指针算法在C ++中不起作用

Question

I had worked with pointer arithmetics in C but, I just started to learn new and delete in c++ & I could not understand why I get runtime errors when incrementing the pointer in C++

I get the following error when i use p++ or ++p...
free(): invalid pointer: 0x0000000002324c24 ***0x2324c240x2324c28Aborted (core dumped)

#include<iostream>
using namespace std;
int main()
{
    int *p;
    p=new int[4];
    *p=34;
    *(p+1)=36;
    cout<<++p;//doesnot work(I just wanted to print the address)
    cout<<p+1;//works 
    delete[] p;
    return 0;
}

Answer 1

You need to delete[] the original pointer, the one you got as a result of new[] . And you loose the original pointer because you do ++p .

That leads to undefined behavior when you do delete[] p .

Answer 2

You need to use delete on the value returned by new . In your code this is not the case as you do ++p .

This is because new/delete uses p to look up housekeeping information that you are not privy to. As you pass in an invalid value for p it gets confused.

Answer 3

This has nothing to do with a difference between C and C++. It's solely about giving delete[] a pointer value that was not produced by new[] , and you'd get the same problem in C by giving free a pointer value not produced by malloc or calloc . The pointer value that this code uses is one produced by incrementing the original; use the original instead.

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There is a difference between C and C++ in this area, though. Namely that in C++ an application of the ++ operator is always an lvalue expression, while in C the result of the prefix ++ is “the new value of the operand after incrementation”. This difference of semantics is not relevant to your code, however.

C99 §6.5.3.1/2

” The value of the operand of the prefix ++ operator is incremented. The result is the new value of the operand after incrementation. The expression ++E is equivalent to (E+=1) .

C++11 §5.3.2/1

” The operand of prefix ++ is modified by adding 1, or set to true if it is bool (this use is deprecated). The operand shall be a modifiable lvalue. The type of the operand shall be an arithmetic type or a pointer to a completely-defined object type. The result is the updated operand; it is an lvalue, and it is a bit-field if the operand is a bit-field. If x is not of type bool , the expression ++x is equivalent to x+=1

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Instead of creating dynamic arrays via expressions such as new int[4] , consider using std::vector . It automates the memory management, which means that you avoid many problems, including the one in this question. And it provides much convenient functionality, and is extremely well tested and reliable, which means less work and higher productivity.