MIPS - How does MIPS allocate memory for arrays in the stack?
I'm quite new to the MIPS assembly language and am currently taking a class on computer architecture which has a large section on MIPS coding. I've studied several other high-level programming languages (C, C#, Python) in the past so have some bases in programming.
My question here specifically asks: How does MIPS allocate memory for arrays in the stack? I'm hoping that answering this question will hopefully give me a better total understanding of MIPS as I'm still a bit lot on conceptualizing the idea of the MIPS language and it's architecture. I don't quite understand how pointers work in this whole regard either...
Would be brilliant if someone could take the time to help out this confused student! :)
Answer
Well.. you should be aware that MIPS, like C, essentially has three different ways of allocating memory.
Consider the following C code:
int arr[2]; //global variable, allocated in the data segment
int main() {
int arr2[2]; //local variable, allocated on the stack
int *arr3 = malloc(sizeof(int) * 2); //local variable, allocated on the heap
}
MIPS assembly supports all these types of data.
To allocate an int array in the data segment you could use:
.data
arr: .word 0, 0 #enough space for two words, initialized to 0, arr label points to the first element
To allocate an int array on the stack you could use:
#save $ra
addi $sp $sp -4 #give 4 bytes to the stack to store the frame pointer
sw $fp 0($sp) #store the old frame pointer
move $fp $sp #exchange the frame and stack pointers
addi $sp $sp -12 #allocate 12 more bytes of storage, 4 for $ra and 8 for our array
sw $ra -4($fp)
# at this point we have allocated space for our array at the address -8($fp)
To allocate space on the heap, a system call is required. In the spim simulator this is system call 9:
li $a0 8 #enough space for two integers
li $v0 9 #syscall 9 (sbrk)
syscall
# address of the allocated space is now in $v0