1 /* |
1 /* |
2 * Copyright (c) 2005, 2010, Oracle and/or its affiliates. All rights reserved. |
2 * Copyright (c) 2005, 2012, Oracle and/or its affiliates. All rights reserved. |
3 * DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER. |
3 * DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER. |
4 * |
4 * |
5 * This code is free software; you can redistribute it and/or modify it |
5 * This code is free software; you can redistribute it and/or modify it |
6 * under the terms of the GNU General Public License version 2 only, as |
6 * under the terms of the GNU General Public License version 2 only, as |
7 * published by the Free Software Foundation. |
7 * published by the Free Software Foundation. |
176 if (res != (uintptr_t)NoBits) { |
176 if (res != (uintptr_t)NoBits) { |
177 // find the position of the 1-bit |
177 // find the position of the 1-bit |
178 for (; !(res & 1); res_offset++) { |
178 for (; !(res & 1); res_offset++) { |
179 res = res >> 1; |
179 res = res >> 1; |
180 } |
180 } |
181 assert(res_offset >= l_offset && |
181 |
182 res_offset < r_offset, "just checking"); |
182 #ifdef ASSERT |
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183 // In the following assert, if r_offset is not bitamp word aligned, |
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184 // checking that res_offset is strictly less than r_offset is too |
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185 // strong and will trip the assert. |
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186 // |
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187 // Consider the case where l_offset is bit 15 and r_offset is bit 17 |
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188 // of the same map word, and where bits [15:16:17:18] == [00:00:00:01]. |
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189 // All the bits in the range [l_offset:r_offset) are 0. |
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190 // The loop that calculates res_offset, above, would yield the offset |
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191 // of bit 18 because it's in the same map word as l_offset and there |
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192 // is a set bit in that map word above l_offset (i.e. res != NoBits). |
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193 // |
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194 // In this case, however, we can assert is that res_offset is strictly |
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195 // less than size() since we know that there is at least one set bit |
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196 // at an offset above, but in the same map word as, r_offset. |
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197 // Otherwise, if r_offset is word aligned then it will not be in the |
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198 // same map word as l_offset (unless it equals l_offset). So either |
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199 // there won't be a set bit between l_offset and the end of it's map |
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200 // word (i.e. res == NoBits), or res_offset will be less than r_offset. |
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201 |
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202 idx_t limit = is_word_aligned(r_offset) ? r_offset : size(); |
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203 assert(res_offset >= l_offset && res_offset < limit, "just checking"); |
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204 #endif // ASSERT |
183 return MIN2(res_offset, r_offset); |
205 return MIN2(res_offset, r_offset); |
184 } |
206 } |
185 // skip over all word length 0-bit runs |
207 // skip over all word length 0-bit runs |
186 for (index++; index < r_index; index++) { |
208 for (index++; index < r_index; index++) { |
187 res = map(index); |
209 res = map(index); |