让一个char *的副本 [英] Make a copy of a char*

问题描述

我有一个接受一个char *作为它的一个参数的函数。我需要对其进行操作，但保留原始的char *完好无损。从本质上讲，我想创建此CHAR的工作副本*。看起来这应该很容易，但我真的很挣扎。

I have a function that accepts a char* as one of its parameters. I need to manipulate it, but leave the original char* intact. Essentially, I want to create a working copy of this char*. It seems like this should be easy, but I am really struggling.

我的第一个（天真）的尝试是创建另一个字符*，并设置它等于原来的：

My first (naive) attempt was to create another char* and set it equal to the original:

char* linkCopy = link;

这是不行的，当然，因为我所做的就是使他们指向同一个地方。

This doesn't work, of course, because all I did was cause them to point to the same place.

我应该用函数strncpy做到这一点？

Should I use strncpy to accomplish this?

我曾尝试以下，但它会导致崩溃：

I have tried the following, but it causes a crash:

char linkCopy[sizeof(link)] = strncpy(linkCopy, link, sizeof(link));

我失去了一些东西明显...？

Am I missing something obvious...?

编辑：我的道歉，我正在努力简化的例子，但我留下了一些在第二个例子中长变量名。固定的。

My apologies, I was trying to simplify the examples, but I left some of the longer variable names in the second example. Fixed.

推荐答案

的的sizeof 会给你指针的大小。这往往是4或8取决于你处理器/编译器，但不包括字符串的大小指出。您可以使用strlen和strcpy：

The sizeof will give you the size of the pointer. Which is often 4 or 8 depending on your processor/compiler, but not the size of the string pointed to. You can use strlen and strcpy:

// +1 because of '\0' at the end
char * copy = malloc(strlen(original) + 1); 
strcpy(copy, original);
...
free(copy); // at the end, free it again.

我见过一些答案建议使用的strdup ，但是这是一个POSIX函数，而不是C的一部分。

I've seen some answers propose use of strdup, but that's a posix function, and not part of C.

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