cout<<使用 char* 参数打印字符串，而不是指针值 [英] cout << with char* argument prints string, not pointer value

问题描述

这个:

const char * terry = "hello";
cout<<terry;

打印hello 而不是'h' 的内存地址.为什么会发生这种情况?

prints hello instead of the memory address of the 'h'. Why is this happening?

推荐答案

这样做的原因是 std::cout 会将 char * 视为指向(第一个字符)一个 C 风格的字符串并打印它.如果你想要地址，你可以把它转换成一个不是那样处理的指针，比如:

The reason for that is that std::cout will treat a char * as a pointer to (the first character of) a C-style string and print it as such. If you want the address instead, you can just cast it to a pointer that isn't treated that way, something like:

cout << (void *) terry;

(或使用 const void * 强制转换，如果您担心丢弃常量，在这种特殊情况下这不是问题).

(or use the const void * cast if you're worried about casting away constness, something that's not an issue in this particular case).

如果您更喜欢纯粹主义者而不是实用主义者，您还可以使用 C++ static_cast，如下所示:

If you're more of a purist than pragmatist, you can also use the C++ static_cast, along the lines of:

cout << static_cast <const void *> (terry);

虽然在这种特殊情况下没有必要，但转换为 void * 会正常工作.以下示例代码显示了所有这些选项的作用:

though it's unnecessary in this particular case, the cast to a void * will work fine. The following sample code shows all these options in action:

#include <iostream>
int main (void) {
    const char *terry = "hello";
    std::cout << terry << '
';
    std::cout << (void *) terry << '
';
    std::cout << (const void *) terry << '
';
    std::cout << static_cast<const void *> (terry) << '
';
    return 0;
}

输出(地址在你的环境中可能不同):

outputting (the address may be different in your environment):

hello
0x8048870
0x8048870
0x8048870

请注意，在使用 static_cast 时，您应该确保不要尝试使用 static_cast 抛弃常量性(这就是 const_cast 是为了).这是由较新的 C++ 强制转换完成的检查之一，旧式强制转换没有此限制.

Note that, when using the static_cast, you should ensure you don't try to cast away the constness with static_cast <void *> (that's what const_cast is for). This is one of the checks done by the newer C++ casts and the old-style cast does not have this limitation.

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