我如何在c++中实现以下(Python伪代码)?
if argv[1].startswith('--foo='):
foo_value = int(argv[1][len('--foo='):])
(例如，如果argv[1]是——foo=98，那么foo_value是98。)
更新:我很犹豫是否要研究Boost，因为我只是想对一个简单的小命令行工具做一个非常小的改变(我宁愿不学习如何链接并使用Boost进行一个小的改变)。
当前回答
还没有人使用STL算法/错配函数。如果返回true, prefix是'toCheck'的前缀:
std::mismatch(prefix.begin(), prefix.end(), toCheck.begin()).first == prefix.end()
完整的例子:
#include <algorithm>
#include <string>
#include <iostream>
int main(int argc, char** argv) {
if (argc != 3) {
std::cerr << "Usage: " << argv[0] << " prefix string" << std::endl
<< "Will print true if 'prefix' is a prefix of string" << std::endl;
return -1;
}
std::string prefix(argv[1]);
std::string toCheck(argv[2]);
if (prefix.length() > toCheck.length()) {
std::cerr << "Usage: " << argv[0] << " prefix string" << std::endl
<< "'prefix' is longer than 'string'" << std::endl;
return 2;
}
if (std::mismatch(prefix.begin(), prefix.end(), toCheck.begin()).first == prefix.end()) {
std::cout << '"' << prefix << '"' << " is a prefix of " << '"' << toCheck << '"' << std::endl;
return 0;
} else {
std::cout << '"' << prefix << '"' << " is NOT a prefix of " << '"' << toCheck << '"' << std::endl;
return 1;
}
}
编辑:
正如@James T. Huggett所建议的，std::equal更适合回答这个问题:a是B的前缀吗?并且是略短的代码:
std::equal(prefix.begin(), prefix.end(), toCheck.begin())
完整的例子:
#include <algorithm>
#include <string>
#include <iostream>
int main(int argc, char **argv) {
if (argc != 3) {
std::cerr << "Usage: " << argv[0] << " prefix string" << std::endl
<< "Will print true if 'prefix' is a prefix of string"
<< std::endl;
return -1;
}
std::string prefix(argv[1]);
std::string toCheck(argv[2]);
if (prefix.length() > toCheck.length()) {
std::cerr << "Usage: " << argv[0] << " prefix string" << std::endl
<< "'prefix' is longer than 'string'" << std::endl;
return 2;
}
if (std::equal(prefix.begin(), prefix.end(), toCheck.begin())) {
std::cout << '"' << prefix << '"' << " is a prefix of " << '"' << toCheck
<< '"' << std::endl;
return 0;
} else {
std::cout << '"' << prefix << '"' << " is NOT a prefix of " << '"'
<< toCheck << '"' << std::endl;
return 1;
}
}
其他回答
我自己使用的代码:
std::string prefix = "-param=";
std::string argument = argv[1];
if(argument.substr(0, prefix.size()) == prefix) {
std::string argumentValue = argument.substr(prefix.size());
}
std::string text = "--foo=98";
std::string start = "--foo=";
if (text.find(start) == 0)
{
int n = stoi(text.substr(start.length()));
std::cout << n << std::endl;
}
还没有人使用STL算法/错配函数。如果返回true, prefix是'toCheck'的前缀:
std::mismatch(prefix.begin(), prefix.end(), toCheck.begin()).first == prefix.end()
完整的例子:
#include <algorithm>
#include <string>
#include <iostream>
int main(int argc, char** argv) {
if (argc != 3) {
std::cerr << "Usage: " << argv[0] << " prefix string" << std::endl
<< "Will print true if 'prefix' is a prefix of string" << std::endl;
return -1;
}
std::string prefix(argv[1]);
std::string toCheck(argv[2]);
if (prefix.length() > toCheck.length()) {
std::cerr << "Usage: " << argv[0] << " prefix string" << std::endl
<< "'prefix' is longer than 'string'" << std::endl;
return 2;
}
if (std::mismatch(prefix.begin(), prefix.end(), toCheck.begin()).first == prefix.end()) {
std::cout << '"' << prefix << '"' << " is a prefix of " << '"' << toCheck << '"' << std::endl;
return 0;
} else {
std::cout << '"' << prefix << '"' << " is NOT a prefix of " << '"' << toCheck << '"' << std::endl;
return 1;
}
}
编辑:
正如@James T. Huggett所建议的，std::equal更适合回答这个问题:a是B的前缀吗?并且是略短的代码:
std::equal(prefix.begin(), prefix.end(), toCheck.begin())
完整的例子:
#include <algorithm>
#include <string>
#include <iostream>
int main(int argc, char **argv) {
if (argc != 3) {
std::cerr << "Usage: " << argv[0] << " prefix string" << std::endl
<< "Will print true if 'prefix' is a prefix of string"
<< std::endl;
return -1;
}
std::string prefix(argv[1]);
std::string toCheck(argv[2]);
if (prefix.length() > toCheck.length()) {
std::cerr << "Usage: " << argv[0] << " prefix string" << std::endl
<< "'prefix' is longer than 'string'" << std::endl;
return 2;
}
if (std::equal(prefix.begin(), prefix.end(), toCheck.begin())) {
std::cout << '"' << prefix << '"' << " is a prefix of " << '"' << toCheck
<< '"' << std::endl;
return 0;
} else {
std::cout << '"' << prefix << '"' << " is NOT a prefix of " << '"'
<< toCheck << '"' << std::endl;
return 1;
}
}
if(boost::starts_with(string_to_search, string_to_look_for))
intval = boost::lexical_cast<int>(string_to_search.substr(string_to_look_for.length()));
这是完全未经测试的。原理与Python相同。需要提高。StringAlgo和boost。lexicalcast。
检查字符串是否以另一个字符串开头，然后获取第一个字符串的子字符串('slice')并使用词法转换。
在c++ 20中，starts_with作为std::string的成员函数可用，定义为:
constexpr bool starts_with(string_view sv) const noexcept;
constexpr bool starts_with(CharT c) const noexcept;
constexpr bool starts_with(const CharT* s) const;
所以你的代码可以是这样的:
std::string s{argv[1]};
if (s.starts_with("--foo="))
