c - TCP 客户端数学运算无法正常工作

Question

用户输入 2 个数字以及 +、-、/、*，应该会得到答案。问题是，它没有正确显示他们输入的内容，也没有计算。为什么是这样？我提供了一个屏幕截图来显示我所得到的内容。我该如何解决？ http://imageshack.us/photo/my-images/404/mathfn.jpg/

#include <iostream>
using namespace std;

#include <stdio.h>
#include <string.h>
#include <winsock.h>

// Function prototype
void StreamClient(char *szServer, short nPort);

// Helper macro for displaying errors
#define PRINTERROR(s)   \
        fprintf(stderr,"\n%s: %d\n", s, WSAGetLastError())

////////////////////////////////////////////////////////////

void main(int argc, char **argv)
{
    WORD wVersionRequested = MAKEWORD(1,1);
    WSADATA wsaData;
    int nRet;
    short nPort;

    //
    // Check for the host and port arguments
    //
    if (argc != 3)
    {
        fprintf(stderr,"\nSyntax: TCPTimeClient ServerName PortNumber\n");
        return;
    }

    nPort = atoi(argv[2]);

    //
    // Initialize WinSock and check the version
    //
    nRet = WSAStartup(wVersionRequested, &wsaData);
    if (wsaData.wVersion != wVersionRequested)
    {   
        fprintf(stderr,"\n Wrong version\n");
        return;
    }

    //
    // Go do all the stuff a datagram client does
    //
    StreamClient(argv[1], nPort);

    //
    // Release WinSock resources
    //
    WSACleanup();

    printf("\nPlease enter a floating point number: ");
float number1;
scanf("%f", &number1);

printf("\nPlease enter a mathematical operator (+,-,*,/): ");
char mathOperator;
scanf("%f", &mathOperator);


printf("\nPlease enter a floating point number: ");
float number2;
scanf("%f", &number2);

printf("\nCalculating %f %f %f...", &number1, &mathOperator, &number2);

printf("\nPlease enter a floating point number: ");
float number3;
scanf("%f", &number2);

}

////////////////////////////////////////////////////////////

void StreamClient(char *szServer, short nPort)
{
    int nRet;                       // return code
    char szBuf[256];                // client buffer area 
    char szSvr[256];                // server name

    LPHOSTENT lpHostEntry;          // host data structure
    SOCKET  theSocket;              // client socket
    SOCKADDR_IN saClient;           // socket address structure

    //
    // Get local machine name
    //
    nRet = gethostname(szSvr, sizeof(szSvr));

    //
    // Check for errors
    //
    if (nRet == SOCKET_ERROR)
    {
        PRINTERROR("gethostname()");
        return;
    }

    // 
    // Display an informational message
    //
    printf("Datagram Client [%s] sending to server [%s] on port %d...\n",
                                szSvr, szServer, nPort);

    //
    // Find the server
    //
    lpHostEntry = gethostbyname(szServer);
    if (lpHostEntry == NULL)
    {
        PRINTERROR("gethostbyname()");
        return;
    }

    //
    // Create a TCP/IP datagram socket
    //
    theSocket = socket(AF_INET,         // Address family
                       SOCK_STREAM,     // Socket type
                       0);              // Protocol

    //
    // Check for errors
    //
    if (theSocket == INVALID_SOCKET)
    {
        PRINTERROR("socket()");
        return;
    }

    //
    // Fill in the address structure of the server
    //
    saClient.sin_family = AF_INET;
    saClient.sin_addr = *((LPIN_ADDR)*lpHostEntry->h_addr_list);
                                        // ^ Client's address
    saClient.sin_port = htons(nPort);   // Port number from command line

    //
    // Connect to the server
    //
    nRet = connect(theSocket, 
            (struct sockaddr *)&saClient, 
            sizeof(saClient));

    //
    // Check for errors
    //
    if(nRet == SOCKET_ERROR)
    {
      PRINTERROR("Connect()");
      return;
    }

    //
    // Prepare some data to send to the server
    //
    sprintf(szBuf, "From the Client [%s]", szSvr);

    //
    // Send data to the server
    //
    nRet = send(theSocket,                  // Socket
                  szBuf,                    // Data buffer
                  (int)strlen(szBuf),       // Length of data
                  0);                       // Flags

    //
    // Check for errors
    //
    if (nRet == SOCKET_ERROR)
    {
        PRINTERROR("send()");
        closesocket(theSocket);
        return;
    }

    //
    // Zero out the incoming data buffer
    //
    memset(szBuf, 0, sizeof(szBuf));

    //
    // Wait for the reply
    //
    nRet = recv(theSocket,                  // Socket
                szBuf,                      // Receive buffer
                sizeof(szBuf),              // Length of receive buffer
                0);                         // Flags

    //
    // Check for errors
    //
    if (nRet == SOCKET_ERROR)
    {
        PRINTERROR("recv()");
        closesocket(theSocket);
        return;
    }

    //
    // Display the data that was received
    //
    printf("\n%s", szBuf);

    //
    // Close the socket
    //
    closesocket(theSocket);
    return;


}

Answer 1

我不明白为什么你无法输入字符，所以我创建了一个测试程序 - 果然有相同的行为!我不经常使用 scanf() ，但我猜想 scanf("%c") 会抓取缓冲区中的任何字符(如果有的话)，并在我们有机会输入任何内容之前继续执行。所以我把它做得完全明确 - 为 fscanf/fgetchar 流指定“stdin”，并执行 fflush() 以确保在读取之前它是空的。另外，我完全停止使用 scanf() 只是为了获取单个字符。就是这样!完整的程序如下:

printf("Please enter a floating point number: ");
float number1;
fscanf(stdin,"%f", &number1);

printf("Please enter a mathematical operator (+,-,*,/): ");
char mathOperator;
//fscanf(stdin,"%f", &mathOperator);

fflush(stdin);
mathOperator = (char)fgetc(stdin);

printf("Please enter a floating point number: ");
float number2;
fscanf(stdin,"%f", &number2);

printf("\nCalculating %f %c %f...", number1, mathOperator, number2);

float result = 0.0F;
switch (mathOperator) {
case '+':
    result = number1 + number2;
    break;
case '-':
    result = number1 - number2;
    break;
case '/':
    result = number1 / number2;
    break;
case '*':
    result = number1 * number2;
    break;
default:
    printf("\nInvalid operator entered!");
}

printf( "\n\nThe answer is %0.1f\n", result );

fflush(stdin);
char ch = getc(stdin);  // wait for keypress

这是测试运行的输出:

Please enter a floating point number: 3.4
Please enter a mathematical operator (+,-,*,/): *
Please enter a floating point number: 2

Calculating 3.400000 * 2.000000...

The answer is 6.8

[结束更新]

您在使用 printf() 时遇到的具体问题是您传递的是浮点值的地址 - 而不是值本身。

printf("\nCalculating %f %f %f...", &number1, &mathOperator, &number2);

应该是:

printf("\nCalculating %f %f %f...", number1, mathOperator, number2);

下一个问题是“mathOperator” - 看起来您要求用户输入一个字符，但 printf() 指定的 int 数据类型是 float。这是行不通的。

所以 printf 真的应该更像:

printf("\nCalculating %f %c %f...", number1, mathOperator, number2);

假设“mathOperator”中有“+”、“-”等。

至于为什么它不计算 - 我没有看到(也许只是为了简洁而省略了)任何实际进行计算的代码。

[计算更新]

由于唯一允许的运算符是“+”、“-”、“/”、“*”，并且操作数始终为 float ，因此计算逻辑非常简单:

float result = 0.0F;
switch (mathOperator) {
case '+':
    result = number1 + number2;
    break;
case '-':
    result = number1 - number2;
    break;
case '/':
    result = number1 / number2;
    break;
case '*':
    result = number1 * number2;
    break;
}

然后您可能想要添加 printf() 来输出结果(和/或通过 TCP 将其发送回):

printf( "The answer is" "%0.1f\n", result );

如果您进行更复杂的计算，您必须担心运算符优先级、类型转换等。简单的“case”是不够的。您将需要查看解析器以从表达式构建解析树并根据某些规则(语法)对其进行评估。但相信我 - 你想首先掌握这个;-)

查看此question了解所涉及的内容。