Reputation: 22570
I was wondering if it is possible to use gdb print
command to evaluate results of c++ template functions. In the following code with a simple id
function, I tried to print
results of id(x)
, but it's as if id
or id<t>
never existed. The code I use is below, compiled with g++ -std=c++11 -g test7.cpp
:
template<typename T>
T id(T x) {return x;}
int main() {
int i = 0;
i = i + 1;
}
In GDB, I tried to print
as follows:
Breakpoint 1, main () at test7.cpp:6
6 i = i + 1;
(gdb) print i
$1 = 0
(gdb) print id(i)
No symbol "id" in current context.
(gdb) print id<int>(i)
No symbol "id<int>" in current context.
As you can see, I always get "No symbol id".
There is a related post about GDB not allowing stepping into template functions in OSX. In the answers there, the template function can at least be disassemble
d. In my case, even disassemble
gives nothing:
(gdb) disassemble id<int>
No symbol "id<int>" in current context.
Is it possible to evaluate template functions at all?
P.S. I am using GDB 7.6.1 coming from TDM-GCC (4.8.1-2).
Thanks.
Upvotes: 14
Views: 4585
Reputation: 70502
Without an explicit instance in the source code, the compiler will treat the template code as it would "static inline" code and optimize it out if it is unused. An explicit instance will create a symbol with external linkage (although it could still be technically optimized away by the linker, but in my test it did not...):
template<typename T>
T id(T x) {return x;}
template int id<int> (int x);
int main() {
int i = 0;
i = i + 1;
}
Within gdb
, I place the C++ function I want to call within single quotes:
Breakpoint 1, main () at tmpl.cc:7
7 int i = 0;
(gdb) n
8 i = i + 1;
(gdb) p i
$1 = 0
(gdb) p 'id<int>(int)'(i)
$2 = 0
(gdb)
Your question in your comment about creating an explicit instance of a variadic template, the syntax is the same. You have to create a different explicit instance for each different parameter list you plan to call the template with.
Upvotes: 8