char * - 为什么指针中没有地址？ [英] char* - Why is there no address in the pointer?

问题描述

我有一个关于 char * 的基本问题我不明白

I have a basic question about char* I dont understand

char* aString = "Hello Stackoverflow";

指针指向字符链的第一个字符。

The Pointer points at the first character of the character chain.

cout << *aString; // H

但是为什么整个字符串保存在指针中？

but why is the whole string saved in the pointer?

cout << aString //Hello Stackoverflow

我希望一个地址，不是地址保存在指针？ Hello Stackoverflow的地址在哪里？

I would expect an adress, aren´t adresses saved in pointers? Where is the adress of "Hello Stackoverflow"?

任何帮助非常感谢

推荐答案

有一个运算符<<（ostream& char const *）的重载，它输出从该指针开始的以null结束的字符串， ostream :: operator 将输出地址。

There is an overload for operator<<(ostream&, char const*) which output the null-terminated string starting at that pointer and which is preferred to the operator ostream::operator<<(void*) which would have output the address.

想要的地址，转换指针 void * 。

If you want the address, cast the pointer to void*.

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