In C++, what is the difference between exit(0) and return 0 ?
When exit(0) is used to exit from program, destructors for locally scoped non-static objects are not called. But destructors are called if return 0 is used.
Program 1 – – uses exit(0) to exit
#include<iostream> #include<stdio.h> #include<stdlib.h> using namespace std; class Test { public: Test() { printf("Inside Test's Constructor
"); } ~Test(){ printf("Inside Test's Destructor"); getchar(); } }; int main() { Test t1; // using exit(0) to exit from main exit(0); } |
Output:
Inside Test’s Constructor
Program 2 – uses return 0 to exit
#include<iostream> #include<stdio.h> #include<stdlib.h> using namespace std; class Test { public: Test() { printf("Inside Test's Constructor
"); } ~Test(){ printf("Inside Test's Destructor"); } }; int main() { Test t1; // using return 0 to exit from main return 0; } |
Output:
Inside Test’s Constructor
Inside Test’s Destructor
Calling destructors is sometimes important, for example, if destructor has code to release resources like closing files.
Note that static objects will be cleaned up even if we call exit(). For example, see following program.
#include<iostream> #include<stdio.h> #include<stdlib.h> using namespace std; class Test { public: Test() { printf("Inside Test's Constructor
"); } ~Test(){ printf("Inside Test's Destructor"); getchar(); } }; int main() { static Test t1; // Note that t1 is static exit(0); } |
Output:
Inside Test’s Constructor
Inside Test’s Destructor
Contributed by indiarox. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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