Prolog: control flow compiler

Here we look at a small Prolog compiler for control flow.

This small example shows how Prolog can be used to prototype various aspects of control flow in a compiler that generates code. Here is the Prolog code.

prints([]):- nl.
prints([X|Rest]):- print(X), prints(Rest).

testCode(sequence(
[

declare("x", 2),
declare("y", 3),
declare("z", 5)
]
)).

testCode(ifthen("x == 2", assign("y","2*y"))).

testCode(ifthen("z == 1", assign("x","z"),assign("x","2*z"))).

testCode(sequence(
[

assign("x", 0),

while("x != 10",

sequence([

assign("x","x+1"),
assign("z","z+1")
])
)
]
)).

nextLabel(N1):-
	retract(labelPos(N)),
	N1 is N + 1,
	assert(labelPos(N1)).

goto(L):-
	prints(['goto L',L,';\n']).

ifnotgoto(E,L):-
	prints(['if (! (',E,')) goto L',L,';\n']).

label(L):-
	prints(['L',L,':\n']).

sequence([]).
sequence([Head|Tail]):-
	Head,
	sequence(Tail).

printf(S):-
	prints(['printf("', S, '");']).

printf(S,Xs):-
	prints(['printf("', S, '",', Xs, ');']).

assign(V,E):-
	prints([V,' = ',E,';\n']).

declare(V):-
	prints(['int ',V,';\n']).

declare(V,E):-
	prints(['int ',V,' = ',E,';\n']).

while(B,S) :-
	nextLabel(L1),
	nextLabel(L2),
	label(L1),
	ifnotgoto(B,L2),
	S,
	goto(L1),
	label(L2).

ifthen(B,S1):-
	nextLabel(L1),
	ifnotgoto(B,L1),
	S1,
	label(L1).

ifthen(B,S1,S2):-
	nextLabel(L1),
	nextLabel(L2),
	ifnotgoto(B,L1),
	S1,
	goto(L2),
	label(L1),
	S2,
	label(L2).

run(Code):- Code, !.

go:-
	prints(['#include <stdio.h>;\n']),
	prints(['\n']),
	prints(['int main() {\n']),
	assert(labelPos(0)),
	testCode(Code),
	run(Code),
	run(printf("x=%d y=%d z=%d\n","x,y,z")),
	fail.

go:-
	prints(['return 0;\n']),
	prints(['}\n']).

:- solve(go).

Here is the output of the Prolog code.

#include <stdio.h>;\n
\n
int main() {\n
int x = 2;\n
int y = 3;\n
int z = 5;\n
printf("x=%d y=%d z=%d\n",x,y,z);
if (! (x == 2)) goto L1;\n
y = 2*y;\n
L1:\n
printf("x=%d y=%d z=%d\n",x,y,z);
if (! (z == 1)) goto L2;\n
x = z;\n
goto L3;\n
L2:\n
x = 2*z;\n
L3:\n
printf("x=%d y=%d z=%d\n",x,y,z);
x = 0;\n
L4:\n
if (! (x != 10)) goto L5;\n
x = x+1;\n
z = z+1;\n
goto L4;\n
L5:\n
printf("x=%d y=%d z=%d\n",x,y,z);
return 0;\n
}\n

Here is the C code.

#include <stdio.h>

#include <stdio.h>

int main() {
	int x = 2;
	int y = 3;
	int z = 5;
	printf("x=%d y=%d z=%d\n",x,y,z);
	if (! (x == 2)) goto L1;
	y = 2*y;
L1:
	printf("x=%d y=%d z=%d\n",x,y,z);
	if (! (z == 1)) goto L2;
	x = z;
	goto L3;
L2:
	x = 2*z;
L3:
	printf("x=%d y=%d z=%d\n",x,y,z);
	x = 0;
L4:
	if (! (x != 10)) goto L5;
	x = x+1;
	z = z+1;
	goto L4;
L5:
	printf("x=%d y=%d z=%d\n",x,y,z);
	return 0;
	}

Here is the output of the C code.

x=2 y=3 z=5
x=2 y=6 z=5
x=10 y=6 z=5
x=10 y=6 z=15