为什么在返回类型中使用 char * 而不是 const char * 会导致崩溃？

发布于 2024-08-04 12:07:21 字数 264 浏览 2 评论 0原文

我在某处读到，如果您希望 C/C++ 函数返回字符数组（而不是 std::string），则必须返回 const char* 而不是 char*。执行后者可能会导致程序崩溃。

有人能解释一下这是否属实吗？如果这是真的，为什么从函数返回 char* 如此危险？谢谢。

const char * my_function()
{
    ....
}

void main(void)
{
    char x[] = my_function();
}

原文

I read somewhere that if you want a C/C++ function to return a character array (as opposed to std::string), you must return const char* rather than char*. Doing the latter may cause the program to crash.

Would anybody be able to explain whether this is true or not? If it is true, why is returning a char* from a function so dangerous? Thank you.

const char * my_function()
{
    ....
}

void main(void)
{
    char x[] = my_function();
}

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ぺ禁宫浮华殁 2024-08-11 12:07:21

如果您有一个返回“字符串文字”的函数，那么它必须返回 const char*。这些不需要由 malloc 在堆上分配，因为它们被编译成可执行文件本身的只读部分。

例子：

const char* errstr(int err)
{
    switch(err) {
        case 1: return "error 1";
        case 2: return "error 2";
        case 3: return "error 3";
        case 255: return "error 255 to make this sparse so people don't ask me why I didn't use an array of const char*";
        default: return "unknown error";
    }
}

If you have a function that returns "string literals" then it must return const char*. These do not need to be allocated on the heap by malloc because they are compiled into a read-only section of the executable itself.

Example:

const char* errstr(int err)
{
    switch(err) {
        case 1: return "error 1";
        case 2: return "error 2";
        case 3: return "error 3";
        case 255: return "error 255 to make this sparse so people don't ask me why I didn't use an array of const char*";
        default: return "unknown error";
    }
}

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囚我心虐我身 2024-08-11 12:07:21

您被告知的情况不是真的。

返回 const char * 可以改进函数的语义（即不要混淆我给你的内容），但返回 char * 就完全没问题了。

但是，无论哪种情况，您必须确保返回一个 char * 或 const char *在 my_function 的堆上分配（即使用 malloc 或 new 分配），否则每当 my_function 返回时，内存因为 [const] char * 将被释放，并且您将访问无效的指针。

最后，您必须记住释放或删除[const] char *一旦你完成它，它就会返回给你，否则你会泄漏内存。 C/C++ 不是很棒的语言吗？

所以，在 C 语言中，你会得到

const char *my_function() {
    const char *my_str = (const char *)malloc(MY_STR_LEN + 1);  // +1 for null terminator.
    /* ... */
    return my_str;
}

int main() {
    const char *my_str = my_function();
    /* ... */
    free(my_str);
    /* ... */
    return 0;
}

What you were told is not true.

Returning a const char * can improve the semantics of a function (i.e. don't mess with what I'm giving you) but returning a char * is perfectly fine.

However, in either case, you must make sure that you return a char * or const char * that was allocated on the heap in my_function (i.e. allocated using malloc or new), otherwise whenever my_function returns, the memory for the [const] char * will be deallocated, and you will be accessing an invalid pointer.

And finally you must remember to free or delete the [const] char * that's been returned to you once you're done with it, or you will leak memory. Aren't C/C++ such great languages?

So, in C, you would have

const char *my_function() {
    const char *my_str = (const char *)malloc(MY_STR_LEN + 1);  // +1 for null terminator.
    /* ... */
    return my_str;
}

int main() {
    const char *my_str = my_function();
    /* ... */
    free(my_str);
    /* ... */
    return 0;
}

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