我很困惑如何解析多个分隔符使用getline和strtok在c ++ [英] I am very confused on how to parse multiple delimiters using getline and strtok in c++

问题描述

我的代码发布在下面。

My code is posted below. I want to be able to parse using the delimiters " ()," and convert the strings into integers in cpp.

while(getline(fin, line))
{
    x  = atoi((strtok(line.c_str(),'(,)'));
    xx = atoi((strtok(NULL,"(),"));
    xxx = atoi((strtok(NULL,"(),")));
    cout << x << "    " << xx << "    " << xxx << "\n";
}

但由于某种原因，我得到以下错误

but for some reason I get the following errors

GraphTest.cpp：134：错误：从'const char *'到'char *'的无效转换

GraphTest.cpp:134: error: invalid conversion from ‘const char*’ to ‘char*’

GraphTest.cpp：134：error：初始化'char * strtok（char *，const char *）'的参数1

GraphTest.cpp:134: error: initializing argument 1 of ‘char* strtok(char*, const char*)’

.c_str应该将我的字符串转换为ac类型的字符串，允许我使用atoi和strtok函数。我很困惑，并会感谢任何帮助。

The .c_str should convert my string into a c type string allowing me to use the atoi and strtok functions. I am very confused and would appreciate any help.

推荐答案

它不会编译，因为 c_str（）返回 const char * ，它应该是一个不可修改的内部 string 缓冲区的常量指针。另一方面， strtok（）接受 char * ，因为它修改其输入字符串。

It doesn't compile because c_str() returns a const char*, it's supposed to be a constant pointer to not modifiable internal string buffer. On the other hand strtok() accepts a char* because it modifies its input string.

现在你有两个选择：从 strtok（）获取一个C字符串，或者将所有的内容重写为C ++。

Now you have two options: get a C string usable from strtok() or rewrite everything to be C++.

从C ++字符串中创建一个新的可修改的C字符串：

Create a new modifiable C string from your C++ string:

char* modifiableLine = strdup(line.c_str());

x  = atoi((strtok(modifiableLine, "(,)"));
// Other processing

free(modifiableLine);

如果你必须保持一个大函数/类更好的解决方案是使用C ++标准库提供的（如果C ++ 11），还要删除 atoi（） C函数让我们先写一个帮助函数： p>

You can do it if you have to keep a big amount of C code wrapped inside a C++ function/class. A better solution is to use what C++ Standard Library offers (also dropping atoi() C function if C++ 11). Let's first write an helper function:

int readNextInt(istringstream& is, const string& delimiters) {
    string token;

    if (getline(is, token, delimiters))
        return stoi(token);

    return 0; // End of stream?
}

使用像这样：

istringstream is(line)
x = readNextInt(is, "(),");
xx = readNextInt(is, "(),");
xxx = readNextInt(is, "(),");

请注意，标准C ++函数 getline / code>不接受分隔符参数的字符串，但只有一个 char 编写自己的重载版本。看看这篇文章是否有好的可能实现（你也可以简单地替换 getline ）之后的是>>标记示例）。

Please note that standard C++ function getline() doesn't accept a string for delimiters parameter but a single char only then you have to write your own overloaded version. Take a look to this post for good and nice possible implementation (you may also simply replace getline() with is >> token after is.imbue(), see given example).

好吧，如果你已经使用Boost，那么你可以使用 boost :: tokenizer 。

Well...if you're already using Boost then you may simply use boost::tokenizer.

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