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

Question

I've asked the same question about strncpy, 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

For simplicity I've also included the live example link above inline in the code:

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

    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);
}

If I call this with: foo("lorem ipsum %d", 13) the output I get is:

lorem ipsum 1

Where as I would have expected: lorem ipsum 13

Can anyone explain the discrepancy? When I debug I get a buf_size of 14 which should be enough to contain the entire string, yet it does not :(

Answer 1

Can anyone explain the discrepancy?

Because their documented behavior is different.

strncpy()

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

but 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.

emphasis is mine.