Answer: B, 2.
Answer analysis:
To answer this question, we need to analyze the execution flow and conditional jump logic of the program.
First, the program starts executing from the LAB1 label and then sets the value of register R1 to 1 (SET R1,1). Then, the program jumps to the LAB2 label to continue execution.
In LAB2, the program first executes two MOVJ instructions (these instructions may be used for moving joints in a robot or some control system, but in the context of this question, their specific execution content has no effect on the number of loops). Then, the value of R1 is increased by 1 (INC R1).
After that, the program checks whether the value of R1 is less than 4 (JUMP LAB2, IF R1<4). If the condition is true (that is, the value of R1 is less than 4), the program jumps back to LAB2 to continue execution; if the condition is false (that is, the value of R1 is not less than 4), the program continues to execute the code after the JUMP instruction (but in this question, the JUMP instruction is the last instruction of the program, so in fact the program will stop when the condition is false).
Now, let's analyze the number of loops based on the value of R1:
1. Initially, R1 is set to 1.
2. After entering LAB2 for the first time, R1 becomes 2, which is less than 4, so it jumps back to LAB2.
3. After entering LAB2 for the second time, R1 becomes 3, which is less than 4, so it jumps back to LAB2 again.
4. After entering LAB2 for the third time, R1 becomes 4, which is not less than 4, so it does not jump back to LAB2 and the loop ends.
Therefore, the program repeats (i.e. loops) 2 times (counting starts from the second entry into LAB2, because the first entry is a normal entry after the initial setting, which is not considered a "repeat"). But according to the question asking "how many times the program repeats", we are actually counting how many times R1 jumps back before it increases from 1 to 4 (excluding 4), in addition to the first normal entry into LAB2. If calculated in this way, it is 2 times.
So the correct answer is B, 2.