如何在 C++ 中将字符串转换为双精度值？
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How can I convert string to double in C++?
提问by Alessandro Jacopson
How can I convert string to double in C++? I want a function that returns 0 when the string is not numerical.
如何在 C++ 中将字符串转换为双精度值？我想要一个当字符串不是数字时返回 0 的函数。
回答by Alessandro Jacopson
See C++ FAQ Lite How do I convert a std::string to a number?
请参阅 C++ FAQ Lite如何将 std::string 转换为数字？
See C++ Super-FAQ How do I convert a std::string to a number?
请参阅 C++ Super-FAQ如何将 std::string 转换为数字？
Please note that with your requirements you can't distinguish all the the allowed string representations of zero from the non numerical strings.
请注意，根据您的要求，您无法将所有允许的零字符串表示与非数字字符串区分开来。
// the requested function
#include <sstream>
double string_to_double( const std::string& s )
{
std::istringstream i(s);
double x;
if (!(i >> x))
return 0;
return x;
}
// some tests
#include <cassert>
int main( int, char** )
{
// simple case:
assert( 0.5 == string_to_double( "0.5" ) );
// blank space:
assert( 0.5 == string_to_double( "0.5 " ) );
assert( 0.5 == string_to_double( " 0.5" ) );
// trailing non digit characters:
assert( 0.5 == string_to_double( "0.5a" ) );
// note that with your requirements you can't distinguish
// all the the allowed string representation of zero from
// the non numerical strings:
assert( 0 == string_to_double( "0" ) );
assert( 0 == string_to_double( "0." ) );
assert( 0 == string_to_double( "0.0" ) );
assert( 0 == string_to_double( "0.00" ) );
assert( 0 == string_to_double( "0.0e0" ) );
assert( 0 == string_to_double( "0.0e-0" ) );
assert( 0 == string_to_double( "0.0e+0" ) );
assert( 0 == string_to_double( "+0" ) );
assert( 0 == string_to_double( "+0." ) );
assert( 0 == string_to_double( "+0.0" ) );
assert( 0 == string_to_double( "+0.00" ) );
assert( 0 == string_to_double( "+0.0e0" ) );
assert( 0 == string_to_double( "+0.0e-0" ) );
assert( 0 == string_to_double( "+0.0e+0" ) );
assert( 0 == string_to_double( "-0" ) );
assert( 0 == string_to_double( "-0." ) );
assert( 0 == string_to_double( "-0.0" ) );
assert( 0 == string_to_double( "-0.00" ) );
assert( 0 == string_to_double( "-0.0e0" ) );
assert( 0 == string_to_double( "-0.0e-0" ) );
assert( 0 == string_to_double( "-0.0e+0" ) );
assert( 0 == string_to_double( "foobar" ) );
return 0;
}
回答by Evgeny Lazin
Most simple way is to use boost::lexical_cast:
最简单的方法是使用boost::lexical_cast：
double value;
try
{
value = boost::lexical_cast<double>(my_string);
}
catch (boost::bad_lexical_cast const&)
{
value = 0;
}
回答by jmucchiello
atof and strtod do what you want but are very forgiving. If you don't want to accept strings like "32asd" as valid you need to wrap strtod in a function such as this:
atof 和 strtod 做你想做的事，但非常宽容。如果您不想接受像“32asd”这样的字符串作为有效字符串，则需要将 strtod 包装在这样的函数中：
#include <stdlib.h>
double strict_str2double(char* str)
{
char* endptr;
double value = strtod(str, &endptr);
if (*endptr) return 0;
return value;
}
回答by Colin
If it is a c-string (null-terminated array of type char), you can do something like:
如果它是一个 c 字符串（字符类型的空终止数组），您可以执行以下操作：
#include <stdlib.h>
char str[] = "3.14159";
double num = atof(str);
If it is a C++ string, just use the c_str() method:
如果是C++字符串，只需使用c_str()方法：
double num = atof( cppstr.c_str() );
atof() will convert the string to a double, returning 0 on failure. The function is documented here: http://www.cplusplus.com/reference/clibrary/cstdlib/atof.html
atof() 会将字符串转换为双精度值，失败时返回 0。该函数记录在此处：http: //www.cplusplus.com/reference/clibrary/cstdlib/atof.html
回答by Bad
#include <iostream>
#include <string>
using namespace std;
int main()
{
cout << stod(" 99.999 ") << endl;
}
Output: 99.999(which is double, whitespace was automatically stripped)
输出：（99.999这是双倍，空格被自动剥离）
Since C++11 converting string to floating-point values (like double) is available with functions:
stof- convert str to a float
stod- convert str to a double
stold- convert str to a long double
由于 C++11 将字符串转换为浮点值（如 double）可用于以下函数：
stof- 将 str 转换为浮点数
stod- 将 str 转换为 double
stold- 将 str 转换为 long double
I want a function that returns 0 when the string is not numerical.
我想要一个当字符串不是数字时返回 0 的函数。
You can add try catch statement when stodthrows an exception.
您可以在stod抛出异常时添加 try catch 语句。
回答by martiert
Must say I agree with that the most elegant solution to this is using boost::lexical_cast. You can then catch the bad_lexical_cast that might occure, and do something when it fails, instead of getting 0.0 which atof gives.
必须说我同意最优雅的解决方案是使用 boost::lexical_cast。然后，您可以捕获可能发生的 bad_lexical_cast，并在失败时执行某些操作，而不是获得 atof 给出的 0.0。
#include <boost/lexical_cast.hpp>
#include <string>
int main()
{
std::string str = "3.14";
double strVal;
try {
strVal = boost::lexical_cast<double>(str);
} catch(bad_lexical_cast&) {
//Do your errormagic
}
return 0;
}
回答by hyperboreean
One of the most elegant solution to this problem is to use boost::lexical_cast as @Evgeny Lazin mentioned.
这个问题最优雅的解决方案之一是使用 boost::lexical_cast 正如@Evgeny Lazin 提到的那样。
回答by Kyle Cronin
I think atofis exactly what you want. This function parses a string and converts it into a double. If the string does not start with a number (non-numerical) a 0.0 is returned.
我认为atof正是您想要的。此函数解析一个字符串并将其转换为双精度值。如果字符串不以数字（非数字）开头，则返回 0.0。
However, it does try to parse as much of the string as it can. In other words, the string "3abc" would be interpreted as 3.0. If you want a function that will return 0.0 in these cases, you will need to write a small wrapper yourself.
但是，它确实尝试尽可能多地解析字符串。换句话说，字符串“3abc”将被解释为 3.0。如果您想要一个在这些情况下返回 0.0 的函数，您将需要自己编写一个小包装器。
Also, this function works with the C-style string of a null terminated array of characters. If you're using a string object, it will need to be converted to a char* before you use this function.
此外，此函数可与空终止字符数组的 C 样式字符串一起使用。如果您使用的是字符串对象，则需要在使用此函数之前将其转换为 char*。
回答by Ron Savage
There is not a single function that will do that, because 0 is a valid number and you need to be able to catch when the string is not a valid number.
没有一个函数可以做到这一点，因为 0 是一个有效数字，当字符串不是一个有效数字时，您需要能够捕捉到。
You will need to check the string first (probably with a regular expression) to see if it contains only numbers and numerical punctuation. You can then decide to return 0 if that is what your application needs or convert it to a double.
您需要首先检查字符串（可能使用正则表达式）以查看它是否仅包含数字和数字标点符号。如果这是您的应用程序所需要的，您可以决定返回 0 或将其转换为双精度值。
After looking up atof()and strtod()I should rephrase my statement to "there shouldn't be" instead of "there is not" ... hehe
在查找atof()和strtod() 之后，我应该将我的陈述改写为“不应该有”而不是“没有”......呵呵
