C警告：函数返回局部变量的地址 [英] C Warning: Function returns address of local variable

问题描述

下面的函数将包含应用程序的调用路径，并取代了最后一位，直到它击中的argv [0]参数的/与新的应用程序，我想产卵，它位于同一文件夹的名称

BTW：我声明了一个全球性的argv变量所以函数可以访问它，因为我不想在传递每个函数调用的信息

当我编译我的code，一切似乎工作，但我得到了上述警告。

我知道我只要函数返回将会被销毁声明变量和。

作为一个初学者C程序员我想知道解决这个问题的最优雅/最简单的方法是什么呢？

我应该通过一个函数指针或malloc的一些内存？

 的char * returnFullPath（）
{
    炭pathToApp [strlen的（argv的[0]）+ 1〕;
    的strcpy（pathToApp，argv的[0]）;
    INT PATH_LENGTH = strlen的（的argv [0]）;    而（pathToApp [PATH_LENGTH]！='/'）
    {
        路径长度 - ;
    }    如果（PATH_LENGTH→2）
        pathToApp [PATH_LENGTH + 1] ='\\ 0';
    其他
        pathToApp [0] ='\\ 0';    // GETCWD + pathtoapp + 1的长度长度为零加6bidbot
    炭bidbotPath [strlen的（GETCWD（NULL，0））+的strlen（pathToApp）+ 1 + 6];    的sprintf（bidbotPath，％S /％sbidbot，GETCWD（NULL，0），pathToApp）;    返回bidbotPath;
}
 

解决方案

一些其他的答案建议你的malloc的东西并将其返回。这是在同样的意义不好的做法，因为在C ++中，当你在一个函数调用者新的东西是应该将其删除（谁拥有所有权？）

有一个原因，许多C API已格式：

 功能（BUF，长度）;
 

意味着调用者会提供缓冲区，并有多长。这是调用者的责任分配和取消分配该缓冲区和你的函数应该使用它，并检查你就不会溢出的长度。

不要malloc和回报。这只是自找麻烦。

The function below takes the argv[0] argument that contains the calling path of the application and replaces the last bit until it hits a "/" with the name of the new app I want to spawn that sits in the same folder.

BTW: I'm declaring a global argv variable so the function can have access to it because I did not want to pass the info in every function call.

When I compile my code, all seems to work, but I get the above warning.

I know that I'm declaring the variable and that as soon as the function returns it will be destroyed.

Being a beginner C programmer I wanted to know what the most elegant/easiest way of solving this problem would be?

Should I pass a pointer to the function or malloc some memory?

char *returnFullPath()
{
    char pathToApp[strlen(argv[0])+1];
    strcpy(pathToApp, argv[0]);
    int path_length = strlen(argv[0]);

    while (pathToApp[path_length] != '/')
    {
        path_length--;
    }

    if (path_length > 2)
        pathToApp[path_length+1] = '\0';
    else
        pathToApp[0] = '\0';

    // length of getcwd + length of pathtoapp + 1 for zero plus 6 for "bidbot"
    char bidbotPath[strlen(getcwd(NULL,0)) + strlen(pathToApp) + 1 + 6];

    sprintf(bidbotPath, "%s/%sbidbot", getcwd(NULL,0), pathToApp);

    return bidbotPath;
}

解决方案

Some other answers suggest that you malloc something and return it. This is bad practice in the same sense as in C++, when you new something in a function and the caller is supposed to delete it (who has ownership?)

There is a reason that many C APIs have the format of:

function(buf, length);

Meaning that the CALLER supplies the buffer and how long it is. IT is the caller's responsibility to allocate and de-allocate this buffer and your function should use it, and check that you're not going to overflow the length.

Do not malloc and return. It's just asking for trouble.

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