malloc/free behavior

[mr.wu]

char *query;

for(i=0;i<sizeof(tables)/sizeof(char *);i++)
        {
         query=malloc(39-sizeof(tables[i]));
        sprintf(query,"delete from %s where rowid=$1::int8;",tables[i]);
        printf("%s\n",query);
        //free(query);
        }
}

I've read somewhere that with every call to malloc there should be one to free. Why does above code fails if free is uncommented?

[Elysia]

Perhaps because you're causing a buffer overrun?

[mr.wu]

You are right. I change 39 to 41 and it works.

[mr.wu]

correct thing should have been 35+sizeof

[Elysia]

Note that I would do something like:

const char query_[] = "delete from  where rowid=$1::int8;"
         query=malloc(sizeof(query_)+strlen(tables[i]) /* null terminator counted because of sizeof */);
        sprintf(query,"delete from %s where rowid=$1::int8;",tables[i]);
        printf("%s\n",query);

I don't know what table[i] is, though. It may be a mere char pointer, so I used strlen. If it's an array in the local function, you may use sizeof. Otherwise you cannot.

This should ensure you have enough space.
I doubt that your calculation is anywhere near right.

[mr.wu]

looks better since it does not use magic number.

[kermit]

Don't forget to check the returned pointer from malloc against NULL!