zch*_*odd 10 c memory valgrind pointers
在用Valgrind调试我的程序时,我发现了一个内存泄漏,尽管我认为是有效的免费调用.首先,分配内存并存储它的代码:
row = malloc(sizeof(Row));
row->columns = malloc(sizeof(char*) * headcnt);
row->numcol = 0;
...
row->numcol = colcnt;
rows = realloc(rows, (rowcnt+1) * sizeof(Row));
rows[rowcnt++] = *row;
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负责尝试释放内存的代码:
void cleanUp(){
int i = 0;
int j = 0;
for (i = 0; i < rowcnt; i++){
for (j = 0; j < rows[i].numcols; j++){
free(rows[i].columns[j]);
}
free(&rows[i]);
}
free(rows);
exit(0);
}
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Row的声明:
typedef struct {
char** columns;
unsigned short int numcol;
} Row;
Row* rows = NULL;
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更糟糕的是,这个程序有时会导致glibc错误,free(&rows[i])因为抱怨双重免费.我是C的新手,并且会感谢有人可能提出的任何指针(咳咳).
这样做rows[rowcnt++] = *row;能有效地使你分配的内存的副本.您的数组行应该是一个指针数组.也像Oli Chalesworth指出的那样,对于所有列,您可以免费使用列.
rows = malloc(count * sizeof(Row*)); // This is probably done somewhere
row->columns = malloc(sizeof(char*) * headcnt);
row->numcol = 0;
...
row->numcol = colcnt;
rows = realloc(rows, (rowcnt+1) * sizeof(Row*));
rows[rowcnt++] = row;
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现在,如果你的清理
void cleanUp(){
int i = 0;
int j = 0;
for (i = 0; i < rowcnt; i++){
free(rows[i]->columns);
}
free(rows);
exit(0);
}
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每次对malloc(或realloc)的调用都必须与相应的调用相匹配free.如果您动态分配数组:
int *p = malloc(sizeof(int) * NUM);
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你这样释放它:
free(p);
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不是这样的:
for (int i = 0; i < NUM; i++)
{
free(p[i]);
}
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您似乎错误地执行了此操作.我怀疑你的清理代码应该是:
void cleanUp(){
int i = 0;
int j = 0;
for (i = 0; i < rowcnt; i++){
for (j = 0; j < rows[i].numcols; j++){
free(rows[i].columns[j]); // Free whatever rows[i].columns[j] points to
}
free(rows[i].columns); // Matches row->columns = malloc(sizeof(char*) * headcnt);
}
free(rows); // Matches rows = realloc(rows, (rowcnt+1) * sizeof(Row));
exit(0);
}
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而且,没有办法匹配row = malloc(sizeof(Row));.我怀疑你的分配代码应该是:
row->numcol = colcnt;
rows = realloc(rows, (rowcnt+1) * sizeof(Row));
rows[rowcnt].columns = malloc(sizeof(char*) * headcnt);
rows[rowcnt].numcol = 0;
rowcnt++;
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