标签: syntax

我在这里的第二行到最后一行遇到了语法错误,但不知道为什么.它看起来与它之前的第2行相同,但由于某种原因我得到了语法错误.我已经尝试过它和它之前的线和它之间的空行,结果相同.

## numlist = some list
array_size = 20
for row in xrange(array_size):
    for col in xrange(array_size):
        if(col<=(array_size-4)):
            check(sum(numlist[row][col:col+4])
        if(row<=(array_size-4)):
            check(sum([numlist[row+i][col] for i in range(4)]))

if语句有一定的条件然后我在if语句中使用while循环和另一个条件,if语句中的条件是否会在每次迭代中被测试？

LIST-SEARCH'(L,k)
x=L.nil.next
if x!=L.nil
   while x.key!=k
         x=x.next
return x

我一直试图运行这个程序一段时间但我似乎无法找出在尝试运行它时导致错误的原因.

这是我收到错误的代码行:

from math import *
from myro import *
init("simulator")

def rps(score):
    """ Asks the user to input a choice, and randomly assigns a choice to the computer."""
    speak("Rock, Paper, Scissors.")
    computerPick = randint(1,3)
    userPick = raw_input("Please enter (R)ock, (P)aper, or (S)cissors.")
    if userPick = R <#This line is where the error shows up at>
        print "You picked rock."
    elif userPick = P
        print "You picked paper."
    else
        print "You picked Scissors."
    score = outcome(score, userPick, computerPick)
    return score

string sqlQueryString = " SELECT g.code AS GoodCode, g.name AS GoodName, " +
"msr.name AS MsrName, sm.min_quan AS KolMin, sm.max_quan AS KolMax, " +
"sm.quan AS KolNal, ord.prc AS EdPrice, s.name AS Sklad, m.name AS Mol, " +
"k.code AS KodDost, k.name AS NameDost " +
"FROM N_GOODS_{0} AS g INNER JOIN " +
"G_SMGS_{0} AS sm ON g.id = sm.good_id INNER JOIN " +
"N_KNTRS_{0} AS k ON g.id = k.id INNER JOIN " +
"N_PRC_LISTS_{0} AS pr ON …

如何用DOG语言编写IF条件语法.我不知道如何编写条件语法.有任何想法吗.IF

如果我想要一个while循环:虽然变量不等于几个不同的值,这是否有效？:

while (uInput != ('a' || 'b' || 'c')){
    //do something to make it equal one of those
}

或者我是否必须单独将uInput与a,uInput与b进行比较,依此类推？

我有这个Python代码:

for name, age in read(file, ('name','age')):

有人可以解释一下它的意思吗？

大家好昨天我发了一个问题:像c#allow; ; ; ; ; ; ; ; ; ; ; ; (C#空声明)

为什么C#允许这个有趣？; ; ; ;

并找到答案:https://stackoverflow.com/a/20551989/2218635

但是今天我又看到了一个有趣的(bug)

见下图

为什么visual studio允许这个？代码构建成功.为什么要构建？

另一个有趣的代码总是运行

private void install()
    {
        http://www.stackoverflow.com
        return;
    }

编辑:

看上面的方法是一个void方法,我知道void方法不返回.但为什么没有显示任何错误？而且我没有为" http://www.stackoverflow.com " 指定任何变量.但为什么它不会给我一个错误？

我尝试运行脚本时收到错误

Error:"IndentationError: unindent does not match any outer indentation"

抛出错误的代码snipet:

def update():
    try:
        lines = open("vbvuln.txt", "r").readlines()
    except(IOError): 
         print "[-] Error: Check your phpvuln.txt path and permissions"
         print "[-] Update Failed\n" 
        sys.exit(1)
    try:

这是出现错误的实际行:

print "[-] Update Failed\n" 

// operator_overloading.cpp
// compile with: /EHsc
#include <iostream>
using namespace std;

struct Complex {
   Complex( double r, double i ) : re(r), im(i) {} // what is this syntax?
   Complex operator+( Complex &other );
   void Display( ) {   cout << re << ", " << im << endl; }
private:
   double re, im;
};

// Operator overloaded using a member function
Complex Complex::operator+( Complex &other ) {
   return Complex( re + other.re, im + other.im );
}

int main() { …