MIPS assembly language

Question

Consider the function with header: int foo(int s, int n, int a, int b).

Write MIPS assembly code for the following statement located inside the above function:

return(s+n);

I was told the answer is

add $v0, $a0, $a1

jr $ra

but I don’t understand why... I know

add $v0, $a0, $a1 means

1. Register $a0 stores the first argument i.e., a0= s
2. & Register $a1 stores the second argument i.e, a1=n
3. add instruction adds the values in $a0 and $a1 and saves the result in register $v0.
4. The value stored in register $v0 is the one to be returned from the function.

but what about the

jr $ra

Answer 1

The jr (Jump Register) instruction causes the PC to jump to the contents of the first source register, in this case - $ra, which holds the return address of the location following where we jumped to the foo function. This has to be placed into ra so that the program knows where to return to after executing the foo function.

Also of note, the v0 register contains the returned value of a subroutine; if the value is 1 word in size only $v0 is significant, otherwise $v1 is as well.