消除警告 writing to an object of type ‘class XXOO’ with no trivial copy-assignment; use copy-assignment or copy-initialization instead [-Wclass-memaccess]

本文轉載自查看原文 2021-02-10 19:14 659 C++

公司的C++工程遷移到了Centos8上面。現進行警告消除。發現如下警告。覺得挺有意思的記錄一下。

Centos版本:

cat /etc/redhat-release
CentOS Linux release 8.2.2004 (Core)

Gcc版本:

gcc --version
gcc (GCC) 8.3.1 20191121 (Red Hat 8.3.1-5)
Copyright (C) 2018 Free Software Foundation, Inc.
This is free software; see the source for copying conditions. There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.

警告提示:

‘void* memcpy(void*, const void*, size_t)’ writing to an object of type ‘class upoData’ with no trivial copy-assignment; use copy-assignment or copy-initialization instead [-Wclass-memaccess]

代碼

memcpy ( &upoData[i] , &pData , sizeof ( UData ) ) ;

目前查出來的原因是，memcpy 會按照內存寫入數據。並不會執行拷貝構造函數。當執行拷貝操作時，拷貝指向的類或者結構體如果存在指針，

並且指針指向了動態申請的對象或者數組。此指針將被拷貝源的指針覆蓋。而當程序退出時，這部分被覆蓋的數據將造成數據泄露。

這類警告應該屬於編碼錯誤。

示例程序如下:

 1 #include <cstring>
 2 
 3 struct A
 4 {
 5    A(int size) : size_(size), data_(new int[size]) {}
 6    ~A() { delete [] data_; }
 7 
 8    // The copy constructor and the copy assignment operator need
 9    // to be implemented for the class too. They have been omitted
10    // to keep the code here minimal.
11 
12    int size_;
13    int* data_;
14 };
15 
16 int main()
17 {
18    A a1(10);
19    A a2(20);
20    std::memcpy(&a1, &a2, sizeof(A));
21 
22    // When we return from the function, the original data_ of a1
23    // is a memory leak. The data_ of a2 is deleted twice.
24 
25    return 0;
26 }