将字符串转换为int [英] convert a string to int

问题描述

我有一个大文件，其中每行包含空格分隔的整数。任务是逐行稀疏这个文件。对于字符串到int转换我有三个解决方案：

I have a large file where each line contains space-separated integers. The task is to sparse this file line-by-line. For the string to int conversion I have three solutions:

static int stringToIntV1(const string& str) {
    return (atoi(str.c_str()));
}

但是，如果我传递一个格式错误的字符串， 。例如，字符串123error转换为123.

However, if I pass a malformed string, it doesn't produce any error. For instance the string "123error" is converted to 123.

第二个解决方案：

static int stringToIntV2(const string& str)
{
    int result;
    istringstream myStream(str);

    if (myStream >> result) {
        return result;
    }
    // else
    throw domain_error(str + " is not an int!");
}

我有同样的问题，格式错误的字符串不会引发错误。

I have the same problem here, malformed strings don't raise an error.

使用Boost的第三个解决方案（位于 http：// stackoverflow。 com / questions / 149268 / boost-library ）：

Third solution with Boost (found at http://stackoverflow.com/questions/149268/boost-library):

static int stringToIntV3(const string& str)
{
    int iResult = 0;
    try {
        iResult = lexical_cast<int>(str);
    }
    catch(bad_lexical_cast &) {
        throw domain_error(str + " is not an int!");
    }
    return iResult;
}

这个结果正确。

但是，执行时间有显着差异。在大型文本文件（32 MB）上进行测试时，我获得了以下时间：

However, there is a significant difference in the execution time. Testing on a large text file (32 MB), I got the following times:

• （1） atoi：4.522s（胜出者）


• （2） with istringstream：15.303s（very slow）


• （3） with lexical_cast：10.958s（between both）

• (1) with atoi: 4.522s (winner)
• (2) with istringstream: 15.303s (very slow)
• (3) with lexical_cast: 10.958s (between the two)

我的问题：你知道如何注意atoi的格式化字符串吗？它会给最快的解决方案。或者你知道更好的解决方案吗？

My question: do you know how to notice malformed strings with atoi? It would give the fastest solution. Or do you know a better solution?

更新：感谢您的答案。根据提示，我想出了这个解决方案：

Update: Thanks for the answers. Following the tips, I came up with this solution:

static int stringToIntV4(const string& str)
{
    char * pEnd;
    const char * c_str = str.c_str();
    int result = strtol(c_str, &pEnd, 10);
    if (pEnd == c_str+str.length()) {
        return result;
    }
    // else
    throw domain_error("'" + str + "'" + " is not an int!");
}

好消息是，如果有问题，

The good news is that it yields if there is a problem and as efficient as the atoi version.

推荐答案

我有效率高达 atoi strtol 。它需要一个参数，它设置指向它不能转换的第一个字符，所以你可以使用它来确定整个字符串是否被转换。

I'd use strtol. It takes a parameter that it sets to point at the first character it couldn't convert, so you can use that to determine whether the entire string was converted.

编辑：就速度而言，我预计它会比 atoi 略慢，但比你尝试的其他人快。

as far as speed goes, I'd expect it to be slightly slower than atoi, but faster than the others you tried.

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