Computer-Organization
August 1, 2024Machine-Instructions
August 1, 2024Machine-Instructions
| Question 47 |
In a 10 bit computer instruction format, the size of address field is 3 bits. Thecomputer uses expanding OPcode technique and has 4 two address instructions and 16
one address instructions. The number of zero address instructions it can support is
one address instructions. The number of zero address instructions it can support is
| 256 | |
| 356 | |
| 640 | |
| 56 |
Question 47 Explanation:
No. of possible instruction encoding =2 10 =1024
No. of encoding taken by two-address instructions =4×2 3 ×2 3 =256
No. of encoding taken by one-address instructions =16×2 3 =128
So, no. of possible zero-address instructions =1024−(256+128)=640
No. of encoding taken by two-address instructions =4×2 3 ×2 3 =256
No. of encoding taken by one-address instructions =16×2 3 =128
So, no. of possible zero-address instructions =1024−(256+128)=640
Correct Answer: C
Question 47 Explanation:
No. of possible instruction encoding =2 10 =1024
No. of encoding taken by two-address instructions =4×2 3 ×2 3 =256
No. of encoding taken by one-address instructions =16×2 3 =128
So, no. of possible zero-address instructions =1024−(256+128)=640
No. of encoding taken by two-address instructions =4×2 3 ×2 3 =256
No. of encoding taken by one-address instructions =16×2 3 =128
So, no. of possible zero-address instructions =1024−(256+128)=640
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