1/*
2 * Copyright (c) 1999-2008 Mark D. Hill and David A. Wood
3 * All rights reserved.
4 *
5 * Redistribution and use in source and binary forms, with or without
6 * modification, are permitted provided that the following conditions are
7 * met: redistributions of source code must retain the above copyright
8 * notice, this list of conditions and the following disclaimer;

--- 20 unchanged lines hidden (view full) ---

29#include <cmath>
30#include <iomanip>
31
32#include "base/intmath.hh"
33#include "mem/ruby/common/Histogram.hh"
34
35using namespace std;
36
37Histogram::Histogram(int binsize, int bins)
37Histogram::Histogram(int binsize, uint32_t bins)
38{
39 m_binsize = binsize;
40 m_bins = bins;
41 clear();
40 clear(bins);
41}
42
43Histogram::~Histogram()
44{
45}
46
47void
49Histogram::clear(int binsize, int bins)
48Histogram::clear(int binsize, uint32_t bins)
49{
50 m_binsize = binsize;
51 clear(bins);
52}
53
54void
56Histogram::clear(int bins)
55Histogram::clear(uint32_t bins)
56{
58 m_bins = bins;
57 m_largest_bin = 0;
58 m_max = 0;
61 m_data.resize(m_bins);
62 for (int i = 0; i < m_bins; i++) {
59 m_data.resize(bins);
60 for (uint32_t i = 0; i < bins; i++) {
61 m_data[i] = 0;
62 }
63
64 m_count = 0;
65 m_max = 0;
67
66 m_sumSamples = 0;
67 m_sumSquaredSamples = 0;
68}
69
70void
71Histogram::doubleBinSize()
72{
73 assert(m_binsize != -1);
74 uint32_t t_bins = m_data.size();
75
76 for (uint32_t i = 0; i < t_bins/2; i++) {
77 m_data[i] = m_data[i*2] + m_data[i*2 + 1];
78 }
79 for (uint32_t i = t_bins/2; i < t_bins; i++) {
80 m_data[i] = 0;
81 }
82
83 m_binsize *= 2;
84}
85
86void
87Histogram::add(int64 value)
88{
89 assert(value >= 0);
90 m_max = max(m_max, value);
91 m_count++;
92
93 m_sumSamples += value;
94 m_sumSquaredSamples += (value*value);
95
83 int index;
96 uint32_t index;
97
98 if (m_binsize == -1) {
99 // This is a log base 2 histogram
100 if (value == 0) {
101 index = 0;
102 } else {
103 index = floorLog2(value) + 1;
104 if (index >= m_data.size()) {
105 index = m_data.size() - 1;
106 }
107 }
108 } else {
109 // This is a linear histogram
96 while (m_max >= (m_bins * m_binsize)) {
97 for (int i = 0; i < m_bins/2; i++) {
98 m_data[i] = m_data[i*2] + m_data[i*2 + 1];
99 }
100 for (int i = m_bins/2; i < m_bins; i++) {
101 m_data[i] = 0;
102 }
103 m_binsize *= 2;
104 }
110 uint32_t t_bins = m_data.size();
111
112 while (m_max >= (t_bins * m_binsize)) doubleBinSize();
113 index = value/m_binsize;
114 }
107 assert(index >= 0);
115
116 assert(index < m_data.size());
117 m_data[index]++;
118 m_largest_bin = max(m_largest_bin, index);
119}
120
121void
113Histogram::add(const Histogram& hist)
122Histogram::add(Histogram& hist)
123{
115 assert(hist.getBins() == m_bins);
116 assert(hist.getBinSize() == -1); // assume log histogram
117 assert(m_binsize == -1);
124 uint32_t t_bins = m_data.size();
125
119 for (int j = 0; j < hist.getData(0); j++) {
120 add(0);
126 if (hist.getBins() != t_bins) {
127 fatal("Histograms with different number of bins cannot be combined!");
128 }
129
123 for (int i = 1; i < m_bins; i++) {
124 for (int j = 0; j < hist.getData(i); j++) {
125 add(1<<(i-1)); // account for the + 1 index
130 m_max = max(m_max, hist.getMax());
131 m_count += hist.size();
132 m_sumSamples += hist.getTotal();
133 m_sumSquaredSamples += hist.getSquaredTotal();
134
135 // Both histograms are log base 2.
136 if (hist.getBinSize() == -1 && m_binsize == -1) {
137 for (int j = 0; j < hist.getData(0); j++) {
138 add(0);
139 }
140
141 for (uint32_t i = 1; i < t_bins; i++) {
142 for (int j = 0; j < hist.getData(i); j++) {
143 add(1<<(i-1)); // account for the + 1 index
144 }
145 }
146 } else if (hist.getBinSize() >= 1 && m_binsize >= 1) {
147 // Both the histogram are linear.
148 // We are assuming that the two histograms have the same
149 // minimum value that they can store.
150
151 while (m_binsize > hist.getBinSize()) hist.doubleBinSize();
152 while (hist.getBinSize() > m_binsize) doubleBinSize();
153
154 assert(m_binsize == hist.getBinSize());
155
156 for (uint32_t i = 0; i < t_bins; i++) {
157 m_data[i] += hist.getData(i);
158
159 if (m_data[i] > 0) m_largest_bin = i;
160 }
161 } else {
162 fatal("Don't know how to combine log and linear histograms!");
163 }
164}
165
166// Computation of standard deviation of samples a1, a2, ... aN
167// variance = [SUM {ai^2} - (SUM {ai})^2/N]/(N-1)
168// std deviation equals square root of variance
169double
170Histogram::getStandardDeviation() const

--- 37 unchanged lines hidden (view full) ---

208 if (m_count == 0) {
209 out << "average: NaN |";
210 out << "standard deviation: NaN |";
211 } else {
212 out << "average: " << setw(5) << ((double) m_sumSamples)/m_count
213 << " | ";
214 out << "standard deviation: " << getStandardDeviation() << " |";
215 }
180 for (int i = 0; i < m_bins && i <= m_largest_bin; i++) {
216
217 for (uint32_t i = 0; i <= m_largest_bin; i++) {
218 if (multiplier == 1.0) {
219 out << " " << m_data[i];
220 } else {
221 out << " " << double(m_data[i]) * multiplier;
222 }
223 }
224 out << " ]";
225}
226
227bool
228node_less_then_eq(const Histogram* n1, const Histogram* n2)
229{
230 return (n1->size() > n2->size());
231}