我可以在C ++中使用这个C风格的字符串获得一些帮助吗？

Question

I'm trying to understand un-managed code. I come from a background of C# and I'm playing around a little with C++.

Why is that this code:

#include <iostream>

using namespace std;

int main()
{
    char s[] = "sizeme";

    cout << sizeof(s);

    int i = 0;
    while(i<sizeof(s))
    {
        cout<<"\nindex "<<i<<":"<<s[i];
        i++;
    }
    return 0;
}

prints out this:

7
index 0:s
index 1:i
index 2:z
index 3:e
index 4:m
index 5:e
index 6:

????

Shouldn't sizeof() return 6?

Answer 1

C strings are "nul-terminated" which means there is an additional byte with value 0x00 at the end. When you call sizeof(s) , you are getting the size of the entire buffer including the nul terminator. When you call strlen(s) , you are getting the length of the string contained in the buffer, not including the nul.

Note that if you modify the contents of s and put a nul terminator somewhere other than at the end, then sizeof(s) would still be 7 (because that's a static property of how s is declared) but strlen(s) could be somewhat less (because that's calculated at runtime).

Answer 2

No, all trings in C are terminated by the null character (ascii 0). So s is actually 7 bytes

s i z e m e \0

Answer 3

这是因为C字符串包含值0 （或'\\0' ）作为标记字符串结尾的最后一个字符。

Answer 4

s是7个字节，6个用于字符串，1个用于空终止。