使用指针的strcpy [英] strcpy using pointers

问题描述

我试着写我自己的使用指针的strcpy，我在运行时出现错误。

I'm trying to write strcpy on my own using pointers and I get an error during runtime.

void str_cpy(char **destination, const char *source) {
//    char *s1 = *destination;

   while (*source != '\0') {
      **destination++ = *source++; //Get an error here
   }
   **destination = '\0';
}

我调用该函数如下：

I call the function as follows:

char *str = NULL;
str_cpy(&str, "String");

这难道不好吗？

谢谢！

推荐答案

没有，这是不行的。为什么？因为 STR 是 NULL 指针。它指向什么。当您尝试将值写入它，在那里他们将何去何从？它不指向任何分配的内存！

No, it's not okay. Why? Because str is a NULL pointer. It's pointing to nothing. When you try to write values into it, where will they go? It's not pointing to any allocated memory!

您必须先分配给 STR 内存。你可以这样做：

You first have to allocate memory for str. You can do:

char *str = malloc(strlen("String") + 1); // + 1 for the '\0' character at the end of C-style strings

或者，你可以这样做：

Or you can do:

char str[256]; // make str large enough to hold 256 chars. Note that this is not as safe as the above version!

此外，目标应该是一个指针，而不是一个双指针。好吧，这不是技术上的错误使用双指针，它只是不必要的。

Also, destination should be a single pointer, not a double pointer. Well, it's not technically wrong to use a double pointer, it's just unnecessary.

或者，您可以分配在 STR_CPY 函数的记忆，像这样：

Optionally, you can allocate the memory in the str_cpy function, like so:

void str_cpy(char **destination, const char *source) {
    *destination = malloc(strlen(source) + 1);
    // ... continue as normal

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