是否可以在 C 中将 char[] 转换为 char*? [英] Is it possible to convert char[] to char* in C?

问题描述

我正在做一个任务，我们必须将一系列字符串从一个文件读入一个数组.我必须在数组上调用密码算法(密码转置二维数组).因此，起初我将文件中的所有信息放入一个二维数组中，但我在其余代码中遇到了很多冲突类型的问题(特别是尝试将 char[] 设置为 char*).所以，我决定切换到一个指针数组，这使得我的大部分代码中的一切都变得更加容易.

I'm doing an assignment where we have to read a series of strings from a file into an array. I have to call a cipher algorithm on the array (cipher transposes 2D arrays). So, at first I put all the information from the file into a 2D array, but I had a lot of trouble with conflicting types in the rest of my code (specifically trying to set char[] to char*). So, I decided to switch to an array of pointers, which made everything a lot easier in most of my code.

但现在我需要将 char* 转换为 char[] 并再次转换回来，但我无法弄清楚.我一直无法在谷歌上找到任何东西.我开始怀疑这是否可能.

But now I need to convert char* to char[] and back again, but I can't figure it out. I haven't been able to find anything on google. I'm starting to wonder if it's even possible.

推荐答案

听起来你对指针和数组感到困惑.指针和数组(在本例中为 char * 和 char [])是 不是一回事.

It sounds like you're confused between pointers and arrays. Pointers and arrays (in this case char * and char []) are not the same thing.

• 一个数组char a[SIZE]表示a位置的值是一个长度为SIZE
的数组
• 指针char *a; 表示a 位置的值是指向char 的指针.这可以与指针算法相结合，使其表现得像一个数组(例如，a[10] 是 a 指向的 10 个条目)

• An array char a[SIZE] says that the value at the location of a is an array of length SIZE
• A pointer char *a; says that the value at the location of a is a pointer to a char. This can be combined with pointer arithmetic to behave like an array (eg, a[10] is 10 entries past wherever a points)

在记忆中，它看起来像这样(示例取自 常见问题解答):

In memory, it looks like this (example taken from the FAQ):

 char a[] = "hello";  // array

   +---+---+---+---+---+---+
a: | h | e | l | l | o | |
   +---+---+---+---+---+---+

 char *p = "world"; // pointer

   +-----+     +---+---+---+---+---+---+
p: |  *======> | w | o | r | l | d | |
   +-----+     +---+---+---+---+---+---+

指针和数组之间的区别很容易混淆，因为在许多情况下，数组引用衰减"到指向它的第一个元素的指针.这意味着在许多情况下(例如传递给函数调用时)数组成为指针.如果您想了解更多信息，C 常见问题解答的这一部分详细描述了不同之处.

It's easy to be confused about the difference between pointers and arrays, because in many cases, an array reference "decays" to a pointer to it's first element. This means that in many cases (such as when passed to a function call) arrays become pointers. If you'd like to know more, this section of the C FAQ describes the differences in detail.

一个主要的实际区别是编译器知道数组的长度.使用上面的例子:

One major practical difference is that the compiler knows how long an array is. Using the examples above:

char a[] = "hello";  
char *p =  "world";  

sizeof(a); // 6 - one byte for each character in the string,
           // one for the '' terminator
sizeof(p); // whatever the size of the pointer is
           // probably 4 or 8 on most machines (depending on whether it's a 
           // 32 or 64 bit machine)

如果没有看到您的代码，很难推荐最佳行动方案，但我怀疑更改为在任何地方使用指针将解决您当前遇到的问题.请注意:

Without seeing your code, it's hard to recommend the best course of action, but I suspect changing to use pointers everywhere will solve the problems you're currently having. Take note that now:

• 你需要在数组曾经所在的地方初始化内存.例如，char a[10]; 将变为 char *a = malloc(10 * sizeof(char));，然后检查 a !=空.请注意，在这种情况下，您实际上不需要说 sizeof(char)，因为 sizeof(char) 被定义为 1.为了完整起见，我将其保留.

• You will need to initialise memory wherever the arrays used to be. Eg, char a[10]; will become char *a = malloc(10 * sizeof(char));, followed by a check that a != NULL. Note that you don't actually need to say sizeof(char) in this case, because sizeof(char) is defined to be 1. I left it in for completeness.

您以前使用 sizeof(a) 表示数组长度的任何地方都需要替换为您分配的内存长度(如果您使用的是字符串，则可以使用 strlen()，直到 '').

Anywhere you previously had sizeof(a) for array length will need to be replaced by the length of the memory you allocated (if you're using strings, you could use strlen(), which counts up to the '').

您需要对 free() 用于每次调用 malloc().这告诉计算机您已完成使用 malloc() 请求的内存.如果您的指针是 a，只需在代码中您知道不再需要任何 a 点的位置写入 free(a);到.

You will need a make a corresponding call to free() for each call to malloc(). This tells the computer you are done using the memory you asked for with malloc(). If your pointer is a, just write free(a); at a point in the code where you know you no longer need whatever a points to.

正如另一个答案所指出的，如果你想获取一个数组的起始地址，你可以使用:

As another answer pointed out, if you want to get the address of the start of an array, you can use:

char* p = &a[0] 

可以理解为字符指针p变成a的元素[0]的地址".

You can read this as "char pointer p becomes the address of element [0] of a".

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