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copied!<p>A.x = 5</p> <p>A.y = 7</p> <p>B.x = 3</p> <p>Calls P(3, 7, Q) - Inside P, y is a ref to B.x, which value is 3, z is 7. Due to dynamic scope, inside P, x is B.x.</p> <p>Inside P, in the first instruction, we have the post-decrement y-- evaluated first, so the evaluated value would be 3 and B.x will change to 2.</p> <p>R is called by name, so the return (w + x++ + y++) will become return ++x + ++y + x++ + y++. The entire first expression is expanded to something equivalent to z = y-- + (++x + ++y + x++ + y++).</p> <p>In the expression ++x + ++y + x++ + y++, we have a pre-increment ++x, which will change B.x to 3 and give 3 as result. The ++y will change B.x to 4 and evaluate 4. The x++ will evaluate to 4 and change to 5 and the y++ will evaluate to 5 and change to 6.</p> <p>So the returned value from Q is 3 + 4 + 4 + 5 = 16. This will be added to the previous z-- value which is 3, so 19 will be assigned to z and B.x will be 6.</p> <p>The write will print x which is due to dynamic scope B.x, so will print 6. The y is a reference to the same B.x, so will print 6. The z is 19. <strong>So it will print 6, 6, 19</strong>.</p> <p>In the third instruction of the P, we have z = R(z++) which due to call by name will expand to something equivalent to z = (z++ + x++ + y++).</p> <p>So we get the z value, which is 19 and increment to 20 (post-increment). Add it to the value of B.x (6) and change B.x to 7. Add the 7 from y and increment y (which is B.x) to 8. So z = 19 + 6 + 7 = 32. B.x is 8.</p> <p>Due to the valueresult of z, the A.y will be changed to 32.</p> <p>Outside P, the write(y++, x++) will get the A.y value, which is 32 and print it. Then A.y will change to 33. B.x which is 8 will be printed and changed to 9. <strong>So it will print 32, 8</strong>.</p> <p>Finally, the A.y will be printed (its value is 33). The A.x will be printed, which is 5. <strong>So it will print 33, 5</strong>.</p> <p>Concluding, the program prints <strong>6, 6, 19, 32, 8, 33, 5</strong>.</p>

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