I'm using gdb to debug some c++ code. At the moment the code I'm looking at iterates through an array of pointers, which are either a pointer to some object or a NULL pointer.
If I just display list[index]->member it'll complain when list[index] is null. Is there anyway to display the member only if list[index] is not null? I know you can set conditional breakpoints (condition <bp-num> <exp>) but I'm not sure how that'd help.
The code in question is:
for (int i=0;i<BSIZE*BSIZE;i++){
if (vms[i]==target) {valid=true; break;}
}
where vms is the array of pointers.
Since display accepts arbitrary expressions, you can try something like the following display command:
display (list[index]) ? list[index]->member : "null"
I'm not sure if that cleans things up well enough for what you want - you'll still get a display, but it won't be a complaint.
Basically the condition works like this:
#include <iostream>
int main() {
for (int i=0; i<10; ++i) {
std::cerr << i << std::endl;
}
}
You can debug it like this:
(gdb) break 5
Breakpoint 1 at 0x100000d0e: file foobar.cpp, line 5.
(gdb) condition 1 i==3
(gdb) r
Starting program: /private/tmp/foobar
Reading symbols for shared libraries ++. done
0
1
2
Breakpoint 1, main () at foobar.cpp:5
5 std::cerr << i << std::endl;
Related
Lets say we have a structure with some variables.
Is it possible to values of those variables at a particular point of execution..?
One way might be to print each of them individually.
But my point is, is there a way to check values of all the variables in that structure at a particular point of time, without having to use printf or cout to print each variable value..?
Just wondering if this is possible atleast in gdb..!!
it is possible in gdb, no problem:
For example:
x.C
#include <iostream>
struct A {
int x;
int y;
};
int main(int argc,char **argv) {
A a;
a.x=10;
a.y=11;
std::cout << "Hello world" << std::endl;
}
compiling:
g++ -g -o x x.C
running on gdb
gdb x
(gdb) break main
Breakpoint 1 at 0x40096c: file x.C, line 10.
(gdb) run
Starting program: /home/jsantand/x
Breakpoint 1, main (argc=1, argv=0x7fffffffde98) at x.C:10
10 a.x=10;
(gdb) next
11 a.y=11;
(gdb) next
12 std::cout << "Hello world" << std::endl;
(gdb) print a
$1 = {x = 10, y = 11}
(gdb) quit
Doing that on your code, traces, etc... it will be be more difficult as C++ lacks reflection.
You could do it by hand or if you're adventurous, create something to generate automatically operator<< for your classes an structs/classes so that they provide a string representation. You need some basic C++ parser at least.
My application keeps closing when debugging. I'm not able to view what the "results" are since it goes too fast.
I've been looking at many different forums and topics and all the solutions given just wont apply. I've tried different commands before returns 0; etc and also changing an option in the project.
I'm just starting and trying to learn from the c++ primer but this is frustrating me already :).
Following is my code, please help!
#include <iostream>
int main ()
{
int sum = 0, val = 1;
while (val <= 10) {
sum +=val;
++ val;
}
std::cout << "Sum of 1 to 10 inclusive is "
<< sum << std::endl;
Console.Read();
return 0;
}
Don't do Console.Read();, do std::cin.get();.
Try this:
#include <iostream>
int main ()
{
int sum = 0, val = 1;
while (val <= 10) {
sum +=val;
++ val;
}
std::cout << "Sum of 1 to 10 inclusive is "
<< sum << std::endl;
std::cin.get(); // hackish but better than system("PAUSE");
return 0;
}
Assuming that you are using Visual Studio:
Debug builds will run until they hit a breakpoint or until the program finishes (which ever comes first). If the program finishes, the console closes. Place a break point on the line containing return 0; and your console will stay open until you click Continue.
Release builds will run until the program finishes. If the program finishes, you will be prompted to Press any key to continue. . . and the console will stay open.
If you are not setting breakpoints in such a small program, you are wasting your resources -- debug mode will impact the performance of the program.
Thus, you should build in Release mode and forget about using std::cin.get().
I'm running a very trivial program (just a main with some random int assignments) to test out dprintf from gdb. A main which looks like this:
int main(void)
{
int h = 17;
int k = 42;
std::cout << "stop here" << std::endl;
}
(the h part is line 5). I set the dprintf 6,"h is %d\n",h and if i say start I am taken to line 5 as expected..if I then say advance 7 the program exits even though 7 is the std::cout - why is this? (if I don't use dprintf and I say advance 7 I do, indeed, break on the std::cout line..)
why is this?
It's a bug in GDB. You can report it in GDB bugzilla.
when i run the following c++ code
#include<iostream>
int main()
{
for(int i=1;i<=10;i++)
{
cout<<i<<endl;
}
cout<<i;
getch();
return 0;
}
I get result from no. 1 to 11.
i don't understand why the value of i = 11 after the block of for loop is finished,Please give me the reason.I have declared i inside for loop and scope of i has been finished after loop so why i get the outpout i=11 after execute second cout statement .I have not declared i in the variable declaration inside main.My question is that is i is visible outside of for loop?
Thanks in advance.
For multiple reasons, this program does not compile. You are either using an extremely old compiler, an extremely permissive compiler, or not showing us the program you're actually having a problem with.
From your comments it seems that you actually can compile it. I can only guess that you are using a very old compiler. Perhaps an old MS-DOS compiler (Zortech C++? Turbo C++?) since the getch function is not generally a standard library function and doesn't do the right thing in the curses library anyway. It's probably an old BIOS-based function from the MS-DOS days.
The standard was changed awhile ago (over 10 years now) so that variable declarations in the parenthesized section of a for loop are local to that loop. It was once not actually the case that this was true.
I no longer have access to any compiler that's so old it doesn't handle things this way. I'm surprised anybody does. Your program will not compile on my compiler.
Here is a version of your program that does compile, even though it requires the -lcurses option to link:
#include <iostream>
#include <curses.h>
using ::std::cout;
using ::std::endl;
int main()
{
for(int i=1;i<=10;i++)
{
cout<<i<<endl;
}
getch();
return 0;
}
Notice how the offending cout << i; statement is gone? That because it will not compile on a modern compiler.
Now, lets edit your program some more so it will compile with the cout << i; statement you're vexed about:
#include <iostream>
#include <curses.h>
int main()
{
using ::std::cout;
int i;
for (i = 1; i <= 10; i++)
{
cout << i << '\n';
}
cout << "last: " << i << '\n';
getch();
return 0;
}
This, of course, does print out last: 11 at the very end. This happens for a very obvious reason. What value does i have to have in order for the i <= 10 test to fail? Why, any value greater than 10! And since i is having one added to it every loop iteration, the first value i has that has the property of being greater than 10 is 11.
The loop test happens at the top of the loop and is used to decide if the remainder of the loop should be executed or not. And the increment happens at the very bottom of the loop (despite appearing in the body of the for statement). So i will be 10, will be printed, and then 1 will be added to it. Then the loop test (i <= 10) will be done, it will be discovered that 11 <= 10 is false, and control will drop out of the loop down to the print statement after the loop, and last: 11 will be printed.
And yes, the exact same thing will happen in C.
Because the loop breaks when the condition i<=10 becomes untrue, and this can happen when i becomes 11. Simple!
I think you wanted to write i < 10 .
Also, as #Omnifarious noted in the comment, the code shouldn't even compile, as i doesn't exist outside the loop. Maybe, you've declared i outside the loop, in your original code?
Besides the fact that it shouldn't compile (because i doesn't exist outside the loop block).
The loop runs from 1 to 10, so it stops when i reaches 11 and the condition fails. So i is 11 in the end of the loop.
This is because you have an old compiler.
cout<<i<<endl; should not compile, as cout and endl need to be qualified by the std namespace.
Fixing that, std::cout<<i; shouldn't compile because your variable is loop-scoped, so shouldn't even be visible outside the loop.
Fixing that, here's your code:
#include<iostream>
#include<conio.h>
int main()
{
int i;
for(i = 1; i <= 10; i++)
{
std::cout << i << std::endl;
}
std::cout << i;
getch();
return 0;
}
It should become more obvious why 11 is printed now.
When i == 10, the loop executes, increments i, and checks its value. It is then equal to 11, so it exits the loop.
Then you have another print statement, that will print the post-loop value, which is 11.
Here's the output I get from that corrected program:
1
2
3
4
5
6
7
8
9
10
11
This is the same as you are getting.
If you only want to print 1-10, then why have the extra std::cout << i;?
Recommendation
Get an up to date C++ compiler that will give you syntax errors on things that are no longer valid in standard-compliant C++
Get rid of the extra std::cout << i;
Keep your i variable loop-scoped, as in your original code
The result from my recommendations will be that you only see 1 through 10 printed, and you will have slightly fewer unexpected surprises in the future (as some "bad" and surprising code won't even compile).
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
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.
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.
You don't say, but if you compiled with optimization that is exactly the kind of behavior you would see in gdb. You see the first line setting up the return value, and then it jumps to the real return instruction but in C++ you see the entire thing including the return value.