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MIPS 32 ASM to C code, did i do this right? |
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| Sep13-06, 07:24 PM | #1 |
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MIPS 32 ASM to C code, did i do this right?
have the following MIPS 32 asm code:
Assuming B is an array of 10 words whose base address is in register $s0, andvariable c and i are in $s1 and $s2, respectively. What is the C statement implemented by the below MIPS assembly code? add $t0, $s2, $s2 add $t0, $t0, $t0 lw $t1, 0($s0) lw $t2, 8($s0) sub $s1, $t2, $t1 In C what does it look like? i said c = B[i+2] - B[i] but the other optpoins are: c = B[i] - B[i+2] c = B[i] - B[i+8] c = B[i+8] - B[i] Can someone see if i'm correct? lw $t2,8($s0); I said was really index B[2] because, in mips memory is set up by 4's. meaning If i have memory set up like this: [0x10ff00ff] 32 [0x00ff00ff] 28 [0x0000ffff] 24 [0xffff0000] 20 [0x000011ff]16 [0x10ffff00] 12 [0xff00ffff] 8 [0xff0000ff] 4 [0x0000ff44] 0 32 bit data and the numbers are word address I'm confused on what the beginning codes prupose is for, the 2 add statements. It basically does this i think: $t0 = i + i; $t0 = (i+i) + (i+i)? |
| Sep13-06, 11:40 PM | #2 |
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Yep it was correct
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