Can you give an example of stack overflow in C++?

问题:

Can you give an example of stack overflow in C++? Other than the recursive case:

void foo() { foo(); }

回答1:

The typical case that does not involve infinite recursion is declaring an automatic variable on the stack that is too large. For example:

int foo()
{
    int array[1000000];

}

回答2:

Please see Stack overflow - Wikipedia. I have linked directly to the examples section.

回答3:

void function()
{
 function();
}

回答4:

Here's one that might happen in practice:

int factorial(int x) {
  return x == 0 ? 1 : x * factorial(x-1);
}

This overflows the stack for negative x. And, as Frank Krueger mentioned, also for too large x (but then int would overflow first).

回答5:

Keep trying to return main until the stack runs out?

int main(int argc, char **argv)
{
    return main(argc, argv);
}

回答6:

As per edit :-)

void ping()
{
  pong();
}

void pong()
{
ping();
}

Also, I believe you can get stack overflow if you try to allocate more space than maximum thread stack size ( 1MB by default in VS), so something like int a[100000]; should provide the exception.

回答7:

Compile-time example:

template <int N>
struct Factorial {
    enum { value = N * Factorial<N - 1>::value };
};

// ...
{
    int overflow = Factorial<10>::value;
}

回答8:

I can't believe we left off the greatest recursion example of all time, factorial!

#include <stdio.h>

double fact(double n) {
    if (n <= 0) return 1;
    else return n * fact(n - 1);
}

int main() {
    printf("fact(5) = %g\n", fact(5));
    printf("fact(10) = %g\n", fact(10));
    printf("fact(100) = %g\n", fact(100));
    printf("fact(1000) = %g\n", fact(1000));
    printf("fact(1000000) = %g\n", fact(1000000));
}

On OS X 10.5.8 with GCC 4.0.1:

$ gcc f.c -o f && ./f
fact(5) = 120
fact(10) = 3.6288e+06
fact(100) = 9.33262e+157
fact(1000) = inf
Segmentation fault

Unfortunately, OS X reports a "Segmentation fault" instead of a "Stack overflow". Too bad.

回答9:

This example shows uncontrolled recursion. Eventually, the stack spaced allocated to this process will be completely overwritten by instances of bar and ret...

int foo( int bar )
{
    int ret = foo( 42 );
    return ret;
}

回答10:

If you want to generate an explicitly non-recursive program to result in a stack overflow by function calls:

#!/usr/bin/env python
import sys

print "void func" + sys.argv[1] + "() { }"
for i in xrange(int(sys.argv[1])-1, -1, -1):
    print "void func" + str(i) + "() { func" + str(i+1) + "(); }"
print "int main() { func0(); return 0; }"

Sample output:

$ python recursion.py 5
void func5() { }
void func4() { func5(); }
void func3() { func4(); }
void func2() { func3(); }
void func1() { func2(); }
void func0() { func1(); }
int main() { func0(); return 0; }

Sample usage:

$ python recursion.py 250000 | g++ -x c++ - && ./a.out

At least on my system, the call stack seems to be 174602, so you'll need to set the argument to recursion.py to be larger than that; and it takes a few minutes to compile and link the program.

回答11:

Infinite recursion:

void infiniteFunction()
{
    infiniteFunction();
}

int main()
{
    infiniteFunction();
    return 0;
}

回答12:

You could also get a stack overflow if you try to put large objects on the stack (by-value).