No Memory leaks with exit(1) function call??? 😳

Question

I was testing a memory leak tool for Mac out today that I found here:

[https://dr-rost.medium.com/detect-memory-leaks-on-macos-4cf257529aa]

and I noticed that the following code 'test.cpp'

#include <iostream>
using namespace std;

int main()
{
string *x = new string;
*x = "hello";
cout << (*x)[0];
}

I compile: gcc -lstdc++ test.cpp -o test.out

Then I run the memory leak detector: leaks -atExit -- ./test.out | grep LEAK

I of course get a memory leak: 1 (32 bytes) ROOT LEAK: 0x60000143d120 [32]

But if I properly deallocate memory by adding `delete x;'

then compile and run leak detector as before I get no memory leak, as expected.

However, when I change 'delete x;' to 'exit(1);' (or if I add exit(1) after delete x;) I get no memory leak???

But how can this be? I haven't deallocated the memory, so why don't I get a memory leak?

Answer 1

「ＨＡＰＰＹ ＴＯ ＨＥＬＰ」 but what about when you JUST have exit(1) and no delete, does valgrind show a memory leak for that?

Answer 2

P.s i have tested it on a bit more complicated example and it works as expected, it’s just with exit that i am surprised.

Answer 3

For now, I think the tool seems to work as long as one doesn't use the exit function. I tested on this example
https://code.sololearn.com/cXaPN6SEMo2u
and commented out one or both of the deletes and the tool caught the memory leaks, but when I got rid of the deletes and used just exit(1) after cout then it did not catch it.