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// Copyright 2010 The Go Authors. All rights reserved.
// Use of this source code is governed by a BSD-style
// license that can be found in the LICENSE file.
// gcc '-std=c99' cmplxdivide.c && a.out >cmplxdivide1.go
#include <complex.h>
#include <math.h>
#include <stdio.h>
#include <string.h>
#define nelem(x) (sizeof(x)/sizeof((x)[0]))
double f[] = {
0,
1,
-1,
2,
NAN,
INFINITY,
-INFINITY,
};
char*
fmt(double g)
{
static char buf[10][30];
static int n;
char *p;
p = buf[n++];
if(n == 10)
n = 0;
sprintf(p, "%g", g);
if(strcmp(p, "-0") == 0)
strcpy(p, "negzero");
return p;
}
int
iscnan(double complex d)
{
return !isinf(creal(d)) && !isinf(cimag(d)) && (isnan(creal(d)) || isnan(cimag(d)));
}
double complex zero; // attempt to hide zero division from gcc
int
main(void)
{
int i, j, k, l;
double complex n, d, q;
printf("// skip\n");
printf("// # generated by cmplxdivide.c\n");
printf("\n");
printf("package main\n");
printf("var tests = []Test{\n");
for(i=0; i<nelem(f); i++)
for(j=0; j<nelem(f); j++)
for(k=0; k<nelem(f); k++)
for(l=0; l<nelem(f); l++) {
n = f[i] + f[j]*I;
d = f[k] + f[l]*I;
q = n/d;
// BUG FIX.
// Gcc gets the wrong answer for NaN/0 unless both sides are NaN.
// That is, it treats (NaN+NaN*I)/0 = NaN+NaN*I (a complex NaN)
// but it then computes (1+NaN*I)/0 = Inf+NaN*I (a complex infinity).
// Since both numerators are complex NaNs, it seems that the
// results should agree in kind. Override the gcc computation in this case.
if(iscnan(n) && d == 0)
q = (NAN+NAN*I) / zero;
printf("\tTest{complex(%s, %s), complex(%s, %s), complex(%s, %s)},\n",
fmt(creal(n)), fmt(cimag(n)),
fmt(creal(d)), fmt(cimag(d)),
fmt(creal(q)), fmt(cimag(q)));
}
printf("}\n");
return 0;
}
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