为什么我需要的char [K + 1]而不是CHAR [K]与长度为k的字符串？ [英] Why do I need char[k + 1] instead of char[k] for a string with length k?

问题描述

所以我有一个简单的设置code的：

So I have a simple set of code:

#include <stdio.h>

int main()
{
  char x[3] = "ABC"; // (*)
  puts(x);
  return 0;
}

它返回一个奇怪的输出：

It returns a strange output:

ABC¬ a

使用从上面的答案<一个href=\"http://stackoverflow.com/questions/31057309/why-is-there-a-question-mark-at-the-end-of-the-string\">this问题，我发现，当我换 X [3] 到 X [4] 一切都正常运行

Using the top answer from this question, I found that when I change x[3] to x[4] everything runs fine.

但是，为什么？我为什么会在 X奇怪的输出[3] ，为什么 X [4] 罚款？

But why? Why do I get a strange output on x[3], and why is x[4] fine?

推荐答案

既然你问为什么这个收益率 ABC -a ，这里是一个解释：你的字符X [3] =ABC不适合看跌。 看跌预计到零结束的字符串。然而，你的 X 基本上是：

Since you've asked "why" this yields ABC -a, here's an explanation: your char x[3] = "ABC" isn't well suited for puts. puts expects a string terminated by zero. However, your x is basically:

char x[3] = {'A', 'B', 'C'};

如你所知，有没有办法得到一个（动态）数组的长度：

As you know, there's no way to get the length of a (dynamic) array:

char * allocate(){
   return malloc(rand() + 1);
}

char * mem = allocate(); // how large is mem??

有没有办法让你知道它有多长。但是，打印字符串，它不是别的，只是在内存中的字符的连续序列，函数需要知道当字符串（又名字符序列）结束。

There's no way for you to know how long it is. However, to print a string which is nothing else than a continuous sequence of characters in memory, a function needs to know when the string (aka the character sequence) ends.

这就是为什么美国标准code为信息交换（ASCII）和许多其他字符集包含空字符。它基本上字符值为 0 ：

That's why the American Standard Code for Information Interchange (ASCII) and many other character sets contain the null character. It's basically char with value 0:

char wrong_abc[3]   = {'A', 'B', 'C'};     // when does it end?
char correct_abc[4] = {'A', 'B', 'C', 0 }; // oh, there's a zero!

现在的功能就像看跌可以简单地检查 0 ：

Now functions like puts can simply check for 0:

// Simplified, actual "puts" checks for errors and returns
// EOF on error or a non-negative int on succes.
void puts(const char * str){
   int i = 0;

   while(str[i] != 0){
      putchar(str[i]);
      i++;
   }

   putchar('\n');
}

这就是为什么你

• 需要为在该字符序列+1所有字符存储器，


• 得到了一个未定义的行为，当你忘记了 0 。

• need memory for all characters in the character sequence +1,
• get undefined behaviour when you forget the 0.

看跌期权的执行上面再也找不到 0 时，不慎留下你自己的内存（或访问其他数据），这通常会导致一个段错误或其他错误（或者更糟的是，没有得到检测相当长的时间，然后产生关键错误）。在这种情况下，实际的行为是不确定的。

The implementation of puts above would never find 0 and accidentally leave the memory you own (or access other data), which usually leads to a segfault or other errors (or worse, doesn't get detected for a long time and then yields critical errors). The actual behaviour in such a situation is undefined.

注意字符串（例如ABC）在结束时自动有一个'\\ 0'。此外，编译器是足够聪明图文字的长度为你，让你可以简单地使用

Note that string literals (e.g. "ABC") automatically have a '\0' at the end. Also, the compiler is smart enough to figure the length of the literal for you, so you can simply use

char x[] = "ABC";

这样，您不必如果你改变了文字后的后顾之忧。

That way, you don't have to worry if you change the literal later.

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