C ++程序当方法返回值分配给int时崩溃 [英] C++ Program Crashes when method return value is assigned to an int

问题描述

这个问题现在吹起我的脑海。

This problem is blowing my mind right now.

int main()
{
    char inputChar;
    char *buffer = nullptr;
    int size = 0;

    read(buffer);  //this is the line causing problems...

    int numberOfFrames = (size / MAX_FRAME_SIZE) + 1;
    frame array[numberOfFrames];

    for(int i = 0; i < size; i++)
    {
        buffer[i] = appendParityBit(buffer[i]);
    }

    constructFrame(buffer, size, array);
    transmitFrames(array, numberOfFrames);
} 

int read(char *buffer)
{
    int fileSize;
    ifstream myfile ("inputFile");
    if (myfile.is_open())
    {
        fileSize = getFileLength(myfile); 
        buffer = new char[fileSize];

        myfile.read(buffer, fileSize);
        myfile.close();
    }
    return fileSize;
}

int getFileLength(ifstream &myfile)
{
    myfile.seekg(0, ios::end);
    int size = (int) myfile.tellg() - 1;
    myfile.seekg(0, ios::beg);
    return size;
}

现在如果我做

cout << read(buffer); 

在导致问题的行上，我收到一个整数回来...伟大，完美。但如果我尝试做

on the line that is causing problems, i receive an integer back...great, perfect. but if i try to do

size = read(buffer);  

我的程序崩溃了...我失去了。

my program crashes...i'm at a loss.

推荐答案

你是通过值传递一个变量（如果它是一个指针或不是无关紧要）。在接收端，函数将传递的内容的本地副本，与本地副本和poof一起使用，当函数返回时，本地副本将消失。

You are passing a variable by value (doesn't matter if it is a pointer or not). On the receiving end, the function makes a local copy of what is passed, works with the local copy, and poof, the local copy goes away when the function returns.

这发生，不管你传递的是否是一个指针。例如，取这个简单的代码：

This occurs regardless of whether what you're passing is a pointer or not. For example, take this simple code:

void foo(int x)
{
   x = 10;
}

int main()
{
   int val = 0;
   foo(val);
   cout << val; // how come val is still 0 and not 10?
}

注意 val 仍为0，即使函数正在更改要传递的参数。要解决此问题，请将引用传递给将要更改的值：

Note that val is still 0, even though the function is changing the parameter that is being passed. To fix this problem, you pass a reference to the value that will be changed:

void foo(int& x)
{
   x = 10;
}

int main()
{
   int val = 0;
   foo(val);
   cout << val; // now val is 10
}

使用指针，规则不会改变。您需要传递一个指针的引用，以使更改反映回调用者：

With pointers, the rules don't change. You need to pass a reference to the pointer to have the change reflect back to the caller:

int read(char*& buffer)
{
    int fileSize;
    ifstream myfile ("inputFile");
    if (myfile.is_open())
    {
        fileSize = getFileLength(myfile); 
        buffer = new char[fileSize];

        myfile.read(buffer, fileSize);
        myfile.close();
    }
    return fileSize;
}

现在在该函数不是本地副本，而是对您传递的变量的引用。

Now the buffer in that function is not a local copy, but a reference to the variable you passed.

另一种方法（更多的是C风格）是传递一个指针到你想改变的东西。您要更改指针值，因此您将指针传递给指针：

The other method (which is more "C" style) is to pass a pointer to the thing you want to change. You want to change the pointer value, so you pass a pointer to the pointer:

int read(char** buffer)
{
    int fileSize;
    ifstream myfile ("inputFile");
    if (myfile.is_open())
    {
        fileSize = getFileLength(myfile); 
        *buffer = new char[fileSize];

        myfile.read(buffer, fileSize);
        myfile.close();
    }
    return fileSize;
}

// the caller
char *buffer;
//...
read(&buffer);

当然，我们必须更改语法，因为它是一个正在传递的指针，因此我们需要解除引用。

Of course, we have to change the syntax since it is a pointer that is being passed, thus we need to dereference it.

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