C++ 将char数组转换为单个int？
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原文地址: http://stackoverflow.com/questions/6093414/
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Convert char array to single int?
提问by IsThisTheEnd
Anyone know how to convert a char array to a single int?
有人知道如何将char数组转换为单个int吗？
char hello[5];
hello = "12345";
int myNumber = convert_char_to_int(hello);
Printf("My number is: %d", myNumber);
回答by Alok Save
There are mulitple ways of converting a string to an int.
有多种方法可以将字符串转换为 int。
Solution 1: Using Legacy C functionality
解决方案 1：使用传统 C 功能
int main()
{
//char hello[5];
//hello = "12345"; --->This wont compile
char hello[] = "12345";
Printf("My number is: %d", atoi(hello));
return 0;
}
Solution 2: Using lexical_cast(Most Appropriate & simplest)
解决方案 2：使用lexical_cast（最合适和最简单）
int x = boost::lexical_cast<int>("12345");
Solution 3: Using C++ Streams
解决方案 3：使用 C++ Streams
std::string hello("123");
std::stringstream str(hello);
int x;
str >> x;
if (!str)
{
// The conversion failed.
}
回答by Rick Smith
If you are using C++11, you should probably use stoibecause it can distinguish between an error and parsing "0".
如果您正在使用C++11，您可能应该使用，stoi因为它可以区分错误和解析"0"。
try {
int number = std::stoi("1234abc");
} catch (std::exception const &e) {
// This could not be parsed into a number so an exception is thrown.
// atoi() would return 0, which is less helpful if it could be a valid value.
}
It should be noted that "1234abc" is implicitly convertedfrom a char[]to a std:stringbefore being passed to stoi().
应当指出的是，“1234abc”被隐式转换从char[]一个std:string被传递到前stoi()。
回答by Rick Smith
回答by Holly
I'll just leave this here for people interested in an implementation with no dependencies.
我将把这个留在这里给那些对没有依赖关系的实现感兴趣的人。
inline int
stringLength (char *String)
{
int Count = 0;
while (*String ++) ++ Count;
return Count;
}
inline int
stringToInt (char *String)
{
int Integer = 0;
int Length = stringLength(String);
for (int Caret = Length - 1, Digit = 1; Caret >= 0; -- Caret, Digit *= 10)
{
if (String[Caret] == '-') return Integer * -1;
Integer += (String[Caret] - '0') * Digit;
}
return Integer;
}
Works with negative values, but can't handle non-numeric characters mixed in between (should be easy to add though). Integers only.
适用于负值，但不能处理中间混合的非数字字符（虽然应该很容易添加）。仅整数。
回答by Tr?n Hi?p
I use :
我用 ：
int convertToInt(char a[1000]){
int i = 0;
int num = 0;
while (a[i] != 0)
{
num = (a[i] - '0') + (num * 10);
i++;
}
return num;;
}
回答by Reno
Long story short you have to use atoi()
长话短说你必须使用atoi()
ed:
编：
If you are interested in doing this the right way:
如果您有兴趣以正确的方式执行此操作：
char szNos[] = "12345";
char *pNext;
long output;
output = strtol (szNos, &pNext, 10); // input, ptr to next char in szNos (null here), base
回答by harper
Ascii string to integer conversion is done by the atoi()function.
Ascii 字符串到整数的转换是由atoi()函数完成的。
