misunderestanding of logical operator of c language

I have such these questions in my book that I have do it for checking my self but I don't know the correct answer and there is no any ex开发者_Go百科planation.

it has two parts "a" and "b"

a) A = B & C[0]
b) A = A ? B : C[0]

C[0]  = 0x00001234
A = 0x00000000
B = 0x00002222

it want from me that

1)what is the result of A 
2) the MIPS instruction for each part
3)show the bit level representation of each in structure.

thanks

& is a bit-wise AND. Each bit of the output is 1 if both corresponding input bits are 1:

0x00001234 = .... 0001 0010 0011 0100
0x00002222 = .... 0010 0010 0010 0010

AND result = .... 0000 0010 0010 0000 = 0x00000220

I don't know MIPS but the instruction is probably just called and.

? is the ternary operator: your example means "if A is a true boolean expression then return B else return C[0]". Generally in C false is zero and true non-zero (I'm not 100% sure if this is standardised - I don't think it was in the earlier standards, but it's common uses) so

• if A = 0x0000 then (bool)A == false and the result is C[0].
• if A = 0x0220 (carrying forward the result from part a) then (bool)A == true and the result is B.

& is the bit-wise AND operator in C. eg. 0000 & 0001 = 0000

? is another way of writing if

A = A ? B : C[0]

says that A gets the value = (if A is true then B ELSE C[0])

Convert for Hex to Binary and do the exercise.

edit : clarify : made "AND" to -> "bit-wise AND"

the mips:

a) A = B & C[0]

$t1 = C

$t2 = B

$t3 = A

lw $t0, 0($t1) ;move c into temp reg

and $t3, $t0, $t2 ;perform bitwise and

b) A = A ? B : C[0]

$t1 = C

$t2 = B

$t3 = A

beq $0, $t3, 3 ;if A is 0 go 3 lines down

add $t3, $0, $t2 ;move B to A

j end ;goto the end

lw $t0, 0($t1) ;load c into a temp reg

add $t3, $0, $t0 ;move temp value into a

end: