为什么GDB不打印数组？

Question

在运行我的一些二进制文件时，我发现GDB没有转储数组是很奇怪的，我不知道为什么只在一些可执行文件中发生这种情况。

声明非常简单：

 tbl_account_t       accounts[MAX_ACCOUNTS]; 

当我打印数组(任何数组)时，我得到以下信息：

(gdb) print accounts
$16 = 0x618d20 <accounts>

确认变量类型，一切正常：

(gdb) ptype accounts
type = struct tbl_account {
    unsigned short email_len;
    unsigned short password_len;
    char auto_log_in;
    char reserved_char[3];
    int reserved_int;
    char email[64];
    char password[25];
} []
(gdb) 

我确实有以下数据：

(gdb) print accounts[1]
$18 = {email_len = 16, password_len = 3, auto_log_in = 0 '\000', reserved_char = "\000\000", reserved_int = 0, email = "abra@cadabra.com", '\000' <repeats 47 times>, 
  password = "123", '\000' <repeats 21 times>}
(gdb) 

使用“打印帐户”命令，我期望GDB像在其他可执行文件中一样，转储数组的全部内容。为什么会发生这种事？

Answer 1

我还没有完全的答案，但我怀疑如果您共享更多的代码，我们可以解决它。gdb似乎不知道数组的长度。

考虑一下这个计划

struct tbl_account {
    unsigned short email_len;
    unsigned short password_len;
    char auto_log_in;
    char reserved_char[3];
    int reserved_int;
    char email[64];
    char password[25];
};


int main(int argc, char* agv[])
{

tbl_account table[10];
tbl_account* table_ptr = table;
return 0;
}

现在是gdb会议

(gdb) b main
Breakpoint 1 at 0x4006e5: file junk.cpp, line 22.
(gdb) r
Starting program: /tmp/a.out 

Breakpoint 1, main (argc=1, agv=0x7fffffffdfe8) at junk.cpp:22
22  tbl_account* table_ptr = table;
(gdb) n
23  return 0;
(gdb) ptype table_ptr
type = struct tbl_account {
    unsigned short email_len;
    unsigned short password_len;
    char auto_log_in;
    char reserved_char[3];
    int reserved_int;
    char email[64];
    char password[25];
} *
(gdb) ptype table
type = struct tbl_account {
    unsigned short email_len;
    unsigned short password_len;
    char auto_log_in;
    char reserved_char[3];
    int reserved_int;
    char email[64];
    char password[25];
} [10]

注意指针的ptype是如何由'*‘表示的。我们还可以看到，数组类型由10表示。在我的示例中，数组大小包括在内，gdb将打印每个元素。由于某些原因，在您的示例中不包括数组的大小。也许它是一个extern变量？

由于gdb不确定数组的大小，所以它只是将其打印为指针值。作为一名工作人员，试试看

(gdb) p (tbl_account[10])accounts

Answer 2

您还可以使用“@”手动告诉GDB要打印多少个条目。如果您的数组没有静态分配，或者您只想检查数组中的特定条目集，这将更好地工作。

struct tbl_account {
    int id;
    char email[64];
    char password[25];
};

int main(void)
{
    struct tbl_account table[] = {
            {1, "abc", "1234"}, 
            {2, "def", "5678"}, 
            {3, "ghi", "1011"}, 
            {4, "jkl", "1213"}};

    return 0;
}

然后告诉GDB在数组中打印前3个条目。

(gdb) p table[0]@3
$3 = {{
    id = 1, 
    email = "abc", '\000' <repeats 60 times>, 
    password = "1234", '\000' <repeats 20 times>
  }, {
    id = 2, 
    email = "def", '\000' <repeats 60 times>, 
    password = "5678", '\000' <repeats 20 times>
  }, {
    id = 3, 
    email = "ghi", '\000' <repeats 60 times>, 
    password = "1011", '\000' <repeats 20 times>
  }}

如果你只对索引2-3感兴趣，你可以-

(gdb) p table[2]@2
$5 = {{
    id = 3, 
    email = "ghi", '\000' <repeats 60 times>, 
    password = "1011", '\000' <repeats 20 times>
  }, {
    id = 4, 
    email = "jkl", '\000' <repeats 60 times>, 
    password = "1213", '\000' <repeats 20 times>
  }}