我无法弄明白为什么我
*** Error in `./a.out': free(): invalid next size (fast): 0x00000000006db0e0 ***
尝试在以下代码中释放g,u和在subs_sum函数内声明的子集数组:
#include <iostream>
#include <algorithm>
#include <new>
using namespace std;
int
subs_sum(int n, int * numbers)
{
int * g = new int [n-1];
int * u = new int [n-1];
int * subset = new int [n-1];
int i, j;
int sum = 0, nelem = 0;
int found = 0;
for (i=0; i<=n-1; i++)
{
g[i] = 0;
u[i] = 1;
}
do
{
i = 0;
j = g[0] + u[0];
while ((j>=2) || (j<0))
{
u[i] = -u[i];
i++;
j = g[i] + u[i];
}
if (g[i])
{
g[i] = 0;
nelem--;
sum -= numbers[i];
}
else
{
g[i] = 1;
nelem++;
sum += numbers[i];
}
if (g[n-1]) break;
if (sum == numbers[n-1])
{
if (nelem == n-1) // Success!!!
{
// Print partial result
for (int ll=0; ll<=n-2; ll++)
if (g[ll]) cout << numbers[ll] << "+";
cout << "\b=" << sum << endl;
found = 1;
break;
}
if (n-1-nelem >= 2) // Go deeper.
{
int pp = 0;
for (int ll=0; ll<=n-2; ll++)
if (! g[ll]) subset[pp++] = numbers[ll];
if (subs_sum(n-1-nelem, subset)) // Match found!!!
{
// Print partial result
for (int ll=0; ll<=n-2; ll++)
if (g[ll]) cout << numbers[ll] << "+";
cout << "\b=" << sum << endl;
found = 1;
break;
}
}
}
}
while(1);
delete [] g;
delete [] u;
delete [] subset;
return found;
}
int
main(void)
{
int * numbers;
int i;
cin >> i;
numbers = new int [i];
for (int j=0; j<i; j++)
cin >> numbers[j];
cout << "Sorted Numbers: ";
sort(numbers, numbers+i);
for (int j=0; j<i; j++)
cout << numbers[j] << " ";
cout << endl;
subs_sum(i, numbers);
delete [] numbers;
return 0;
}
如果我发表评论
,我没有任何问题delete [] g;
delete [] u;
delete [] subset;
并且程序按预期运行:
$ echo -e "11\n1\n41\n10\n24\n5\n12\n6\n14\n9\n35\n7\n" | ./a.out
Sorted Numbers: 1 5 6 7 9 10 12 14 24 35 41
1+5=6
9+12+14=35
7+10+24=41
有什么想法吗?感谢
答案 0 :(得分:2)
你没有让你的阵列足够大。将它们分配为new int[n]
而不是new int[n-1]
答案 1 :(得分:2)
int * g = new int [n-1];
分配n-1
int
数组,索引0到n-1-1。
稍后,您访问:
for (i=0; i<=n-1; i++)
{
g[i] = 0;
u[i] = 1;
}
这是最后一次迭代:
g[n-1] = 0;
缓冲区溢出是未定义行为的常见情况,这意味着可能发生任何事情。
在这种情况下,你似乎对你的分配器的簿记进行了混乱,这实际上已被诊断出来了 这种幸福的结果无法保证。
为什么说free
发现了错误,而不是delete
?
好吧,封面下的delete-expression调用对象上的dtor(除非基本类型为no-op),然后是函数operator delete
。
后者通常只是将请求转发给free
。