通过函数返回指向静态常量字符串的指针

I have a file which has a static constant string and a function which will return pointer to that string. File looks like this:

我有一个文件,它有一个静态常量字符串和一个函数,它将返回指向该字符串的指针。文件看起来像这样:

typedef unsigned char   BOOLEAN;

#define TRUE    1
#define FALSE   0


static const unsigned char MAL_Version[8] = "2.001";

/* Function to return Version string */
BOOLEAN GetVersion ( unsigned char* pu8Version )
{
    BOOLEAN success = FALSE;

    if(pu8Version != NULL)
    {
        pu8Version = &MAL_Version[0];
        success = TRUE;
        printf("\r\nTRUE");
        printf("\r\n%s", pu8Version);
    }

    return success;
}

and in main(), I declare an array and pass it's address to GetVersion function. When I do this, I am getting random characters.

在main()中,我声明一个数组并将其地址传递给GetVersion函数。当我这样做时,我会得到随机字符。

int main() {
   unsigned char buffer[10];

   GetVersion(buffer);
   printf("\r\n%s", buffer);
}

Output is:

TRUE

2.001

D�3�

What I am missing? The pointer in function is correctly printing the string, but when it returns, it prints garbage.

我错过了什么?函数中的指针正确打印字符串,但是当它返回时,它会打印垃圾。

3 个解决方案

#1

This statement

pu8Version = &MAL_Version[0];

only modifies the local pointer pu8Version in GetVersion() and that doesn't change buffer in main().

只修改GetVersion()中的本地指针pu8Version,并且不会更改main()中的缓冲区。

Instead of:

pu8Version = &MAL_Version[0];

you can copy the MAL_Version to buffer with:

你可以将MAL_Version复制到缓冲区:

strcpy(pu8Version, MAL_Version);

If you really don't need a copy of MAL_Version, you can also return the pointer to MAL_Version directly. Something like:

如果你真的不需要MAL_Version的副本,你也可以直接返回指向MAL_Version的指针。就像是:

/* Function to return Version string */
const char *GetVersion(void)
{
    return MAL_Version;
}

int main(void) {
   const char *version = GetVersion();
   printf("\n%s", version);
}

Note that you don't define a "BOOLEAN" yourself. bool (from <stdbool.h> header) type is available in C since C99.

请注意,您自己没有定义“BOOLEAN”。自C99起,bool(来自 header)类型在C中可用。

#2

The function parameter pu8Version is the local copy of the pointer passed.

函数参数pu8Version是传递的指针的本地副本。

You change pu8Version to point to the static string MAL_Version which prints correctly. On returning fom the function, the changed version of pu8Version is forgotten.

您将pu8Version更改为指向正确打印的静态字符串MAL_Version。在返回功能时,pu8Version的更改版本被遗忘了。

The original unsigned char buffer[10]; is uninitialised and remains so, so rubbish is printed.

原始的unsigned char缓冲区[10];是未经初始化的,仍然如此,所以打印垃圾。

Note you cannot copy a C string with the = operator. What that does is to change the pointer, but not what it is pointing to. You should use strcpy.

请注意,您无法使用=运算符复制C字符串。这样做是为了改变指针,而不是它指向的指针。你应该使用strcpy。

#3

Or you can pass a pointer to a pointer:

或者您可以将指针传递给指针:

/* Function to return Version string */
BOOLEAN GetVersion ( unsigned char **pu8Version )
{
    BOOLEAN success = FALSE;

    if(*pu8Version != NULL)
    {
        *pu8Version = MAL_Version;
        success = TRUE;
        printf("\r\nTRUE");
        printf("\r\n%s", *pu8Version);
    }

    return success;
}

int main() {
   unsigned char *buffer;

   GetVersion(&buffer);
   printf("\r\n%s", buffer);
}

#1