C++ 逐个字符遍历字符串
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原文地址: http://stackoverflow.com/questions/14668326/
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Iterate through string char by char
提问by user1988385
I try to iterate through a string char by char. I tried something like this:
我尝试逐个字符地遍历字符串。我试过这样的事情：
void print(const string& infix)
{
char &exp = infix.c_str();
while(&exp!='h
e
l
l
o
')
{
cout<< &exp++ << endl;
}
}
So this function call print("hello"); should return:
所以这个函数调用 print("hello"); 应该返回：
for(unsigned int i = 0; i<infix.length(); i++) {
char c = infix[i]; //this is your character
}
I try using my code, but it doesn't work at all. btw the parameter is a reference not a pointer. Thank you
我尝试使用我的代码，但它根本不起作用。顺便说一句，参数是引用而不是指针。谢谢
采纳答案by Dhaivat Pandya
void print(const std::string& infix)
{
for(auto c : infix)
std::cout << c << std::endl;
}
That's how I've done it. Not sure if that's too "idiomatic".
我就是这样做的。不知道这是否太“惯用”了。
回答by Mark Tolonen
Your code needs a pointer, not a reference, but if using a C++11 compiler, all you need is:
您的代码需要一个指针，而不是一个引用，但如果使用 C++11 编译器，您只需要：
for (auto i = inflix.begin(); i != inflix.end(); ++i) std::cout << *i << '\n';
回答by 0x499602D2
If you're using std::string, there really isn't a reason to do this. You can use iterators:
如果您正在使用std::string，真的没有理由这样做。您可以使用迭代器：
void print(const string& infix)
{
for (auto c = infix.begin(); c!=infix.end(); ++c)
{
std::cout << *c << "\n";
}
std::cout << std::endl;
}
As for your original code you should have been using char*instead of charand you didn't need the reference.
至于你的原始代码，你应该一直使用char*而不是char你不需要参考。
回答by billz
std::string::c_str()returns const char*, you can't use char&to hold it. Also exp is pointer already, you don't need reference:
std::string::c_str()返回const char*，你不能char&用来保持它。exp 已经是指针了，你不需要引用：
Better use iterator though:
最好使用迭代器：
void print(const string& infix)
{
const char *exp = infix.c_str();
while(*exp!='##代码##')
{
cout << *exp << endl;
exp++;
}
}
To fix your original code, try:
要修复您的原始代码，请尝试：
##代码##