我曾经写过一种编程语言，它有一个“strfry”操作符:

"hello world"?
# => "wdo rlholle"

有用的,是吗?

С#:

var a = Double.Parse("10.0", CultureInfo.InvariantCulture); // returns 10
var b = Double.Parse("10,0", CultureInfo.InvariantCulture); // returns 100

在不变区域性中，逗号不是小数点符号，而是组分隔符。

据我所知，对于一些地区的新手程序员来说，这是一个常见的错误。

在fortran中(当然是77，可能在95中也是如此)，未声明的变量和以I到N开头的参数(“In”组)将是INTEGER，所有其他未声明的变量和参数将是REAL(源)。这与“在某些情况下可选的空白”相结合，导致了最著名的错误之一。

正如弗雷德·韦伯在1990年的《另类民间传说:计算机》一书中所说:

I worked at Nasa during the summer of 1963. The group I was working in was doing preliminary work on the Mission Control Center computer systems and programs. My office mate had the job of testing out an orbit computation program which had been used during the Mercury flights. Running some test data with known answers through it, he was getting answers that were close, but not accurate enough. So, he started looking for numerical problems in the algorithm, checking to make sure his tests data was really correct, etc. After a couple of weeks with no results, he came across a DO statement, in the form: DO 10 I=1.10 This statement was interpreted by the compiler (correctly) as: DO10I = 1.10 The programmer had clearly intended: DO 10 I = 1, 10 After changing the . to a , the program results were correct to the desired accuracy. Apparently, the program's answers had been "good enough" for the sub-orbital Mercury flights, so no one suspected a bug until they tried to get greater accuracy, in anticipation of later orbital and moon flights. As far as I know, this particular bug was never blamed for any actual failure of a space flight, but the other details here seem close enough that I'm sure this incident is the source of the DO story.

我认为这是一个很大的WTF，如果DO10I被作为DO10I，并且反过来，因为隐式声明被认为是类型REAL。这是个很棒的故事。

在C。

int;

(&a)[0] = 10;/*将值10赋给*/

[0]等于*(&a +0)得到*(&a)也就是a。

Java泛型 都是WTF:

List<String> ls = new ArrayList<String>(); //1
List<Object> lo = ls; //2

2:是非法的(??)这是令人费解的，但你必须想想接下来会发生什么:

lo.add(new Object());
String s = ls.get(0);

我们将对象赋值给字符串引用，哦不!就像这样，他们周围有很多陷阱。

在Python中:

i = 1
++i
print i

输出“1”。行'++i'的计算结果为+(+i) (Python不支持增量操作符)