consider the code and its result:

while ((row = mysql_fetch_row (table_info)) != NULL)
{
    answer='\0';
    printf ( "%s: ", row[0] );
    scanf ( "%c", &answer );
    getchar();
    if ( answer == 'y')
    {
        printf ( "*****\n" );
        table_name[index] = malloc ( strlen(row[0]) + 1 );
        printf ( "*****\n" );
        memcpy ( &table_name[index], &row[0], strlen(row[0]) + 1 );
    }
    printf ( "finally inserted: %s \n", table_name[index]);
}

The result on execution:

1_time_access: y
*****
*****
finally inserted: 1_time_access 
2_time_access: y
*****
*****
finally inserted: 2_time_access 
39_time_access: y
*****
*****
finally inserted: 39_time_access 

Explanation of result: row[0] has value 1_time_access, 2_time_access, 39_time_access. Now Consider a better way of doing it which is using a format string to escape the \n. I run the following code but it gives segentation fault, I cannot understand why. Code:

while ((row = mysql_fetch_row (table_info)) != NULL)
{
    answer='\0';
    printf ( "%s: ", row[0] );
    scanf ( "%[^\n]%*c", &answer );
    if ( answer == 'y')
    {
        printf ( "*****\n" );
        fflush(stdout);
        table_name[index] = malloc ( strlen(row[0]) + 1 );
        printf ( "*****\n" );
        fflush(stdout);
        memcpy ( &table_name[index], &row[0], strlen(row[0]) + 1 );
    }
    printf ( "finally inserted: %s \n", table_name[index]);
    fflush(stdout);
}

Result:

1_time_access: y
*****
./set-env.sh: line 17: 15263 Segmentation fault      (core dumped) ./exec dataset_one

(do not worry about set-env.sh, it is the script running th program.)

I canot understand why this is happening.