GATE CS 2006 Q56 — Programming in C

To understand the behavior of the code, let's analyze how pass-by-reference handles different types of arguments (variables vs. expressions):

1. Handling of Arguments in Pass-by-Reference

• Variables (L-values): When a variable like ia is passed, its actual memory address is passed to the subroutine. Any modification to the corresponding parameter inside the subroutine directly affects the original variable.
• Expressions/Constants (R-values): When an expression like ib + 5 is passed, it does not have a persistent memory address. To handle this, the compiler allocates a temporary nameless memory cell, evaluates the expression (which is $8 + 5 = 13$), stores this value in the temporary cell, and passes the address of this temporary cell to the subroutine. This validates statement S1.

2. Step-by-Step Execution Trace

Initially, we have:

• ia = 3
• ib = 8

When call swap(ia, ib + 5) is executed:

1. A temporary cell (let's call it temp) is created with value 13.
2. The address of ia is bound to parameter ix.
3. The address of temp is bound to parameter iy.

Inside the subroutine swap(ix, iy):

• it = ix: The local variable it gets the value of ix (which is ia, i.e., $3$). So, it = 3.
• L1: ix = iy: The variable ix (which is ia) is assigned the value of iy (which is temp, i.e., $13$). Thus, ia becomes $13$.
• L2: iy = it: The variable iy (which is temp) is assigned the value of it (which is $3$). Thus, temp becomes $3$.

3. Final Output

After returning from the subroutine:

• ia has been updated to $13$.
• ib remains unchanged at $8$ (since only the expression ib + 5 was passed using a temporary cell, not ib itself).

Thus, the print statement print *, ia, ib outputs 13 and 8. This validates statement S4.

Therefore, the correct statements are S1 and S4, which corresponds to option B.