C/C ++中的sizeof char *数组 [英] sizeof char* array in C/C++

问题描述

有很多类似的查询，但就我而言，我不明白什么是行不通的:

There are plenty of similar inquires, but in my case I don't understand what isn't working:

int mysize = 0;
mysize = sizeof(samplestring) / sizeof(*samplestring);
std::cout << mysize << '\n' << samplestring;

这将输出:

4

Press 'q' to quit.

怎么可能?4绝对不是此字符串的大小.我什至尝试了以下方法，结果相同:

How is it possible? 4 definitely isn't the size of this string. I even tried the following, with the same result:

mysize = sizeof(samplestring)/sizeof(samplestring [0]);

好的，这是声明:

char *samplestring = "Start."; 

我使用的是C ++，但是我需要使用仅接受char *的函数.稍后在代码中，我将新字符串分配给该变量，例如:

I'm on C++, but I need to use functions that only accept char *. Later in the code I assign new strings to that variable, like:

samplestring = "Press 'r' for red text.";

是的，编译器会向我发出警告，但是我不知道如果无法覆盖它们，该如何使用不同的字符串...

Yes, the compiler gives me warnings, but I have no idea how can I use different strings if I can't overwrite them...

推荐答案

4 不是字符串的大小，因为 samplestring 不是字符串.这是一个 char * ，其大小(在您的平台上)为4除以1( char 的大小)正确为4.

4 isn't the size of the string, because samplestring isn't a string. It's a char*, whose size is (on your platform) 4, divided by 1 (size of char) is, correctly, 4.

在C ++中，您将使用 std :: string 和 length()方法.

In C++, you'd use std::string and the length() method.

在C语言中，您将使用 strlen ，它将以NULL终止的char指针作为参数.

In C, you'd use strlen which takes as parameter a NULL-terminated char pointer.

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