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The following code segment is executed on a processor which allows only register operands in its instructions. Each instruction can have atmost two source operands and one destination operand. Assume that all variables are dead after this code segment.

c = a + b;
d = c * a;
e = c + a;
x = c * c;
if (x > a) {
y = a * a;
}
else {
d = d * d;
e = e * e;
}

What is the minimum number of registers needed in the instruction set architecture of the processor to compile this code segment without any spill to memory? Do not apply any optimization other than optimizing register allocation.

1. 3
2. 4
3. 5
4. 6
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this is still in syllabus asked in 2017
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If there are not enough registers to hold all the variables, some variables may be moved to and from RAM. This process is called "spilling" the registers. Over the lifetime of a program, a variable can be both spilled and stored in registers: this variable is then considered as "split".

Here, we are told not to do code motion. So, we start with 3 registers

c = a + b; //a, c in register
d = c * a; //a, c, d in register
e = c + a; //a, c, e in register, d spilled.


So, now we try with 4 registers

c = a + b; //a, c in register
d = c * a; //a, c, d in register
e = c + a; //a, c, d, e in register
x = c * c; //a, x, d, e in register
if (x > a) {
y = a * a;
}
else {
d = d * d;
e = e * e;
} 

No spilling. So, 4 is the minimum number of registers needed for avoiding spilling. (If code motion was allowed, we need only 3 registers for avoiding spilling).

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$\text{sir , why you have not considered registers(?) below the code }$

if (x > a) {
y = a * a;
}
else {
d = d * d;
e = e * e;
} 
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Thanks sir  for this wonderful explanation.

Here 4 registers are needed

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