我想在程序中回答是的时候去切换并做我在那儿写的工作，但不能 [英] I want in my program when the answer is yes go to switch and do a work that i write there but it cannot

问题描述

<pre lang="xml">#include "StdAfx.h"
#include <iostream>
#include <sstream>
#include <algorithm>
#include <string>
#include <vector>
using namespace std;
int main()
{
char str[] ="10 input x";
char * pch;
pch = strtok (str," ,.-");
int i = 0;
while (pch != NULL)
{
  printf ("%s\n",pch);
  if((i == 1) && !strcmp( pch, "input"))
  {
    int accumulator =0;
    int memory[100];
    int counter=0;
    int opCode ;
    int operand ;
    int instructionRegister;
    for(int i=0;i<counter;i++){
        instructionRegister=memory[i];
        opCode =instructionRegister/100;
        operand =instructionRegister%100;
        switch ( opCode){
            case 10:
            cout<<"plz enter a number:\n";
            cin>>memory[operand];
            break;
        }
    }
  }else{
      cout<<"no"<<endl;
  }
  pch = strtok (NULL, " ,.-");
  i++;
}
    return 0;
}

:doh:

:doh:

推荐答案

进行测试.

目前，您开始令牌化:

Move your test.

At present, you start the tokenising:

char str[] ="10 input x";
char * pch;
pch = strtok (str," ,.-");

然后遍历每个令牌:

Then you run through each token:

while (pch != NULL)
{
  printf ("%s\n",pch);
  pch = strtok (NULL, " ,.-");
}

然后尝试检查第二个:

Then you try to check the second one:

if(strcmp( pch++, "input"))
{
    cout<<"yes";
}

问题是，到检查的时间，您已经用完了令牌...
试试这个:

The problem is that by the time to check, you have run out of tokens completely...
Try this:

char str[] ="10 input x";
char * pch;
pch = strtok (str," ,.-");
int i = 0;
while (pch != NULL)
{
  printf ("%s\n",pch);
  if((i == 1) && strcmp( pch, "input"))
  {
    cout<<"yes";
  }
  pch = strtok (NULL, " ,.-");
  i++;
}

这样，您就可以对照匹配字符串检查第二个标记，同时仍然可以找到它！ :laugh:

That way, you check the second token against your match string while you can still get at it! :laugh:

这是使用标准C ++库的解决方案.

And here is a solution using The Standard C++ Library.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <string>
#include <vector>

int main()
{
    std::vector<std::string> vtokens;
    std::string str("10 input x");
    
    // Extract the tokens from string
    std::istringstream iss(str);
    std::copy(std::istream_iterator<std::string>(iss),
              std::istream_iterator<std::string>(),
              std::back_inserter< std::vector<std::string> >(vtokens));
    
    // Now "vtokens" contains all of the tokens from the initial string.
    // Your specific requirement was to check if the second token is "input",
    // Here you go:
    bool banswer = (vtokens.size() > 1) ? vtokens[1] == "input" : false;
    std::string str_answer(banswer ? "yes" : "no");
    std::cout << "Answer: " << str_answer << std::endl;
    
    // Optional: hold the screen to see the results
    std::cin.get();
    return 0;
}

我给你这作为你C风格代码的替代方法.
这是一个演示，说明如何避免使用strtok并在C ++中使用标准算法，容器和流.

:)

[更新]
OP稍微改变了他的问题，答案(来自我和来自OriginalGriff)是针对原始问题的.

到OP:
您正在从未初始化的内存中读取(请参见int memory[100];数组)，并且正在使用未初始化的变量(instructionRegister变量).
[/UPDATE]

I give you this as an alternative to your C style code.
It''s a demonstration how can you avoid strtok and use standard algorithms, containers and streams in C++.

:)

[UPDATE]
The OP slightly changed his question and the answers (from me and from OriginalGriff) were for the original question.

To OP:
You are reading from uninitialized memory (see your int memory[100]; array) and you are using uninitialized variables (instructionRegister variable).
[/UPDATE]

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