最直接的答案是：
这里的区别在于
char *s = "Hello world";
将Hello world放置在内存的只读部分并制作
s 一个指向它的指针，在这个内存上进行任何写操作
非法的。做的时候：
char s[] = "Hello world";
将文字字符串放入只读内存并将字符串复制到
堆栈上新分配的内存。从而使
s[0] = 'J';
合法的。
更冗长的解释包括内存存储在哪些段中，以及分配了多少内存：
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
参考文献
-
C 中 char s[] 和 char *s 有什么区别？，访问时间：2014-09-03，
<https://stackoverflow.com/questions/1704407/what-is-the-difference-between-char-s-and-char-s-in-c>
-
声明的字符串和分配的字符串的区别，访问时间：2014-09-03，
<https://stackoverflow.com/questions/16021454/difference-between-declared-string-and-allocated-string>
编辑
为了解决问题中的编辑以及随之而来的评论，我在您的解决方案中添加了注释：
#include <stdio.h>
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，三个主要错误。
- 您正在将
char 值分配给 pointer-to-char 变量。你的编译器应该警告你这一点。 (char *p3 = *p1)。
- 根据您的编译器，您可能必须使用指针
%p 格式说明符而不是%d（整数）格式说明符来打印地址。
- 您正在使用带有
char 数据类型的字符串%s 说明符（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，并且匹配的参数应该是一个字符（即：'c'、char b 等）。如果要打印整个字符串，则使用 %s 格式说明符，并且参数是 pointer-to-char。
示例
#include <stdio.h>
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 - 建议的解决方案
#include <stdio.h>
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