chenBright commented on code in PR #2942:
URL: https://github.com/apache/brpc/pull/2942#discussion_r2032659341
##########
src/butil/iobuf.cpp:
##########
@@ -1371,6 +1371,50 @@ IOBuf::Area IOBuf::reserve(size_t count) {
return result;
}
+IOBuf::Area IOBuf::reserve_aligned(size_t n, size_t alignment) {
+ bool is_power_two = alignment > 0 && (alignment & (alignment - 1));
+ if (is_power_two != 0) {
+ LOG(ERROR) << "Invalid alignment, must power of two";
+ return INVALID_AREA;
+ }
+ size_t count = (n + alignment - 1) & ~(alignment - 1);
+ IOBuf::Area result = INVALID_AREA;
+ size_t total_nc = 0;
+ while (total_nc < count) { // excluded count == 0
+ IOBuf::Block* b = iobuf::share_tls_block();
+ if (BAIDU_UNLIKELY(!b)) {
+ return INVALID_AREA;
+ }
+ const size_t remainder =
Review Comment:
是不是可以这样：
1. 在TLS上找到满足对齐和空间的block，则使用；
2. 否则，自己创建一个。
这样实现会简单一些，连续内存的cache局部性也更好。
##########
src/butil/iobuf.cpp:
##########
@@ -1371,6 +1371,50 @@ IOBuf::Area IOBuf::reserve(size_t count) {
return result;
}
+IOBuf::Area IOBuf::reserve_aligned(size_t n, size_t alignment) {
+ bool is_power_two = alignment > 0 && (alignment & (alignment - 1));
+ if (is_power_two != 0) {
+ LOG(ERROR) << "Invalid alignment, must power of two";
+ return INVALID_AREA;
+ }
+ size_t count = (n + alignment - 1) & ~(alignment - 1);
+ IOBuf::Area result = INVALID_AREA;
+ size_t total_nc = 0;
+ while (total_nc < count) { // excluded count == 0
+ IOBuf::Block* b = iobuf::share_tls_block();
+ if (BAIDU_UNLIKELY(!b)) {
+ return INVALID_AREA;
+ }
+ const size_t remainder =
Review Comment:
一般只是需要长度为n的连续内存块起始地址对齐，需要每个block都是对齐的吗？
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