C++ 将字符串转换为 int 的最有效方法(比 atoi 更快) [英] C++ most efficient way to convert string to int (faster than atoi)

问题描述

正如标题中提到的，我正在寻找比 atoi 更能提供性能的东西.目前，我知道最快的方法是

As mentioned in the title, I'm looking for something that can give me more performance than atoi. Presently, the fastest way I know is

atoi(mystring.c_str())

最后，我更喜欢不依赖 Boost 的解决方案.有没有人有很好的性能技巧来做到这一点?

Finally, I would prefer a solution that doesn't rely on Boost. Does anybody have good performance tricks for doing this?

附加信息:int不会超过20亿，总是正数，字符串中没有小数位.

Additional Information: int will not exceed 2 billion, it is always positive, the string has no decimal places in it.

推荐答案

我尝试了使用查找表的解决方案，但发现它们充满了问题，而且实际上速度不是很快.结果证明，最快的解决方案是最没有想象力的:

I experimented with solutions using lookup tables, but found them fraught with issues, and actually not very fast. The fastest solution turned out to be the least imaginitive:

int fast_atoi( const char * str )
{
    int val = 0;
    while( *str ) {
        val = val*10 + (*str++ - '0');
    }
    return val;
}

使用一百万个随机生成的字符串运行基准测试:

Running a benchmark with a million randomly generated strings:

fast_atoi : 0.0097 seconds
atoi      : 0.0414 seconds

公平地说，我还通过强制编译器不要内联它来测试这个函数.结果还是不错的:

To be fair, I also tested this function by forcing the compiler not to inline it. The results were still good:

fast_atoi : 0.0104 seconds
atoi      : 0.0426 seconds

只要你的数据符合 fast_atoi 函数的要求，这是相当合理的性能.要求是:

Provided your data conforms to the requirements of the fast_atoi function, that is pretty reasonable performance. The requirements are:

1. 输入字符串只包含数字字符，或者为空
2. 输入字符串代表一个从 0 到 INT_MAX
的数字

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