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3b63b69d2f
The various files are confusingly named and their operation not easy to see. Add a comment to cmplxdivide.c, one of the few C files that will endure in the repository, to explain how to build and run the test. Change-Id: I1fd5c564a14217e1b9815b09bc24cc43c54c096f Reviewed-on: https://go-review.googlesource.com/2850 Reviewed-by: Russ Cox <rsc@golang.org>
95 lines
2.3 KiB
C
95 lines
2.3 KiB
C
// Copyright 2010 The Go Authors. All rights reserved.
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// Use of this source code is governed by a BSD-style
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// license that can be found in the LICENSE file.
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// This C program generates the file cmplxdivide1.go. It uses the
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// output of the operations by C99 as the reference to check
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// the implementation of complex numbers in Go.
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// The generated file, cmplxdivide1.go, is compiled along
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// with the driver cmplxdivide.go (the names are confusing
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// and unimaginative) to run the actual test. This is done by
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// the usual test runner.
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//
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// The file cmplxdivide1.go is checked in to the repository, but
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// if it needs to be regenerated, compile and run this C program
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// like this:
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// gcc '-std=c99' cmplxdivide.c && a.out >cmplxdivide1.go
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#include <complex.h>
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#include <math.h>
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#include <stdio.h>
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#include <string.h>
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#define nelem(x) (sizeof(x)/sizeof((x)[0]))
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double f[] = {
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0,
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1,
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-1,
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2,
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NAN,
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INFINITY,
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-INFINITY,
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};
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char*
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fmt(double g)
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{
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static char buf[10][30];
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static int n;
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char *p;
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p = buf[n++];
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if(n == 10)
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n = 0;
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sprintf(p, "%g", g);
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if(strcmp(p, "-0") == 0)
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strcpy(p, "negzero");
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return p;
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}
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int
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iscnan(double complex d)
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{
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return !isinf(creal(d)) && !isinf(cimag(d)) && (isnan(creal(d)) || isnan(cimag(d)));
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}
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double complex zero; // attempt to hide zero division from gcc
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int
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main(void)
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{
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int i, j, k, l;
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double complex n, d, q;
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printf("// skip\n");
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printf("// # generated by cmplxdivide.c\n");
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printf("\n");
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printf("package main\n");
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printf("var tests = []Test{\n");
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for(i=0; i<nelem(f); i++)
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for(j=0; j<nelem(f); j++)
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for(k=0; k<nelem(f); k++)
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for(l=0; l<nelem(f); l++) {
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n = f[i] + f[j]*I;
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d = f[k] + f[l]*I;
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q = n/d;
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// BUG FIX.
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// Gcc gets the wrong answer for NaN/0 unless both sides are NaN.
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// That is, it treats (NaN+NaN*I)/0 = NaN+NaN*I (a complex NaN)
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// but it then computes (1+NaN*I)/0 = Inf+NaN*I (a complex infinity).
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// Since both numerators are complex NaNs, it seems that the
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// results should agree in kind. Override the gcc computation in this case.
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if(iscnan(n) && d == 0)
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q = (NAN+NAN*I) / zero;
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printf("\tTest{complex(%s, %s), complex(%s, %s), complex(%s, %s)},\n",
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fmt(creal(n)), fmt(cimag(n)),
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fmt(creal(d)), fmt(cimag(d)),
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fmt(creal(q)), fmt(cimag(q)));
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}
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printf("}\n");
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return 0;
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}
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