char *p = "hello";
时间: 2024-05-09 22:18:55 浏览: 15
This line declares a pointer variable named "p" that points to a string literal "hello". The string literal is a sequence of characters terminated by a null character '\0' and is stored in read-only memory.
Since the pointer is not declared as const, it is possible to modify the memory that it points to. However, modifying the memory that contains a string literal can lead to undefined behavior. Therefore, it is recommended to declare the pointer as const to prevent accidental modification of the string literal.
Example:
```c
const char *p = "hello";
```
This ensures that the string literal is not modified through the pointer "p".
相关问题
char* p="hello";*p输出多少
在 C 语言中,`char* p = "hello";` 定义了一个指向字符串常量 "hello" 的指针 `p`。当执行 `*p` 时,会输出指针 `p` 所指向的字符,即字符串的第一个字符。
因此,执行 `*p` 将输出字符 "h"。
以下是示例代码:
```c
#include <stdio.h>
int main() {
char* p = "hello";
printf("%c\n", *p); // 输出字符 "h"
return 0;
}
```
输出结果为:
```
h
```
char *s;*p=s;
根据提供的引用内容,`char *s`是一个指向字符的指针,`*p=s`是将`s`指向的字符赋值给指针`p`所指向的内存地址。但是在这个表达式中,指针`p`并没有被初始化,因此这个表达式会导致未定义的行为。正确的做法应该是先为指针`p`分配内存,然后再将`s`指向的字符赋值给`p`所指向的内存地址。
示例代码如下:
```c
char *s = "Hello World"; // 定义一个指向字符串的指针s
char *p = malloc(sizeof(char)); // 为指针p分配内存
*p = *s; // 将s指向的字符赋值给p所指向的内存地址
printf("%c", *p); // 输出:H
free(p); // 释放p所指向的内存
```