使用 snprintf() 的 Fencepost 条件和可移植性？

Question

给出以下代码：

const int size = 20;
char buffer[size];

// From the Linux man page for snprintf():
//
// The 'res' is the number of bytes that would be written to buffer had size been
// sufficiently large excluding the terminating null byte.  Output bytes beyond
// the size-1st are discarded instead of being written to the buffer, and a null
// byte is written at the end of the bytes actually written into the buffer.
int res = snprintf(buffer, size, "some format with %d and %s", 23, "some string");

if (res >= size) {
  cerr << "The buffer was not large enough, we needed " << res
       << " but only had " << size << "." << endl;
} else {
  cout << "The buffer is big enough, we only needed " << res
       << " but had " << size << "." << endl;
}

这是可移植的吗？如果是，我的所有栅栏条件是否正确？

1 将 size 传递给 snprintf()

2 检查 res 是否大于或等于 size

Answer 1

snprintf 严格来说不是可移植的，因为它不是 C/C++ 标准的一部分。 Windows 将其称为 _snprintf - 但其他方面相同。

栅栏条件一切都很好。你的 printf 并不完全正确，在 equals 情况下它将打印

The buffer was not large enough, we needed 215 but only had 215.