字符串和指针用C [英] String and pointers in C

问题描述

 的#include＆LT;＆stdio.h中GT;
＃包括LT＆;＆string.h中GT;
＃包括LT＆;＆stdlib.h中GT;INT主要（无效）
{
        字符* A =的malloc（5 * sizeof的（炭））;
        一个=1;
        自由（一）;
}
 

我有一个关于上述code两个问题：

1. 为什么code给出了一个无差错（一）当我初始化是这样的：

  

在进行=1

和不给任何错误自由（一）当我初始化是这样的：

  

一个[0] ='1'。

<醇开始=2>

当我初始化与
一个[0] ='1'
和使用打印字符串
的printf（％S，一）为什么我得到的结果。

  

1

  
  没有以'\\ 0'，宣布第二个索引？

解决方案

与规则的的malloc 的和的免费的是，你必须只传递给的免费，是由一个呼叫分配到的的malloc 的指针（或类似的的realloc 或释放calloc 的）。你不这样做的：

 的char * A =的malloc（5 * sizeof的（炭））;
一个=1;
自由（一）;
 

当你调用的免费的，你通过的在的这是不通过调用的的malloc分配的。的这会导致你的程序有不确定的行为

---

在您的理解的问题是，我怀疑你有什么不明白的 A =1的一样。

我怀疑你能想象这在字符串中复制到您分配它没有记忆。它改变了指针的在的指向不同的地址。

事实上，分配给的在的，而不是它指向的内存，您正在泄漏分配的内存。您code是完全类似这样的：

  INT I = 1;
我= 2;
 

我敢肯定，你可以看到初始化的 I 的是没有意义的，因为你立即覆盖的的 I 的一个新值。这正是你和你的指针做了什么。

---

您可以复制与的strcpy 的字符串。像这样的：

 的char * A =的malloc（5 * sizeof的（炭））;
的strcpy（一个1）;
自由（一）;
 

同样的，当你写

 的char * A =的malloc（5 * sizeof的（炭））;
一个[0] ='1';
的printf（％S，一）;
自由（一）;
 

您是通过分配给您的呼叫分配的缓冲区的内容的的malloc 的。因此调用免费是正确的。

然而，当你这样做的字符串并不一定是空值终止。你观察到你的的printf 的似乎工作，但那只是偶然。

中的内容的 A [1] 的界定不清，似乎，一个偶然的机会，的 A [1] 的确实包含空当您运行您的程序。

这就是为什么你通常​​使用像一个字符串函数的的strcpy 的进行副本，以确保空终止。你当然也要注意不要溢出你的缓冲区。

---

最后，顾名思义，的sizeof（char）的== 1 的，所以你可以写的的malloc 的是这样的：

 的char * A =的malloc（5）;
 

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int main(void)
{
        char* a = malloc(5 * sizeof(char));
        a = "1";
        free(a);
}

I have two questions regarding the above code:

1. Why does the code gives an error with free(a) when I initialize a like this :

a = "1"

and doesn't give any error with free(a) when I initialize a like this :

a[0] = '1'.

2. When I initialize a with a[0] = '1' and print the string a using printf("%s", a) why do I get the result

   '1'

without declaring the second index with a '\0'?

解决方案

The rules with malloc and free are that you must only pass to free a pointer that was allocated by a call to malloc (or one of its equivalents like realloc or calloc). You do not do that:

char* a = malloc(5 * sizeof(char));
a = "1";
free(a);

When you call free, you pass a which was not allocated by a call to malloc. This results in your program having undefined behaviour.

---

The problem in your understanding is that I suspect you don't understand what a = "1" does.

I suspect that you imagine this to copy the string in to the memory that you allocated it does not. It changes the pointer a to point to a different address.

Indeed, by assigning to a, rather than the memory that it points to, you are leaking the allocated memory. Your code is exactly analogous to this:

int i = 1;
i = 2;

I'm sure you can see that the initialization of i is pointless because you immediately overwrite the value of i with a new value. That's exactly what you did with your pointer.

---

You can copy the string with strcpy. Like this:

char* a = malloc(5 * sizeof(char));
strcpy(a, "1");
free(a);

Similarly, when you wrote

char* a = malloc(5 * sizeof(char));
a[0] = '1';
printf("%s", a);
free(a);

you were assigning the contents of the buffer allocated by your call to malloc. And hence the call to free is correct.

However, the string is not necessarily null-terminated when you do that. You observed that your printf appeared to work but that was just by chance.

The contents of a[1] are ill-defined and it seems that, by chance, a[1] did contain a null when you ran your program.

This is why you typically use a string function like strcpy to perform copies and ensure null-termination. You do of course need to take care not to overrun your buffer.

---

Finally, by definition, sizeof(char) == 1 so you can write the malloc like this:

char* a = malloc(5);

这篇关于字符串和指针用C的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！