It's about time that I met my promise of actually writing a compiler. So in this part of the journey we are going to replace the interpreter in our program with code that generates x86-64 assembly code.
Before we do, it will be worthwhile to revisit the interpreter code
in interp.c:
int interpretAST(struct ASTnode *n) {
int leftval, rightval;
if (n->left) leftval = interpretAST(n->left);
if (n->right) rightval = interpretAST(n->right);
switch (n->op) {
case A_ADD: return (leftval + rightval);
case A_SUBTRACT: return (leftval - rightval);
case A_MULTIPLY: return (leftval * rightval);
case A_DIVIDE: return (leftval / rightval);
case A_INTLIT: return (n->intvalue);
default:
fprintf(stderr, "Unknown AST operator %d\n", n->op);
exit(1);
}
}The interpretAST() function walks the given AST tree depth-first.
It evaluates any left sub-tree, then the right sub-tree. Finally, it
uses the op value at the base of the current tree to operate on
these children.
If the op value is one of the four maths operators, then this maths
operation is performed. If the op value indicates that the node
is simply an integer literal, the literal value is return.
The function returns the final value for this tree. And, as it is recursive, it will calculate the final value for a whole tree one sub-sub-tree at a time.
We are going to write an assembly code generator which is generic. This is, in turn, going to call out to a set of CPU-specific code generation functions.
Here is the generic assembly code generator in gen.c:
// Given an AST, generate
// assembly code recursively
static int genAST(struct ASTnode *n) {
int leftreg, rightreg;
// Get the left and right sub-tree values
if (n->left) leftreg = genAST(n->left);
if (n->right) rightreg = genAST(n->right);
switch (n->op) {
case A_ADD: return (cgadd(leftreg,rightreg));
case A_SUBTRACT: return (cgsub(leftreg,rightreg));
case A_MULTIPLY: return (cgmul(leftreg,rightreg));
case A_DIVIDE: return (cgdiv(leftreg,rightreg));
case A_INTLIT: return (cgload(n->intvalue));
default:
fprintf(stderr, "Unknown AST operator %d\n", n->op);
exit(1);
}
}Looks familar, huh?! We are doing the same depth-first tree traversal. This time:
- A_INTLIT: load a register with the literal value
- Other operators: perform a maths function on the two registers that hold the left-child's and right-child's value
Instead of passing values, the code in genAST() passes around
register identifiers. For example cgload() loads a value into a register and
returns the identity of the register with the loaded value.
genAST() itself returns the identity of the register that holds the final
value of the tree at this point. That's why the code at the top is
getting register identities:
if (n->left) leftreg = genAST(n->left);
if (n->right) rightreg = genAST(n->right);genAST() is only going to calculate the value of the expression given to
it. We need to print out this final calculation. We're also going to need
to wrap the assembly code we generate with some leading code (the
preamble) and some trailing code (the postamble). This is done with
the other function in gen.c:
void generatecode(struct ASTnode *n) {
int reg;
cgpreamble();
reg= genAST(n);
cgprintint(reg); // Print the register with the result as an int
cgpostamble();
}That's the generic code generator out of the road. Now we need to look
at the generation of some real assembly code. For now, I'm targetting
the x86-64 CPU as this is still one of the most common Linux platforms.
So, open up cg.c and let's get browsing.
Any CPU has a limited number of registers. We will have to allocate a register to hold the integer literal values, plus any calculation that we perform on them. However, once we've used a value, we can often discard the value and hence free up the register holding it. Then we can re-use that register for another value.
There are three functions that deal with register allocation:
freeall_registers(): Set all registers as availablealloc_register(): Allocate a free registerfree_register(): Free an allocated register
I'm not going to go through the code as it's straight forward but with some error checking. Right now, if I run out of registers then the program will crash. Later on, I'll deal with the situation when we have run out of free registers.
The code works on generic registers: r0, r1, r2 and r3. There is a table of strings with the actual register names:
static char *reglist[4]= { "%r8", "%r9", "%r10", "%r11" };This makes these functions fairly independent of the CPU architecture.
This is done in cgload(): a register is allocated, then a movq
instruction loads a literal value into the allocated register.
// Load an integer literal value into a register.
// Return the number of the register
int cgload(int value) {
// Get a new register
int r= alloc_register();
// Print out the code to initialise it
fprintf(Outfile, "\tmovq\t$%d, %s\n", value, reglist[r]);
return(r);
}cgadd() takes two register numbers and generates the code to add
them together. The result is saved in one of the two registers,
and the other one is then freed for future use:
// Add two registers together and return
// the number of the register with the result
int cgadd(int r1, int r2) {
fprintf(Outfile, "\taddq\t%s, %s\n", reglist[r1], reglist[r2]);
free_register(r1);
return(r2);
}Note that addition is commutative, so I could have added r2 to r1
instead of r1 to r2. The identity of the register with the final
value is returned.
This is very similar to addition, and again the operation is commutative, so any register can be returned:
// Multiply two registers together and return
// the number of the register with the result
int cgmul(int r1, int r2) {
fprintf(Outfile, "\timulq\t%s, %s\n", reglist[r1], reglist[r2]);
free_register(r1);
return(r2);
}Subtraction is not commutative: we have to get the order correct. The second register is subtracted from the first, so we return the first and free the second:
// Subtract the second register from the first and
// return the number of the register with the result
int cgsub(int r1, int r2) {
fprintf(Outfile, "\tsubq\t%s, %s\n", reglist[r2], reglist[r1]);
free_register(r2);
return(r1);
}Division is also not commutative, so the previous notes apply. On
the x86-64, it's even more complicated. We need to load %rax
with the dividend from r1. This needs to be extended to eight
bytes with cqo. Then, idivq will divide %rax with the divisor
in r2, leaving the quotient in %rax, so we need to copy it
out to either r1 or r2. Then we can free the other register.
// Divide the first register by the second and
// return the number of the register with the result
int cgdiv(int r1, int r2) {
fprintf(Outfile, "\tmovq\t%s,%%rax\n", reglist[r1]);
fprintf(Outfile, "\tcqo\n");
fprintf(Outfile, "\tidivq\t%s\n", reglist[r2]);
fprintf(Outfile, "\tmovq\t%%rax,%s\n", reglist[r1]);
free_register(r2);
return(r1);
}There isn't an x86-64 instruction to print a register out as a decimal
number. To solve this problem, the assembly preamble contains a function
called printint() that takes a register argument and calls printf()
to print this out in decimal.
I'm not going to give the code in cgpreamble(), but it also contains
the beginning code for main(), so that we can assemble our output file
to get a complete program. The code for cgpostamble(), also not given
here, simply calls exit(0) to end the program.
Here, however, is cgprintint():
void cgprintint(int r) {
fprintf(Outfile, "\tmovq\t%s, %%rdi\n", reglist[r]);
fprintf(Outfile, "\tcall\tprintint\n");
free_register(r);
}Linux x86-64 expects the first argument to a function to be in the %rdi
register, so we move our register into %rdi before we call printint.
That's about it for the x86-64 code generator. There is some extra code
in main() to open out out.s as our output file. I've also left the
interpreter in the program so we can confirm that our assembly calculates
the same answer for the input expression as the interpreter.
Let's make the compiler and run it on input01:
$ make
cc -o comp1 -g cg.c expr.c gen.c interp.c main.c scan.c tree.c
$ make test
./comp1 input01
15
cc -o out out.s
./out
15Yes! The first 15 is the interpreter's output. The second 15 is the assembly's output.
So, exactly what was the assembly output? Well, here is the input file:
2 + 3 * 5 - 8 / 3
and here is out.s for this input with comments:
.text # Preamble code
.LC0:
.string "%d\n" # "%d\n" for printf()
printint:
pushq %rbp
movq %rsp, %rbp # Set the frame pointer
subq $16, %rsp
movl %edi, -4(%rbp)
movl -4(%rbp), %eax # Get the printint() argument
movl %eax, %esi
leaq .LC0(%rip), %rdi # Get the pointer to "%d\n"
movl $0, %eax
call printf@PLT # Call printf()
nop
leave # and return
ret
.globl main
.type main, @function
main:
pushq %rbp
movq %rsp, %rbp # Set the frame pointer
# End of preamble code
movq $2, %r8 # %r8 = 2
movq $3, %r9 # %r9 = 3
movq $5, %r10 # %r10 = 5
imulq %r9, %r10 # %r10 = 3 * 5 = 15
addq %r8, %r10 # %r10 = 2 + 15 = 17
# %r8 and %r9 are now free again
movq $8, %r8 # %r8 = 8
movq $3, %r9 # %r9 = 3
movq %r8,%rax
cqo # Load dividend %rax with 8
idivq %r9 # Divide by 3
movq %rax,%r8 # Store quotient in %r8, i.e. 2
subq %r8, %r10 # %r10 = 17 - 2 = 15
movq %r10, %rdi # Copy 15 into %rdi in preparation
call printint # to call printint()
movl $0, %eax # Postamble: call exit(0)
popq %rbp
ret
Excellent! We now have a legitimate compiler: a program that takes an input in one language and generates a translation of that input in another language.
We still have to then assemble the output down to machine code and link it with the support libraries, but this is something that we can perform manually for now. Later on, we will write some code to do this automatically.
Changing from the interpreter to a generic code generator was trivial, but then
we had to write some code to generate real assembly output. To do this,
we had to think about how to allocate registers: for now, we have a naive
solution. We also had to deal with some x86-64 oddities like the idivq
instruction.
Something I haven't touched on yet is: why bother with generating the AST for
an expression? Surely, we could have called cgadd() when we hit a '+'
token in our Pratt parser, ditto for the other operators. I'm going to
leave you to think about this, but I will come back to it in a step or
two.
In the next part of our compiler writing journey, we will add some statements to our language, so that it starts to resemble a proper computer language. Next step