c char字符指针 [英] C Char pointers

问题描述

让我们说我们有一个指针数组：

的char * PTR [30];

现在这个工作得很好，而且似乎并不做任何意外！
我可以输入名字很容易。

  scanf函数（％S，PTR [1]）;
scanf的（％S，PTR [2]）;
scanf函数（％S，PTR [3]）;的printf（％S \\ n，PTR [1]）;
的printf（％S \\ n，PTR [2]）;
的printf（％S \\ n，PTR [3]）;
 

我的问题是，如果一个指针可以通过这种方式来存储姓名结束数使用，那么为什么使用malloc的？在这种情况下ptr的[1]不指向在输入而是以一个新的输入本身的字符。
对于例如，如果PTR有穆库尔，PTR [1]应指向'U'
如果未分配空间时，指针声明如下，有哪些限制？

解决方案

指针的无法的使用这样

。它可能的工作为你有时纯属偶然，但使用指针未分配内存会引起 未定义行为 - 这意味着您的程序可能运行不稳定，给你意想不到的效果。

记住，仅仅因为你的程序编译和运行，并不意味着它是正确的。在C语言中有一个全班同学称为未定义行为的错误，其中许多编译器会很乐意让你做无抱怨。这些措施包括覆盖了缓冲区，使用未初始化变量，或提领的指针不指向一个合法分配的内存块。表现出未定义行为的程序，甚至可能会出现正常工作有时 - 但他们通常很不稳定，容易发生崩溃

Let us say we have a array of pointers:

char *ptr[30];

Now this works fine, and doesn't seems to do anything unexpected! I can input names easily.

scanf("%s", ptr[1]);
scanf("%s", ptr[2]);
scanf("%s", ptr[3]);

printf("%s\n", ptr[1]);
printf("%s\n", ptr[2]);
printf("%s\n", ptr[3]);

My question is if a pointer can be used in this way to store end number of names, then why is malloc used.? and in this case ptr[1] does not point to the character in the input but to a new input itself. for eg if ptr has mukul, ptr[1] should point to 'u' and if space is not allocated when a pointer is declared like this, what are the limits.?

解决方案

The pointer cannot be used that way. It might "work" for you sometimes by sheer chance, but using a pointer without allocating memory for it will cause undefined behavior - meaning that your program may behave erratically and give you unexpected results.

Remember, just because your program compiles and runs doesn't mean it is correct. In the C language there is a whole class of errors known as "undefined behavior", many of which the compiler will happily let you do without complaining. These include overwriting a buffer, using an uninitialized variable, or dereferencing a pointer that doesn't point to a legitimately allocated memory block. Programs that exhibit undefined behavior may even appear to work normally sometimes - but they are usually very unstable and prone to crash.

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