strncpy()字符串长度输出错误

发布于 2024-12-17 07:35:25 字数 907 浏览 1 评论 0原文

我在 strncpy() 复制所需长度的额外字符时遇到问题。我需要将一个字符串的 10 个最高有效位复制到另一个字符串（均为 char* 格式）。

较大变量的大小直观上并不重要，较小变量仅在显着小于我需要的值时才会影响输出（我也不知道为什么会这样）。

我的 strncpy() 中的 n 值可能是导致问题的原因，但我不知道为什么它会这样。如果我没记错的话，目标字符串的长度 n 应该比数据长 1，以考虑末尾的空字符。

我的字符串根本不是这样的。将 n 设置为 11，我收到以下输出：

00000111101100100100001110100000

00000111101

我想象它只会复制 10 个字符，最后一个为空。增加n 会产生相同的结果。

将 n 减少到 10 揭示了问题的核心：

00000111101100100100001110100000

0000011110@

我不知道为什么它会这样做，但随着 n 减少，它会继续这样做，直到 n 变得更小（大约 8）。

这些是我的代码的相关行：

char line[11], hexAddr[8], binAddr[33], lv1Index[11];
    ...
strncpy(lv1Index, binAddr, 10);

其中 lv1Index 在此之前未受影响，并且 binAddr 直接预先显示（在给定的输出中）。

添加的字符始终是 @，所以我不相信它是 lv1Index 预初始化中的垃圾。

原文

I am having a problem with strncpy() copying an extra character at the length I need. I need to copy the 10 most significant bits from one string to another (both in char* format).

The size of the larger variable does not intuitively matter, and the smaller only affects the output when it is significantly smaller than the value I need (I don't know why this is either).

The n value in my strncpy() may be what's causing the problem, but I have no idea why it's acting the way it is. If I'm not mistaken, n, the length of the destination string, should be 1 longer than the data, to account for the null character at the end.

My string is not behaving this way at all. Setting n to 11, I receive the following output:

00000111101100100100001110100000

00000111101

I would imagine it would only copy 10 characters, with the final one for the null. Increasing n produces the same results.

Decreasing n to 10 reveals the heart of the problem:

00000111101100100100001110100000

0000011110@

I don't know why it would do this, but it continues to do this as n decreases, until n is much smaller (around 8).

These are the relevant lines of my code:

char line[11], hexAddr[8], binAddr[33], lv1Index[11];
    ...
strncpy(lv1Index, binAddr, 10);

where lv1Index was untouched before this, and binAddr is displayed (in the given output) directly beforehand.

The added character is always @, so I don't believe it's garbage from lv1Index pre-initialization.

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九歌凝 2024-12-24 07:35:25

来自一个 C++ 参考：

如果在复制 num 个字符之前找到源 C 字符串的末尾（由空字符表示），则用零填充目标，直到总共写入 num 个字符为止。
没有空字符隐式附加到目标末尾，因此只有当源中的 C 字符串长度小于 num 时，目标才会以空字符结尾。

其他文档（MSDN、手册页）有类似的信息。

这意味着您应该将空终止符放在自己身上：

 lv1Index[10] = '\0';

From one C++ reference:

If the end of the source C string (which is signaled by a null-character) is found before num characters have been copied, destination is padded with zeros until a total of num characters have been written to it.
No null-character is implicitly appended to the end of destination, so destination will only be null-terminated if the length of the C string in source is less than num.

Other documentation (MSDN, man page) has similar information.

This means you should put the null terminator on yourself: