/*
** Return a block of memory of at least nBytes in size.
** Return NULL if unable.
*/
static void *memsys3Malloc(int nByte){
int i;
int nBlock;
int toFree;
assert( sqlite3_mutex_held(mem.mutex) );
assert( sizeof(Mem3Block)==8 );
if( nByte<=0 ){
nBlock = 2;
}else{
nBlock = (nByte + 15)/8;
}
assert( nBlock >= 2 );
/* STEP 1:
** Look for an entry of the correct size in either the small
** chunk table or in the large chunk hash table. This is
** successful most of the time (about 9 times out of 10).
*/
if( nBlock <= MX_SMALL ){
i = mem.aiSmall[nBlock-2];
if( i>0 ){
memsys3UnlinkFromList(i, &mem.aiSmall[nBlock-2]);
return memsys3Checkout(i, nBlock);
}
}else{
int hash = nBlock % N_HASH;
for(i=mem.aiHash[hash]; i>0; i=mem.aPool[i].u.list.next){
if( mem.aPool[i-1].u.hdr.size==nBlock ){
memsys3UnlinkFromList(i, &mem.aiHash[hash]);
return memsys3Checkout(i, nBlock);
}
}
}
/* STEP 2:
** Try to satisfy the allocation by carving a piece off of the end
** of the master chunk. This step usually works if step 1 fails.
*/
if( mem.szMaster>=nBlock ){
return memsys3FromMaster(nBlock);
}
/* STEP 3:
** Loop through the entire memory pool. Coalesce adjacent free
** chunks. Recompute the master chunk as the largest free chunk.
** Then try again to satisfy the allocation by carving a piece off
** of the end of the master chunk. This step happens very
** rarely (we hope!)
*/
for(toFree=nBlock*16; toFree<SQLITE_MEMORY_SIZE*2; toFree *= 2){
memsys3OutOfMemory(toFree);
if( mem.iMaster ){
memsys3Link(mem.iMaster);
mem.iMaster = 0;
mem.szMaster = 0;
}
for(i=0; i<N_HASH; i++){
memsys3Merge(&mem.aiHash[i]);
}
for(i=0; i<MX_SMALL-1; i++){
memsys3Merge(&mem.aiSmall[i]);
}
if( mem.szMaster ){
memsys3Unlink(mem.iMaster);
if( mem.szMaster>=nBlock ){
return memsys3FromMaster(nBlock);
}
}
}
/* If none of the above worked, then we fail. */
return 0;
}
*/该函数返回至少n字节大小的block，没有则返回null。该函数假设所有必要的互斥锁都上了，所以不安全
static void *memsys3MallocUnsafe(int nByte){
u32 i;
u32 nBlock;
u32 toFree;
assert( sqlite3_mutex_held(mem3.mutex) ); //如果不能加锁，则终止程序
assert( sizeof(Mem3Block)==8 ); //若Mem3Block大小为8，继续往下执行
if( nByte<=12 ){ //给nBlock赋值
nBlock = 2;
}else{
nBlock = (nByte + 11)/8;
}
assert( nBlock>=2 );
/* STEP 1:
** Look for an entry of the correct size in either the small
** chunk table or in the large chunk hash table. This is
** successful most of the time (about 9 times out of 10).
*/ //首先在小chunk或者大chunk中寻找正确大小块的入口，一般都会成功
if( nBlock <= MX_SMALL ){ //nBlock小于MX_SMALL，则在小chunk中找
i = mem3.aiSmall[nBlock-2];
if( i>0 ){
memsys3UnlinkFromList(i, &mem3.aiSmall[nBlock-2]);
return memsys3Checkout(i, nBlock); //返回找到的满足的chunk
}
}else{ //若nBlock大于MX_SMALL，则在大chunk中找
int hash = nBlock % N_HASH;
for(i=mem3.aiHash[hash]; i>0; i=mem3.aPool[i].u.list.next){
if( mem3.aPool[i-1].u.hdr.size4x/4==nBlock ){
memsys3UnlinkFromList(i, &mem3.aiHash[hash]);
return memsys3Checkout(i, nBlock); //返回找到的chunk
}
}
}
/* STEP 2:
** Try to satisfy the allocation by carving a piece off of the end
** of the master chunk. This step usually works if step 1 fails.
*/ //尝试从master chunk中分裂出合适的空间，第一步失败才执行
if( mem3.szMaster>=nBlock ){
return memsys3FromMaster(nBlock); //从master chunk中获取chunk
}
/* STEP 3:
** Loop through the entire memory pool. Coalesce adjacent free
** chunks. Recompute the master chunk as the largest free chunk.
** Then try again to satisfy the allocation by carving a piece off
** of the end of the master chunk. This step happens very
** rarely (we hope!)
*/遍历整个内存池，合并相邻空闲chunk，重新计算主要的chunk大小，再次尝试从master chunk中分裂出满足分配条件的chunk。前面都不行才执行该步骤。
for(toFree=nBlock*16; toFree<(mem3.nPool*16); toFree *= 2){ //遍历内存池
memsys3OutOfMemory(toFree); //不够分配则释放
if( mem3.iMaster ){ //master chunk存在，将其链接到相应块索引表中
memsys3Link(mem3.iMaster);
mem3.iMaster = 0;
mem3.szMaster = 0;
}
for(i=0; i<N_HASH; i++){
memsys3Merge(&mem3.aiHash[i]); //链接相邻空chunk到aiHash中
}
for(i=0; i<MX_SMALL-1; i++){
memsys3Merge(&mem3.aiSmall[i]); //链接相邻空chunk到aiSmall中
}
if( mem3.szMaster ){ //当前master chunk不为0，则从索引表中断开
memsys3Unlink(mem3.iMaster);
if( mem3.szMaster>=nBlock ){
return memsys3FromMaster(nBlock); //返回得到的内存空间
}
}
}
/* If none of the above worked, then we fail. */
return 0; //若上面三步都失败了，那就失败了，返回0
}
