Question

我正在尝试从.gz文件中进行一些字符串操作。我写了以下代码。

char buffer[1001];
for(;gzeof(f_Handle);){
    gzread(f_Handle, buffer, 1000);
    buffer[1000] = 0;
    char* chars_array = strtok(buffer, " ");

    while(chars_array){
        cout<<chars_array << '\n';
        chars_array = strtok(NULL, " ");
    }
}

但是文件格式（.gz）在

中

A 1 2 3
B 2 3 5
A 4 5 6
B 34 64 123

我想知道它何时是A或B以及A或B中的内容。

目前，它以下列方式打印出来

A
1
2
3
...

想法a）是通过chars_array使用if循环来找出A或B或

b）字符串数组而不是字符指针

Answer 1

以下是使用std::string和函数substr(...)的简单示例。它不会执行整个字符串，但您可以将其放在循环中来执行此操作。

#include <string>
#include <iostream>
#include <vector>

int main()
{
    std::string original = "01234567 89abc defghi j";
    std::vector< std::string > strings;

    // Find space from last index
    int lastSpaceIndex = 0;
    int spaceIndex = original.find( ' ', lastSpaceIndex );

    // Find the number of characters to split
    int numCharacters = spaceIndex - lastSpaceIndex;

    // Split string ( the second argument is the number of characters to splut out)
    std::string tokenizedString = original.substr( lastSpaceIndex, numCharacters );

    // Add to vector of strings
    strings.push_back( tokenizedString);

    // Print result
    std::cout << "Space at : " << spaceIndex << std::endl;
    std::cout << "Tokenized string : " << tokenizedString << std::endl;

    // Find the nextsubstring
    // =========================================================================
    // Need to increase by 1 since we don't want the space 
    lastSpaceIndex = spaceIndex + 1;
    spaceIndex = original.find( ' ', lastSpaceIndex );

    numCharacters = spaceIndex - lastSpaceIndex;
    tokenizedString = original.substr( lastSpaceIndex, numCharacters );

     strings.push_back( tokenizedString);

    std::cout << "Space at : " << spaceIndex << std::endl;
    std::cout << "Tokenized string : " << tokenizedString << std::endl;

    std::cout << "=====================================\n";

    for ( const auto &str : strings )
    {
        std::cout << "String : " << str << std::endl;
    }

}

输出：

Space at : 8
Tokenized string : 01234567
Space at : 14
Tokenized string : 89abc
=====================================
String : 01234567
String : 89abc

如果没有更多空格，original.find( ' ', lastSpaceIndex )将返回std::npos

使用strtok在C ++中进行字符串操作

1 个答案: