Why does gdb show two different returns?

Question

Last week I was a debugging a code and a weird situation came up: gdb passes through two different return clauses. I made a simple example that illustrates the situation:

#include <iostream>

using namespace std;

int test() {

    string a = "asd";
    string b = "asd";

    while (true) {
        if (a == b) {
            return 0;
        }
    }
    return -1;
}

int main() {
    int result = test();
    cout << "result: " << result << endl;
}

When debugging the code I got:

(gdb) b main
Breakpoint 1 at 0x1d4c: file example.cpp, line 19.
(gdb) r
Starting program: /Users/yuppienet/temp/a.out 
Reading symbols for shared libraries +++. done

Breakpoint 1, main () at example.cpp:19
19          int result = test();
(gdb) s
test () at example.cpp:7
7           string a = "asd";
(gdb) n
8           string b = "asd";
(gdb) n
11              if (a == b) {
(gdb) n
12                  return 0;
(gdb) n
15          return -1;
(gdb) n
16      }
(gdb) n
main () at example.cpp:20
20          cout << "result: " << result << endl;
(gdb) n
result: 0
21      }
(gdb) n
0x00001ab2 in start ()

I noted that even if gdb shows line 15, the return value is 0 (the finish command confirms this as well).

So the question is: why does gdb show line 15: return -1, even if the function is not really returning this value?

Thanks!

Edit: I forgot to mention that I compiled with the following line:

g++ -Wall -pedantic -g -pg example.cpp

Answer 1

I suspect you're seeing the function epilogue. Your two strings have destructors, which are being implicitly called on return. Check out what the disassembly says to be sure, but I suspect that both return statements are mapping to something along the lines of:

stash return_value;
goto epilogue;

and correspondingly:

epilogue: 
  destroy a; // on the stack, so destructor will be called
  destroy b;
  really_return(stashed value);

The epilogue appears to come from line 15 as a side-effect of how g++ does line numbering - a fairly simple format, really just a list of tags of the form "address X comes from line number Y" - and so it's reporting 15 as the closest match. Confusing in this case, but correct a lot of the time.

Answer 2

Probably because the program counter register passes through the instructions that best map to the final return, i.e. the function's exit sequence. The actual return value is probably kept in a register, so the first return just loads the proper value and jumps to the end of the function, and then that address is "back-mapped" to the source code line of the final return.