为什么 vsnprintf 没有写出与 strncpy 相同数量的字符？

Question

我问过, but there the string ends up containing the whole input string. When passing a string to vsnprintf the last character always gets chopped off: https://rextester.com/UIQMX91570

为简单起见，我还在代码中内联了上面的实例 link：

void bar(const char* format, va_list vlist) {
    const auto buf_size = vsnprintf(nullptr, 0U, format, vlist);
    string buffer(buf_size, '[=10=]');

    vsnprintf(data(buffer), buf_size, format, vlist);
    cout << data(buffer) << endl;
}

void foo(const char* format, ...) {
    va_list vlist;

    va_start(vlist, format);
    bar(format, vlist);
    va_end(vlist);
}

如果我调用它：foo("lorem ipsum %d", 13) 我得到的输出是：

lorem ipsum 1

如我所料：lorem ipsum 13

任何人都可以解释差异吗？当我调试时，我得到一个 buf_size 的 14，其中应该足以包含整个字符串，但它不 :(

Answer 1

因为手册页清楚地说明了这一点

If the output was truncated due to this limit then the return value is the number of characters (not including the trailing '[=11=]') which would have been written to the final string if enough space had been available.

如果您检查第二个 vsnprintf 调用的 return 值，您会看到 return 值等于大小，如手册页所示：

Thus, a return value of size or more means that the output was truncated.

Answer 2

vsnprintf 的 buf_size 参数指定要写入多少个字符，包括 终止 NUL 字符 . return 值是生成的字符数，不包括 终止 NUL 字符。

你想要

const auto buf_size = vsnprintf(nullptr, 0U, format, vlist) + 1;

Answer 3

Can anyone explain the discrepancy?

因为他们记录的行为不同。

strncpy()

If count is reached before the entire array src was copied, the resulting character array is not null-terminated.

但是vsnprintf()

At most buf_size-1 characters are written. The resulting character string will be terminated with a null character, unless buf_size is zero.

重点是我的。

为什么 vsnprintf 没有写出与 strncpy 相同数量的字符？

Why is vsnprintf Not Writing the Same Number of Characters as strncpy Would?

c++

printf

variadic-functions

string-length

strncpy