C ++中字符的指针 [英] Pointer of a character in C++

问题描述

翻阅书籍，第一行指示行应该向我显示 char变量b 存储位置的地址，对于 int变量a .但是第一个cout语句打印出一个奇怪的'dh ^#'，而第二个语句正确打印出一个十六进制值'ox23fd68'.为什么会这样?

Going by the books, the first cout line should print me the address of the location where the char variable b is stored, which seems to be the case for the int variable a too. But the first cout statement prints out an odd 'dh^#' while the second statement correctly prints a hex value ' ox23fd68'. Why is this happening?

 #include<iostream>
    using namespace std;

    int main()
    {
        char b='d';
        int a=10;
        char *c=new char[10];
        c=&b;
        int *e=&a;
        cout<<"c: "<<c<<endl;
        cout<<"e: "<<e;
    }

推荐答案

实际上，该程序有问题.内存泄漏.

Actually this program has problem. There is a memory leak.

char *c=new char[10];
c=&b;

这会在堆上分配10个字符，但是指向堆的指针会被变量 b 的地址覆盖.

This allocates 10 characters on heap, but then the pointer to heap is overwritten with the address of the variable b.

当使用 operator<< 将 char * 写入 cout 时，则将其视为空终止的C字符串.随着b的地址被初始化为包含 d op<< 的单个字符，继续在堆栈中搜索找到第一个空字符.看来是在几个字符之后找到的，所以写了 dh ^#( d 是变量 b 的值，其余的是只是在第一个 \ 0 char之前在堆栈上找到的一些随机字符).

When a char* is written to cout with operator<< then it is considered as a null terminated C-string. As the address of b was initialized to a single character containing d op<< continues to search on the stack finding the first null character. It seems the it was found after a few characters, so dh^# is written (the d is the value of variable b the rest is just some random characters found on the stack before the 1st \0 char).

如果要获取地址，请尝试使用 static_cast< void *>(c).

If you want to get the address try to use static_cast<void*>(c).

我的例子:

int main() {
  char *c;
  char b = 'd';
  c = &b; 
  cout << c << ", " << static_cast<void*>(c) << endl;
}

输出:

dÌÿÿ, 0xffffcc07

看到'd'之后的奇怪字符.

See the strange characters after 'd'.

我希望这可以有所帮助！

I hope this could help a bit!

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