Prolog: base conversion transformation

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Here is some C code to do a base conversion of the integer 156.

Here is the C code.

#include <stdio.h>

int main() {
	int x1 = 156;
	printf("binary digits for %d (in r , reverse order)\n",x1);
	if (x1 == 0) {
		printf("0\n");
		}
	else {
		while (x1 != 0) {
			int r1 = x1 % 2;
			int q1 = x1 / 2;
			printf("x=%d/2 is q=%d , r=%d\n",x1,q1,r1);
			x1 = q1;
			}
		}
	return 0;
	}

Here is the output of the C code.

binary digits for 156 (in r , reverse order)
x=156/2 is q=78 , r=0
x=78/2 is q=39 , r=0
x=39/2 is q=19 , r=1
x=19/2 is q=9 , r=1
x=9/2 is q=4 , r=1
x=4/2 is q=2 , r=0
x=2/2 is q=1 , r=0
x=1/2 is q=0 , r=1

In the following Prolog program, the above C code is encoded into the term structure of the simple Prolog compiler.

Here is the Prolog code.

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

testCode(

sequence([

declare("x1",156),

printf("binary digits for %d (in r , reverse order)\n","x1"),

ifthen("x1 == 0",

printf("0"),

while("x1 != 0", sequence([

declare("r1", "x1 & 2"),
declare("q1", "x1 / 2"),

printf("x=%d/2 is q=%d , r=%d\n","x1,q1,r1"),

assign("x1","q1")
]))
)
])
).

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),

	fail.

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

:- solve(go).

The output is a C program that converts the while and if statements in the encoded syntax tree of the original C program to goto statements and outputs the resulting C program.

Here is the output of the Prolog code.

#include <stdio.h>;\n
\n
int main() {\n
int x1 = 156;\n
printf("binary digits for %d (in r , reverse order)\n",x1);
if (! (x1 == 0)) goto L1;\n
printf("0");
goto L2;\n
L1:\n
L3:\n
if (! (x1 != 0)) goto L4;\n
int r1 = x1 & 2;\n
int q1 = x1 / 2;\n
printf("x=%d/2 is q=%d , r=%d\n",x1,q1,r1);
x1 = q1;\n
goto L3;\n
L4:\n
L2:\n
return 0;\n
}\n

Here is the output of the above Prolog program that is now run as a C program.

Here is the C code.

#include <stdio.h>

int main() {
	int x1 = 156;
	printf("binary digits for %d (in r , reverse order)\n",x1);
	if (! (x1 == 0)) goto L1;
	printf("0");
	goto L2;
L1:
L3:
	if (! (x1 != 0)) goto L4;
	int r1 = x1 & 2;
	int q1 = x1 / 2;
	printf("x=%d/2 is q=%d , r=%d\n",x1,q1,r1);
	x1 = q1;
	goto L3;
L4:
L2:
	return 0;
	}

The output is the same as the original C program.

Here is the output of the C code.

binary digits for 156 (in r , reverse order)
x=156/2 is q=78 , r=0
x=78/2 is q=39 , r=2
x=39/2 is q=19 , r=2
x=19/2 is q=9 , r=2
x=9/2 is q=4 , r=0
x=4/2 is q=2 , r=0
x=2/2 is q=1 , r=2
x=1/2 is q=0 , r=0

Here is the C code.

#include <stdio.h>

int main() {
	int x1 = 156;
	printf("binary digits for %d (in r , reverse order)\n",x1);
	if (x1 == 0) {
		printf("0\n");
		}
	else {
		while (x1 != 0) {
			int r1 = x1 % 2;
			int q1 = x1 / 2;
			printf("x=%d/2 is q=%d , r=%d\n",x1,q1,r1);
			x1 = q1;
			}
		}
	return 0;
	}

Compare the above code using goto statements with the original code not using goto statements. Which is more concise? Which is more clear? Which might be considered better?