[Tinycc-devel] libtcc bug when evaluating math expressions

[David Wallin]

Hi

I suspect there is a bug somewhere regarding math evaluations in 
libtcc.

Evaluating the following : ((-1.0 * x) > v * fabs(v))
when x &v are declared as:
double int x = -0.65;
double int v = 0.4;

gives different answers if evaluated by gcc or if I dynamically 
compile it using libtcc.

ans1 = 0.000000 (libtcc)
ans2 = 1.000000 (gcc)

The following program shows this, and it has been tried with both 
0.9.19 and 0.9.20:


8<---------------------------------------BEGIN

#include <libtcc.h>
#include <math.h>

//#define TCC19
#define TCC20

int main() {

  char *eval_me =   "double evaluate_me(double x, double v) { return 
(double) ((-1.0 * x) > v * fabs(v)); }\0" ;

  double x = -0.65 ;
  double v = 0.4 ;

  double ans1, ans2, ans3;

#ifdef TCC20
  double (*evaluate_me)(double, double);
  unsigned long val;
#endif

  TCCState *tcc_state;
  tcc_state = tcc_new();
  tcc_set_output_type(tcc_state, TCC_OUTPUT_MEMORY);
  tcc_add_library(tcc_state, "m");

  tcc_compile_string(tcc_state, eval_me);

  tcc_relocate(tcc_state);

#ifdef TCC19
    double (*evaluate_me)(double, double) = (double (*)(double, 
double))tcc_get_symbol(tcc_state, "evaluate_me");
#endif
#ifdef TCC20
    tcc_get_symbol(tcc_state, &val, "evaluate_me");
    evaluate_me = (void *)val;
#endif

  ans1 = evaluate_me(x,v) ;
  ans2 = ((-1.0 * x) > v * fabs(v)) ;

  if(ans1 != ans2) {

    printf("x=%f; v=%f\n", x, v);

    printf("ans1 = %f\n", ans1);
    printf("ans2 = %f\n", ans2);
  }
  return 0;
}

8<-----------------------------------------------------END


cheers,

--david