c++ - 为什么一行输出(在 while 循环中)使不变量为假？

Question

我试图更深入地理解 while 循环。我理解的基础知识是，只要测试条件为真，一段时间就会重复该语句。

在 Accelerated C++ 一书中，作者陈述如下:

//the number of blanks surrounds the greeting
const int pad = 1;

//total number of rows to write
const int rows = pad * 2 + 3;

//we have written r rows so far
int r = 0;
//setting r to 0 makes the invariant true

while (r != rows){
     //we can assume that the invariant is true here

     //writing a row of output makes the invariant false
     std::cout << std::endl;

     //incrementing r makes the invariant true again
     ++r;
}

//we can conclude that the invariant is true here

我不明白，为什么写一行输出不变的 false，然后在递增时再次为 true。这本书的作者是否犯了错误？

编辑 - 该变体只是一种智能工具，可以更轻松地理解 while 循环。在此示例中，变体是“r = 输出中的行数”。

--------------------什么是不变量----------------

What is an invariant?

Answer 1

基本上，您在评论中回答了自己的问题:

They only said "Writing a row of output causes the invariant to become false, because r is no longer the number of rows we have written. However, incrementing r to account for the row that was written will make the invariant true again."

总结:

• 不变量是“r = 写入输出的行数”
• 打印一个空行会使这个不变量无效(因为 r 没有改变，但是写入输出的行数增加了 1)
• 将 r 递增 1 后，不变量再次为真