如果 x 是指针，&x 与 x 有何不同？

发布于 2024-11-09 17:02:31 字数 363 浏览 3 评论 0原文

所以...我有以下代码：

int main(void)
{
const char *s="hello, world";
cout<<&s[7]<<endl;

return 0;
}

它打印“world”...但是我不明白为什么会这样：

int main(void)
{
const char *s="hello, world";
cout<<s[7]<<endl;

return 0;
}

只会打印“w”（我所做的只是去掉了＆符号），但我认为是“地址”运算符......这让我想知道为什么你需要它以及它的功能是什么？

原文

So...I have the following code:

int main(void)
{
const char *s="hello, world";
cout<<&s[7]<<endl;

return 0;
}

and it prints "world"...however I don't understand why this:

int main(void)
{
const char *s="hello, world";
cout<<s[7]<<endl;

return 0;
}

only will print "w" (all I changed is got rid of the ampersand), but I thought that was the "address of" operator...which makes me wonder why you need it and what it's function is?

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一紙繁鸢 2024-11-16 17:02:31

s[7] 是字符 'w'，因此 &s[7] 就变成了地址字符'w'。当您将 char* 类型的地址传递给cout时，它会打印从该字符开始到空字符（即结尾）的所有字符。字符串，即它继续从 'w' 开始打印字符，直到找到 \0 。这就是 world 的打印方式。

就像这样，

const char *s="hello, world";
const char  *pchar = &s[7]; //initialize pchar with the address of `w`
cout<< pchar <<endl; //prints all chars till it finds `\0`

输出：

world

但是，如果你想打印s[7]的地址，即&s[7]的值，那么你必须这样做：

cout << ((void*)&s[7]) << endl; //it prints the address of s[7]

现在我们传递一个 void* 类型的地址，因此 cout 将无法推断出它用来推断 char* 的内容> 输入地址。 void* 隐藏有关地址内容的所有信息。因此，cout 只是打印地址本身。毕竟，这才是它最多得到的。

在线演示：http://www.ideone.com/BMhFy

s[7] is the character 'w', so &s[7] becomes the address of the character 'w'. And when you pass an address of char* type to cout, it prints all characters starting from the character to the null character which is end of the string i.e it continues to print characters from 'w' onward till it finds \0 . That is how world gets printed.

Its like this,

const char *s="hello, world";
const char  *pchar = &s[7]; //initialize pchar with the address of `w`
cout<< pchar <<endl; //prints all chars till it finds `\0`

Output:

world

However, if you want to print the address of s[7], i.e value of &s[7], then you've to do this:

cout << ((void*)&s[7]) << endl; //it prints the address of s[7]

Now we pass an address of void* type, so cout wouldn't be able to infer what it used to infer with char* type address. void* hides all information regarding the content of the address. So cout simply prints the address itself. After all, that is what it gets at the most.

Online demonstration : http://www.ideone.com/BMhFy

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