指针和字符串文字 [英] pointers and string literals

问题描述

很多次我看到以下语句：

Many a times I have seen the following statements:

char* ch = "Hello"
cout<<ch;

我得到的输出是 Hello 。我知道ch指向字符串Hello的第一个字符，Hello是一个字符串字面量并存储在只读存储器中。因为，ch将第一个字符的地址存储在字符串文字中，因此不应该使用语句

The output I get is "Hello". I know that ch points to the first character of the string "Hello" and that "Hello" is a string literal and stored in read only memory. Since, ch stores the address of the first character in the string literal, so shouldn't the statement,

cout<<ch;

给出输出字符地址，一个指针变量？相反，它打印字符串字面量本身。
此外，如果我写这个，

give the output "the address of a character" because, it is a pointer variable? Instead it prints the string literal itself. Moreover, if I write this,

ch++;
cout<<ch;

它提供输出 ello 。类似地，它也发生在更多的连续的ch ++语句。


可以有人告诉我，为什么会发生？

Btw，我已经看到了与字符串字面量有关的其他问题，但是所有这些都解决了为什么我们不能做类似* ch ='a'的事情？

EDIT：我也想参考这个如果我键入，C也会发生，

It gives the output, "ello". And similarly, it happens with more consecutive ch++ statements too.

can anybody tell me, why does it happen?
Btw, I have seen other questions related to string literals, but all of them address the issue of "Why can't we do something like *ch='a'?
EDIT : I also want to ask this in reference to C, it happens in C too, if I type,

printf("%s",ch);

为什么？

推荐答案

有运算符的重载版本

ostream& operator<< (ostream& , const char* );

：

char *p = "Hello";
cout << p;

这个重载定义了如何打印字符串而不是打印地址。但是，如果要打印地址，使用，

This overload defines how to print that string rather than print the address. However, if you want to print the address, you can use,

cout << (void*) p;

，因为这将调用另一个重载，只打印地址流： -

as this would call another overload which just prints the address to stream:-

ostream& operator<< (ostream& , void* );

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