std::lexical_cast - 有这样的事情吗? [英] std::lexical_cast - is there such a thing?

问题描述

C++ 标准库是否定义了这个函数，还是我必须求助于 Boost?

Does the C++ Standard Library define this function, or do I have to resort to Boost?

我在网上搜索，除了 Boost 什么都找不到，但我想我最好在这里问一下.

I searched the web and couldn't find anything except Boost, but I thought I'd better ask here.

推荐答案

仅部分.

C++11 有 std::to_string 用于内置类型:

C++11 <string> has std::to_string for the built-in types:

[n3290: 21.5/7]:

string to_string(int val);
string to_string(unsigned val);
string to_string(long val);
string to_string(unsigned long val);
string to_string(long long val);
string to_string(unsigned long long val);
string to_string(float val);
string to_string(double val);
string to_string(long double val);

返回: 每个函数返回一个 string 对象，其中包含其参数值的字符表示通过使用格式调用 sprintf(buf, fmt, val) 生成"%d"、"%u"、"%ld"、"%lu"的说明符，"%lld", "%llu",分别为 "%f"、"%f" 或 "%Lf"，其中 buf 表示足够大小的内部字符缓冲区.

Returns: Each function returns a string object holding the character representation of the value of its argument that would be generated by calling sprintf(buf, fmt, val) with a format specifier of "%d", "%u", "%ld", "%lu", "%lld", "%llu", "%f", "%f", or "%Lf", respectively, where buf designates an internal character buffer of sufficient size.

还有以下相反的情况:

[n3290: 21.5/1, 21.5/4]:

int stoi(const string& str, size_t *idx = 0, int base = 10);
long stol(const string& str, size_t *idx = 0, int base = 10);
unsigned long stoul(const string& str, size_t *idx = 0, int base = 10);
long long stoll(const string& str, size_t *idx = 0, int base = 10);
unsigned long long stoull(const string& str, size_t *idx = 0, int base = 10);
float stof(const string& str, size_t *idx = 0);
double stod(const string& str, size_t *idx = 0);
long double stold(const string& str, size_t *idx = 0);

然而，没有任何通用的东西可以使用(至少 不是直到 TR2，也许吧！)，而在 C++03 中什么都没有.

However, there's nothing generic that you can use (at least not until TR2, maybe!), and nothing at all in C++03.

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