char *和char 之间的区别

我已经阅读了很多关于它的线索和问题并阅读了很多答案,但仍然难以理解差异以及何时应该使用什么?

我认为你应该在需要存储数据时使用char *而你不知道它的大小,因为它是动态的。 另外我不确定我是对的,但是如果你声明一个char *并且给它分配一个像这样的文字文本,我不明白我的理解:char * ch =“hi”; 它是一个你无法改变的常量,如果你试图改变它,你只需将ch指向另一个分配的存储空间来保存新字符串? 如果这样写:char ch = malloc(20); 然后你可以改变价值,如果你这样做:char ch [] =“hi”; char pch = ch; 您可以更改值,因为您指向数组并且数组指向ch [0]?

所有用粗体写的都是我从阅读中理解的,虽然我可能对我刚才说的大部分内容都是错的,这就是为什么我需要一个非常好的简单解释,所以我可以一劳永逸地理解差异,当我应该用什么。

编辑:

#include  main() { char ch[] = "Hello"; char *p1 = ch; char *p2 = p1; char *p3 = *p1; printf("ch : %s\n", ch); printf("p1 address [%d] value is %s\n", p1, *p1); printf("p2 address [%d] value is %s\n", p2, *p2); printf("p3 address [%d] value is %s\n", p3, *p3); return 0; } 

最直截了当的答案是:

这里的区别在于

 char *s = "Hello world"; 

将Hello世界放在内存的只读部分并制作sa指针,使得对此内存的任何写入操作都是非法的。 做的时候:

 char s[] = "Hello world"; 

将文字字符串放在只读内存中,并将字符串复制到堆栈上新分配的内存中。 从而制作

 s[0] = 'J'; 

法律。

更长的解释将包括存储器存储的段,以及分配的内存量:

 Example: Allocation Type: Read/Write: Storage Location: Memory Used (Bytes): =========================================================================================================== const char* str = "Stack"; Static Read-only Code segment 6 (5 chars plus '\0') char* str = "Stack"; Static Read-only Code segment 6 (5 chars plus '\0') char* str = malloc(...); Dynamic Read-write Heap Amount passed to malloc char str[] = "Stack"; Static Read-write Stack 6 (5 chars plus '\0') char strGlobal[10] = "Global"; Static Read-write Data Segment (R/W) 10 

参考


  1. C中的char s []和char * s有什么区别? ,访问日期2014-09-03,
  2. 声明的字符串和分配的字符串之间的区别 ,访问2014-09-03,

编辑


为了解决问题中的编辑和随之发布的评论,我在您的解决方案中添加了注释:

 #include  int main() { char ch[] = "Hello"; /* OK; Creating an array of 6 bytes to store * 'H', 'e', 'l', 'l', 'o', '\0' */ char *p1 = ch; /* OK; Creating a pointer that points to the * "Hello" string. */ char *p2 = p1; /* OK; Creating a second pointer that also * points to the "Hello" string. */ char *p3 = *p1; /* BAD; You are assigning an actual character * (*p1) to a pointer-to-char variable (p3); * It might be more intuitive if written in * 2 lines: * char* p3; * p3 = *p1; //BAD */ printf("ch : %s\n", ch); /* OK */ printf("p1 address [%d] value is %s\n", p1, *p1); /* Bad format specifiers */ printf("p2 address [%d] value is %s\n", p2, *p2); /* Bad format specifiers */ printf("p3 address [%d] value is %s\n", p3, *p3); /* Bad format specifiers */ return 0; } 

所以,三个主要的错误。

  1. 您正在为pointer-to-char变量赋值char值。 您的编译器应该警告您这一点。 ( char *p3 = *p1 )。
  2. 根据您的编译器,您可能必须使用指针%p格式说明符来打印地址而不是%d (整数)格式说明符。
  3. 您使用字符串%s说明符与char数据类型(即: printf("%s", 'c')是错误的)。 如果要打印单个字符,则使用%c格式说明符,匹配参数应为字符(即:’c’,char b等)。 如果要打印整个字符串,则使用%s格式说明符,参数是指向char指针

例子


 #include  int main(void) { char c = 'H'; // A character char* pC = &c; // A pointer to a single character; IS NOT A STRING char cArray[] = { 'H', 'e', 'l', 'l', 'o' }; // An array of characters; IS NOT A STRING char cString[] = { 'H', 'e', 'l', 'l', 'o', '\0' }; // An array of characters with a trailing NULL charcter; THIS IS A C-STYLE STRING // You could also replace the '\0' with 0 or NULL, ie: //char cString[] = { 'H', 'e', 'l', 'l', 'o', (char)0 }; //char cString[] = { 'H', 'e', 'l', 'l', 'o', NULL }; const char* myString = "Hello world!"; // A C-style string; the '\0' is added automatically for you printf("%s\n", myString); // OK; Prints a string stored in a variable printf("%s\n", "Ducks rock!"); // OK; Prints a string LITERAL; Notice the use of DOUBLE quotes, " " printf("%s\n", cString); // OK; Prints a string stored in a variable printf("%c\n", c); // OK; Prints a character printf("%c\n", *pC); // OK; Prints a character stored in the location that pC points to printf("%c\n", 'J'); // OK; Prints a character LITERAL; Notice the use of SINGLE quotes, ' ' /* The following are wrong, and your compiler should be spitting out warnings or even not allowing the * code to compile. They will almost certainly cause a segmentation fault. Uncomment them if you * want to see for yourself by removing the "#if 0" and "#endif" statements. */ #if 0 printf("%s\n", c); // WRONG; Is attempting to print a character as a string, similar // to what you are doing. printf("%s\n", *pC); // WRONG; Is attempting to print a character as a string. This is // EXACTLY what you are doing. printf("%s\n", cArray); // WRONG; cArray is a character ARRAY, not a C-style string, which is just // a character array with the '\0' character at the end; printf // will continue printing whatever follows the end of the string (ie: // random memory, junk, etc) until it encounters a zero stored in memory. #endif return 0; } 

代码清单 – 提议的解决方案


 #include  int main() { char ch[] = "Hello"; /* OK; Creating an array of 6 bytes to store * 'H', 'e', 'l', 'l', 'o', '\0' */ char *p1 = ch; /* OK; Creating a pointer that points to the * "Hello" string. */ char *p2 = p1; /* OK; Creating a second pointer that also * points to the "Hello" string. */ char *p3 = p1; /* OK; Assigning a pointer-to-char to a * pointer-to-char variables. */ printf("ch : %s\n", ch); /* OK */ printf("p1 address [%p] value is %s\n", p1, p1); /* Fixed format specifiers */ printf("p2 address [%p] value is %s\n", p2, p2); /* Fixed format specifiers */ printf("p3 address [%p] value is %s\n", p3, p3); /* Fixed format specifiers */ return 0; } 

样本输出


 ch : Hello p1 address [0x7fff58e45666] value is Hello p2 address [0x7fff58e45666] value is Hello p3 address [0x7fff58e45666] value is Hello