c++ - 关于c语言指针的一个简单问题

Question

今天碰到了一个简单的题，可是我却有点不理解，代码是这样的：

#include<iostream>
void inc(char *p)
{
    p = p + 1;
}
int main()
{
    char s[] = { '1', '2', '3' };
    inc(s);
    std::cout << *s << std::endl;
    return 0;
}

需要你判断输出的是什么。
我很确定若想输出‘2’，则inc函数内的代码应该是：

*p=*(p+1);

但是为什么源代码就不能正确run了呢？s本身就是指向数组的第一个元素的指针，我的inc中取的是不是p的地址呢？若是，则为什么不能正确运行？
我感觉这个题目很基础了，可是我却似乎没有学透，请问我下一步要怎么加强对指针的理解呢？
谢谢大家。

Answer 1

First, s is immutable;
Second, the C language is transfer by value. The s in your inc is a copy of s and has no impact on s;
Third, are you convinced? The writing does output 2, but you assign s[1] to s[0].

Answer 2

After a slight modification, you know that s is actually an address (pointer), and it points to the first byte, &s[0] == s. Then to change the value inside it, it should eventually be *P = 2; for example, int i =1; when changing this value across functions, you must quote its address, such as int * p = &i, and write *p = 2 like this; here *p==i ; Note that * plus the pointer variable points to the value in this address (pointer). Because changing values ​​across functions usually requires an address to lock the value, and then the value can be changed. So you can change it like this:

#include<iostream>
    void inc(char *p)
    {
        *p = *p + 1;
    }
    int main()
    {
        char s[] = { '1', '2', '3' };
        inc(s);
        std::cout << *s << std::endl;
        return 0;
    }
    

Your mistake is that you directly write p = p+1; you are actually trying to change your newly defined pointer variable. You should have learned C++ directly without learning C, or have you not learned C pointers well?

Answer 3

Mobile version + C4Droid code, please forgive me for the confusing layout.


It can be seen that passing in a pointer can change the object pointed to by the pointer.
The parameter (here a pointer) passed to the function is a copy. After the function is executed, the copy will be destroyed and the original value will be used.
Also, you can try using pointers to pointers.