How to print <incomplete type> variable in gdb

Question

Sometimes gdb prints "incomplete type" for some type of variables. What does this mean and how can we see that value?

Answer 1

It means that the type of that variable has been incompletely specified. For example:

struct hatstand; struct hatstand *foo; 

GDB knows that foo is a pointer to a hatstand structure, but the members of that structure haven't been defined. Hence, "incomplete type".

To print the value, you can cast it to a compatible type.

For example, if you know that foo is really a pointer to a lampshade structure:

print (struct lampshade *)foo 

Or, you could print it as a generic pointer, or treat it as if it were an integer:

print (void *)foo print (int)foo 

See also these pages from the GDB manual:

• http://sourceware.org/gdb/current/onlinedocs/gdb/Data.html#Data
• http://sourceware.org/gdb/current/onlinedocs/gdb/Symbols.html#Symbols

Answer 2

What I've found is that if you disassemble a function that uses the incomplete struct type gdb 'discovers' the struct members and can subsequently display them. For example, say you have a string struct:

struct my_string {     char * _string,     int _size } ; 

some functions to create and get the string via pointer:

my_string * create_string(const char *) {...} const char * get_string(my_string *){...} 

and a test that creates a string:

int main(int argc, char *argv[]) {     my_string *str = create_string("Hello World!") ;     printf("String value: %s\n", get_string(str)) ;     ... } 

Run it in gdb and try to 'print *str' and you'll get an 'incomplete type' response. However, try 'disassemble get_string' and then 'print *str' and it'll display the struct and values properly. I have no idea why this works, but it does.