1、warning: ISO C++ forbids converting a string constant to ‘char*’ [-Wwrite-strings]
char *imgFile = “data/IR_62_78_150_151.jpg”;
分析 ：char *背后的含义是：这个字符串，我要修改它。而传给函数的字面常量是没法被修改的。
warning: non-static data member initializers only available with -std=c++11 or -std=gnu++11warning: non-static data member initializers only available with -std=c++11 or -std=gnu++1...
int a = -...
1.rn32 位机上根据下面的代码，问哪些说法是正确的？rn<em>signed</em> char a = 0xe0;rnun<em>signed</em> int b = a;rnun<em>signed</em> char c = a;rnA. a>0 && c>0 为真 B. a == c 为真 C. b 的十六进制表示是：0xffffffe0 D. 上面都不对rnrn2.rna是八位对么? 有符号1110 0000 ，-6 ?rnb是a带符号扩展32位 ffff ffe0rnc是八位对么? 无符号1110 0000 ，e0 ?rnrn3.rna b c 到底怎么转啊啊啊啊啊？ [img=https://forum.csdn.net/PointForum/ui/scripts/csdn/Plugin/001/face/49.gif][/img]
DescriptionnnArithmetic <em>expressions</em> are usually written with the operators in <em>between</em> the two operands (which is called infix notation). For example, (x+y)*(z-w) is an arithmetic expression in infix notation. However, it is easier to write a program to evaluate an expression if the expression is written in postfix notation (also known as reverse Polish notation). In postfix notation, an operator is written behind its two operands, which may be <em>expressions</em> themselves. For example, x y + z w - * is a postfix notation of the arithmetic expression given above. Note that in this case parentheses are not required.nnTo evaluate an expression written in postfix notation, an algorithm operating on a stack can be used. A stack is a data structure which supports two operations:nnpush: a number is inserted at the top of the stack.npop: the number from the top of the stack is taken out.nDuring the evaluation, we process the expression from left to right. If we encounter a number, we push it onto the stack. If we encounter an operator, we pop the first two numbers from the stack, apply the operator on them, and push the result back onto the stack. More specifically, the following pseudocode shows how to handle the case when we encounter an operator O:nna := pop();nb := pop();npush(b O a);nThe result of the expression will be left as the only number on the stack.nnNow imagine that we use a queue instead of the stack. A queue also has a npush and pop operation, but their meaning is different:nnpush: a number is inserted at the end of the queue.npop: the number from the front of the queue is taken out of the queue.nCan you rewrite the given expression such that the result of the algorithm using the queue is the same as the result of the original expression evaluated using the algorithm with the stack?nnInputnnThe first line of the input contains a number T (T ≤ 200). The following T lines each contain one expression in postfix notation. Arithmetic operators are represented by uppercase letters, numbers are represented by lowercase letters. You may assume that the length of each expression is less than 10000 characters.nnOutputnnFor each given expression, print the expression with the equivalent result when using the algorithm with the queue instead of the stack. To make the solution unique, you are not allowed to assume that the operators are associative or commutative.nnSample Inputnn2nxyPzwIMnabcABdefgCDEFnSample OutputnnwzyxIPMngfCecbDdAaEBF
Function UnsingedAdd (ByVal Start As Long, ByVal Incr As Long) As LongrnrnConst SignBit As Long = &H80000000rnUn<em>signed</em>Add=(Start Xor SignBit) + Incr Xor SignBitrnrnEnd Functionrnrn谁能看懂？ByVal Start As Long, ByVal Incr As Long这两个参数是怎么用的，能举个例子说一下么？
<em>signed</em>表示有符号的，其第一个位表示正负，其余位表示大小，例如<em>signed</em> int 大小区间为-128~127。
un<em>signed</em>表示无符号的，所有位都为大小，没有正负，例如un<em>signed</em> int 大小区间为0~127。
Whether plain chars are <em>signed</em> or un<em>signed</em> is machine-dependent , but printable characters are always positivern出自 the c programming language 2.2倒數第四段話結尾，覺得這句話中“可以打印的字符總是正的”是不是有點問題啊rn請高人多多指教rn
For instance, both Brocade and Cisco layer 3 switches support the Virtual Router Redundancy Protocol (VRRP) as defined in RFC 2338.
Proprietary Protocols - In addition to VRRP, Brocade also supports and extended version of VRRP, called VRRPE. VRRPE functions in a similar way to VRRP, but overcomes many of the limitations of the standards-based protocol. Cisco also provides alternatives to the standard-based protocol; Hot Standby Router Protocol (HSRP) and Gateway Load Balancing Protocol (GLBP).
Problem DescriptionnAs any other marketing company, ACM produces a lot of funky advertising items that may contain a logo and be given to customers and business partners as small gifts. A unique specialty of ACM is a calculator that uses roman numbers.nnRoman numbers are able to express any non-negative <em>integer</em> using uppercase letters:nnNumbers are created by appending several letters together, the letter representing a higher value must always precede letters with lower values. The only exception are the letters “I”, “X”, and “C”, they may be used before higher letters to form values expressed by digits 4 and 9. The only possible combinations are:nnAny roman number must first express thousands, then hundreds, tens, and ones. Therefore, 499 must always be written as “CDXCIX”, not “ID”.nnAlthough not very practical, this gift is considered extremely “cooooool”. Your task is to write software for that calculator.n nnInputnThe input will consist from commands, each written on a separate line. Possible commands arenassignments, “RESET”, and “QUIT”.nAn assignment command starts with a single digit representing one of ten registers that thencalculator has. The register number is followed by an equal sign (“=”) and an expression. Thenexpression will consist only from valid roman numbers, register names (digits), plus (“+”) andnminus (“-”) signs. You may assume that the expression will always be valid and no longer thann10000 characters.n nnOutputnFor each command, output a single line. For assignments, print the register name, equal sign,nand the value that is being as<em>signed</em> to that register. Instead of it, print the word “Error”, if thenexpression contains a reference to a register that has not been as<em>signed</em> before, or if the resultnis negative or larger than 10000. In such cases, no change to register values is made.nFor RESET commands, invalidate all previous register assignments and print the word “Ready”.nThe QUIT command will be the last one. Print the word “Bye” and terminate the program.n nnSample Inputn1=MC+IV-Xn1=1+1nRESETn1=1+Xn1=MMn1=1+1+1+1+1n2=1+1nQUITn nnSample Outputn1=MXCIVn1=MMCLXXXVIIInReadynErrorn1=MMn1=MMMMMMMMMMnErrornByen
请高手看看我下面两个语句：rn语句1rnselect distinct a.str_deductbill_nornfrom [sellrebate].dbo.t_deduct_account_detail arn rnwhere a.str_bill_no <em>between</em> '200812260000' and '200903259999'rnrn and str_deductbill_no='200901070089'rnorder by str_deductbill_norn////rn语句2rnselect distinct a.str_deductbill_nornfrom [sellrebate].dbo.t_deduct_account_detail arn rnwhere a.str_bill_no [color=#FF0000]not[/color] <em>between</em> '200812260000' and '200903259999'rnrn and str_deductbill_no='200901070089'rnorder by str_deductbill_norn以上两个语句的区别是在<em>between</em>前一个有not一个没notrn在本人看来一个两个语句查出的结果是相对的，可是恰恰相反。请问各位这是什么原因？rnrn
DescriptionnnThe most important activity of ACM is the GSM network. As the mobile phone operator, ACM must build its own transmitting stations. It is very important to compute the exact behaviour of electro-magnetic waves. Unfortunately, prediction of electro-magnetic fields is a very complex task and the formulas describing them are very long and hard-to-read. For example, below are the Maxwell's Equations describing the basic laws of electrical engineering. n!(http://poj.org/images/1400_1.jpg)nnACM has de<em>signed</em> its own computer system that can make some field computations and produce results in the form of mathematic <em>expressions</em>. Unfortunately, by generating the expression in several steps, there are always some unneeded parentheses inside the expression. Your task is to take these partial results and make them "nice" by removing all unnecessary parentheses. nnInputnnThere is a single positive <em>integer</em> T on the first line of input. It stands for the number of <em>expressions</em> to follow. Each expression consists of a single line containing only lowercase letters, operators (+, -, *, /) and parentheses (( and )). The letters are variables that can have any value, operators and parentheses have their usual meaning. Multiplication and division have higher priority then subtraction and addition. All operations with the same priority are computed from left to right (operators are left-associative). There are no spaces inside the <em>expressions</em>. No input line contains more than 250 characters.nOutputnnPrint a single line for every expression. The line must contain the same expression with unneeded parentheses removed. You must remove as many parentheses as possible without changing the semantics of the expression. The semantics of the expression is considered the same if and only if any of the following conditions hold: nnThe ordering of operations remains the same. That means "(a+b)+c" is the same as "a+b+c", and "a+(b/c)" is the same as "a+b/c". nThe order of some operations is swapped but the result remains unchanged with respect to the addition and multiplication associativity. That means "a+(b+c)" and "(a+b)+c" are the same. We can also combine addition with subtraction and multiplication with division, if the subtraction or division is the second operation. For example, "a+(b-c)" is the same as "a+b-c". nnYou cannot use any other laws, namely you cannot swap left and right operands and you cannot replace "a-(b-c)" with "a-b+c". nnSample Inputnn8n(a+(b*c))n((a+b)*c)n(a*(b*c))n(a*(b/c)*d)n((a/(b/c))/d)n((x))n(a+b)-(c-d)-(e/f)n(a+b)+(c-d)-(e+f)nSample Outputnna+b*cn(a+b)*cna*b*cna*b/c*dna/(b/c)/dnxna+b-(c-d)-e/fna+b+c-d-(e+f)