In the C programming language and in some others any integer value other than 0 is considered to be a "True", but bit-wise boolean operators such as "&" can provide surprising different results as compared to boolean operators such as: "&&" when used with this definition of truth. The RISCV set conditional instructions are intended to deal with these differences. Please answer, which of these answers you feel is correct:

Consider this C code:
t2=4;
t3=2;
t4=t2 & t3;

And these RISC-V code snippets:

# Code snippet 1
li t2, 4
li t3, 2
and t4, t2, t3

# Code snippet 2
li t2, 4
li t3, 2
and t4, t2, t3
snez t5, t4

# Code snippet 3
li t2, 4
snez t2, t2
li t3, 2
snez t3, t3
and t4, t2, t3

Which code snippet matches the C Program.

Question

In the C programming language and in some others any integer value other than 0 is considered to be a "True", but bit-wise boolean operators such as "&" can provide surprising different results as compared to boolean operators such as: "&&" when used with this definition of truth.  The RISCV set conditional instructions are intended to deal with these differences. Please answer, which of these answers you feel is correct:

Consider this C code:
          t2=4;
          t3=2;
          t4=t2 & t3;

And these RISC-V code snippets:

# Code snippet 1
          li t2, 4
          li t3, 2
          and t4, t2, t3

# Code snippet 2
          li t2, 4
          li t3, 2
          and t4, t2, t3
          snez t5, t4

# Code snippet 3
          li t2, 4
          snez t2, t2
          li t3, 2
          snez t3, t3
          and t4, t2, t3

Which code snippet matches the C Program.

BlackTom AI · Accepted Answer

Let's break down what the C code does versus what each RISCV code snippet does, focusing on the core variable t4 after the operations.

First, the C program:
- t2 = 4; t3 = 2; then t4 = t2 & t3 (bitwise AND).
- Since 4 in binary is 100 and 2 is 010, the bitwise AND 100 & 010 equals 000, which is 0. So, in the C code, t4 ends up being 0.

Now examine Code snippet 1:
- li t2, 4 and li t3, 2 load the same values as the C program.
- and t4, t2, t3 performs a bitwise AND of 4 and 2 with the RISCV instruction format, producing 0 in t4 just like the C program.
- There is no snez (or any other operation) that would alter t4 after the AND, so the result for t4 matches the C code exactly.

Code snippet 2:
- It starts the same as snippet 1 with li t2, 4 and li t3, 2 and then and t4, t2, t3, so t4 is again 0 just as in the C code.
- The additional line snez t5, t4 computes t5 = (t4 != 0). Since t4 is 0, t5 becomes 0. This doesn’t change t4, but it introduces a new register result (t5) that is derived from t4.
- Because the final value of t4 is still 0, snippet 2 reproduces the same t4 value as the C program, though it adds an extra check/side effect on t5.

Code snippet 3:
- li t2, 4 then snez t2, t2 converts t2 to 1 if nonzero, so t2 becomes 1.
- li t3, 2 then snez t3, t3 converts t3 to 1 as well, so t3 becomes 1.
- and t4, t2, t3 then computes 1 & 1, giving t4 = 1.
- This does not match the C program’s t4 value of 0, since the bitwise AND is performed on normalized 1-values rather than the original 4 and 2.

Synthesis and subtlety notes:
- The key discrepancy between the C expression t4 = t2 & t3 and the RISCV snippets is whether the values being ANDed are the literal bit patterns 4 and 2 (as in snippet 1 and snippet 2) or the booleanized 0/1 equivalents produced by snez (as in snippet 3).
- In snippet 2, the snez instruction operates on t4 after the AND, turning 0 into t5 = 0, but it does not modify t4 itself. Therefore, the final t4 value remains identical to the C program, even though there is an extra side effect on t5.
- Snippet 3 shows the effect of converting operands to 0/1 values before the AND, which changes the result of the bitwise operation from 0 (in C) to 1.

In summary, comparing the outcomes relevant to the C program’s t4:
- Snippet 1 yields t4 = 0, matching the C program’s t4.
- Snippet 2 yields t4 = 0 as well, matching the C program’s t4 (though it computes an additional t5 from t4).
- Snippet 3 yields t4 = 1, which does not match the C program’s t4.

CSE-12-02 Quiz 5: ASSEMBLY PROGRAMMING

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