C++ 使用 find(str, index) 计算字符串中某个字符的出现次数
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/8870263/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
C++ To count the occurrences of a char in a string using find(str, index)
提问by delphi316
This is actually my homework and the question states:
这实际上是我的作业，问题指出：
"The program should determine how many times the character is contained in the string. (Hint: Search the string by using the find(str,?ind) method. This method should be used in a loop that starts the index value at 0 and then changes the index value to 1 past the index of where the char was last found.)"
“程序应该确定该字符在字符串中包含多少次。（提示：使用 find(str,?ind) 方法搜索字符串。该方法应该在索引值从 0 开始的循环中使用，并且然后将索引值更改为 1 超过上次找到字符的位置的索引。）”
This is what I've came up with but all it does is count how many character there is in the string. New to C++ so I hope you guys can be patient with me.
这是我想出的，但它所做的只是计算字符串中有多少个字符。刚接触 C++，所以我希望你们能对我有耐心。
#include <string>
#include <iostream>
using namespace std;
int main()
{
string s;
char c;
size_t contain;
int count = 0;
cout << "Enter a string : ";
getline(cin, s);
cout <<"Enter a char : ";
cin >> c;
for(int i = 0; i < s.length(); i++)
{
contain = s.find(c, i);
if (contain =! string::npos )
{
count++;
}
}
cout << count <<endl;
return 0;
}
回答by Oliver Charlesworth
This:
这个：
(contain =! string::npos)
^^
doesn't do what you think it does. Take a look at http://en.wikipedia.org/wiki/Operators_in_C_and_C%2B%2B.
不会做你认为它会做的事情。看看http://en.wikipedia.org/wiki/Operators_in_C_and_C%2B%2B。
Also, this:
还有这个：
contain = s.find(c, i);
doesn't do anything particularly useful (if all you're doing is incrementing ievery iteration). You'll end up counting some occurrences multiple times.
没有做任何特别有用的事情（如果你所做的只是增加i每次迭代）。您最终会多次计算某些出现次数。
[Note:You can solve the actual task much more cleanly by using count = std::count(s.begin(), s.end(), c).]
[注意：您可以使用count = std::count(s.begin(), s.end(), c). ]
回答by ddacot
You can do simply :
你可以简单地做：
Instead of
代替
contain = s.find(c, i);
if (contain =! string::npos )
{
count++;
}
Write
写
if(s[i] == c)
{
count++;
}
also, you can use this.
此外，您可以使用它。
#include <algorithm>
int count = std::count(s.begin(), s.end(), c);
cout<<count;
回答by Martin Kristiansen
I think @parapura's code looks nice'er like this:
我认为@parapura 的代码看起来更像这样：
while((size_t contain = s.find(c,i)) != string::npos){
count++;
i = contain + 1;
}
and it solves the problem nicely ;-)
它很好地解决了这个问题;-)
回答by parapura rajkumar
I think according to your problem statement.
我认为根据你的问题陈述。
changes the index value to 1 past the index of where the char was last found
将索引值更改为 1 超过上次找到字符的位置的索引
Your while loop should be something like
你的while循环应该是这样的
while( 1 )
{
contain = s.find(c, i);
if (contain != string::npos )
{
count++;
i = contain + 1;
}
else
{
break;
}
}
