Why is the code adding 7 if the number is not >= 0

Question

I've got this program in MIPS assembly which comes from a C code that does the simple average of the eigth arguments of the function.

average8:
    addu $4,$4,$5
    addu $4,$4,$6
    addu $4,$4,$7
    lw $2,16($sp)
    #nop
    addu $4,$4,$2
    lw $2,20($sp)
    #nop
    addu $4,$4,$2
    lw $2,24($sp)
    #nop
    addu $4,$4,$2
    lw $2,28($sp)
    #nop
    addu $2,$4,$2
    bgez $2,$L2
    addu $2,$2,7
$L2:
    sra $2,$2,3
    j $31

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When the number is positve, we directly divided by 8 (shift by 3 bits), but when the number is negative, we first addu 7 then do the division.

My question is why do we add 7 to $2 when $2 is not >= 0 ?

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EDIT : Here is the C code :

int average8(int x1, int x2, int x3, int x4, int x5, int x6, int x7, int x8)
{
    return (x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8) / 8;
}

note : the possible loss in the division since we are using ints instead of floats or doubles is not important in this case.

Answer 1

The difference appears to be accounting for the different behaviors of / 8 and >> 3 when negative numbers are involved:

int main() {
    printf("%d\n", (-50) / 8);
    printf("%d\n", (-50) >> 3);
    printf("%d\n", (-50 + 7) >> 3);
}

gives

-6
-7
-6

So, the compiler wants to use the >> 3 optimization, but it's not exactly the same as / 8, so it adds some code to correct for it.