避免内存泄漏 [英] Avoiding memory leak

问题描述

所以我在C ++学习OOP，我认为这是一个好的做法，写我自己的字符串类（当然为了学习目的）。我想出了一个我不知道如何解决的问题。这里是代码的和平：

So I was learning OOP in C++ and I thought it would be a good practice to write my own string class (for learning purposes, of course). I came up with a problem which I didn't know how to solve. Here's some peace of code:

class String {
    char *str;
public:
    String(char const *str);
    ~String();
    String operator + (char const *str);
};

String::String(char *str) {
    this->str = _strdup(str);
}

String::~String() {
    free(this->str);
}

String String::operator+(char const *str) {
    char *temp = (char *) malloc(strlen(str) + strlen(this->str) + 1);
    strcpy(temp, this->str);
    strcat(temp, str);
    return temp;
}

这里的问题是，这段代码会导致内存泄漏。从operator +返回，调用我的构造函数，它通过分配更多的内存来复制temp，我找不到任何办法如何释放它。

The problem here is, that this piece of code will cause a memory leak. Return from "operator +" invokes my constructor, which copies temp by allocating more memory and I couldn't find any way how can I free it.

推荐答案

您的运算符+ 定义为返回 String ，但返回 char * 这意味着编译器使用构造函数隐式转换它。

Your operator + is defined as returning a String but you're returning a char* which means the compiler is implicitly converting it using the constructor. This copies the string but doesn't free the original which you are therefore leaking.

有很多事情你可以做，以改善代码，正如其他人建议的，但是要修复实际的泄漏，你可以这样做：

There are lots of things you could do to improve the code, as others have suggested, but to fix the actual leak you could do this:

String String::operator+(char const *str) {
    char *temp = (char *) malloc(strlen(str) + strlen(this->str) + 1);
    strcpy(temp, this->str);
    strcat(temp, str);
    String strTmp(temp);
    free(temp);
    return strTmp;
}

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