B) Memory write operations are slower than memory read operations: This is a characteristic of memory hardware, but the 8085 instruction's T-states are designed to accommodate these timings. It's a contributing factor but not the fundamental reason for the instruction's internal cycle count difference.
(C) The stack pointer needs to be pre-decremented before writing registers in a PUSH, whereas a POP operation uses the address already in the stack pointer: This accurately describes the SP handling, where PUSH requires an extra decrement step before the write, contributing to more cycles.
(D) Order of registers: The order of bytes (high/low) is fixed and does not involve interchanging registers in a way that explains the cycle difference.Therefore, option (C) provides the most accurate reason related to the stack pointer's operation and the instruction's execution flow.
1.Decrementing the Stack Pointer (SP) by 1.
2.Writing the high-order byte of the register pair to the memory location pointed to by SP.
3.Decrementing the SP by 1 again.
4.Writing the low-order byte of the register pair to the memory location pointed to by SP.
This sequence requires two memory write cycles and involves pre-decrementing the SP before each write.A POP operation involves:1.Reading the low-order byte from the memory location pointed to by SP.
2.Incrementing the SP by 1.
3.Reading the high-order byte from the memory location pointed to by SP.
4.Incrementing the SP by 1 again.
This sequence requires two memory read cycles and involves post-incrementing the SP after each read.The 8085 PUSH instruction typically takes 12 T-states, while the POP instruction takes 10 T-states. The primary reason for this difference lies in the sequence of operations, particularly how the stack pointer is handled. The pre-decrement of the stack pointer in PUSH before writing data adds an operational step that contributes to more clock cycles compared to POP, where the address is already available for the first read.Let's analyze the options:(A) Data transceivers direction: While the direction of data flow changes, this alone does not fully explain the difference in clock cycles for the instruction itself.
(B) Memory write operations are slower than memory read operations: This is a characteristic of memory hardware, but the 8085 instruction's T-states are designed to accommodate these timings. It's a contributing factor but not the fundamental reason for the instruction's internal cycle count difference.
(C) The stack pointer needs to be pre-decremented before writing registers in a PUSH, whereas a POP operation uses the address already in the stack pointer: This accurately describes the SP handling, where PUSH requires an extra decrement step before the write, contributing to more cycles.
(D) Order of registers: The order of bytes (high/low) is fixed and does not involve interchanging registers in a way that explains the cycle difference.Therefore, option (C) provides the most accurate reason related to the stack pointer's operation and the instruction's execution flow.
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