AAL*_*ALC 3 c arrays string pointers char
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是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,但是如果你宣布一个char*并且给它分配一个像这样的文字文本,我会理解我的理解:char*ch ="hi"; 它是一个常量,你无法改变,如果你试图改变它,你只需将ch指向另一个分配的内存空间保存新字符串？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]？
所有写的都是粗体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和简单的解释,所以我可以一劳永逸地理解差异,当我应该用什么.
编辑:
#include <stdio.h>
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;
}
Clo*_*oud 18
最直截了当的答案是:
这里的区别在于
Run Code Online (Sandbox Code Playgroud)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:
Run Code Online (Sandbox Code Playgroud)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
Run Code Online (Sandbox Code Playgroud)s[0] = 'J';法律.
更长的解释包括存储内存的段数,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
<https://stackoverflow.com/questions/1704407/what-is-the-difference-between-char-s-and-char-s-in-c><https://stackoverflow.com/questions/16021454/difference-between-declared-string-and-allocated-string> 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 <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,三大错误.
char为pointer-to-char变量赋值.您的编译器应该警告您这一点.(char *p3 = *p1).%p格式说明符来打印地址而不是%d(整数)格式说明符.%s带有char数据类型的字符串说明符(即:printf("%s", 'c')错误).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的指针.例子
#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