使用snprintf()的Fencepost条件和可移植性？

Question

给定以下代码：

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;
}

这是便携的吗?如果是的话，我是否得到了所有的栅栏条件正确？

1将size传递给snprintf()

2检查res是否大于或等于size

Answer 1

严格地说，snprintf不是可移植的，因为它不是C/C++标准的一部分。Windows将其命名为_snprintf -但在其他方面完全相同。

击剑杆的条件都很好。你的printf并不是完全正确，在equals的情况下，它将被打印出来

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