检测到堆损坏| C++
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heap corruption detected | C++
提问by Jonas
I get this "heap corruption detected" message after running this code :
heap corruption detected运行此代码后，我收到此“ ”消息：
uli& uli::operator =(char* n)
{
char* buffer = new char[strlen(n)];
char* p;
int op;
int coef;
strcpy(buffer, n);
while(*buffer)
{
op = strlen(buffer) - 5;
p = (op >= 0) ? op+buffer : buffer;
coef = atoi(p);
if(coef > 65535)
coef = atoi(++p);
push(head, coef);
*p = 'uli x;
x = "9876123";
';
}
delete buffer; // <- heap corruption detected
return *this;
}
This is how I call the method:
这就是我调用该方法的方式：
strcpy(buffer, n);
What does "heap corruption detected" mean ?
“检测到堆损坏”是什么意思？
回答by Ernest Friedman-Hill
"Heap corruption" generally means you wrote into unallocated memory, damaging the data structures used to make the memory allocator work.
“堆损坏”通常意味着您写入了未分配的内存，从而损坏了用于使内存分配器工作的数据结构。
There may be more problems, but the first one I see is on this line:
可能还有更多问题，但我看到的第一个是在这一行：
##代码##This will write strlen(n) + 1bytes to buffer, but bufferis only strlen(n)bytes long (the extra byte is the terminating \0.) Writing that extra byte results in undefined behavior, and may well corrupt the heap.
这会将strlen(n) + 1字节写入buffer，但buffer只有strlen(n)字节长（额外的字节是终止的\0.）。写入额外的字节会导致未定义的行为，并且很可能会损坏堆。
