C ++使用find（str，index）计算字符串中char的出现次数， [英] C++ To count the occurrences of a char in a string using find(str, index)

问题描述

这实际上是我的家庭作业和问题状态：

This is actually my homework and the question states:

程序应该确定字符的次数
包含在字符串中（提示：使用
find（str，ƒind）方法搜索字符串。这个方法应该在循环中使用，
在0开始索引值，然后更改索引value to 1 past
上次找到字符的位置的索引。）

"The program should determine how many times the character is contained in the string. (Hint: Search the string by using the find(str,ƒind) 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.)"

但是它所做的就是计算字符串中有多少字符。新的C ++，所以我希望你们可以耐心与我。

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.

#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;
}

推荐答案

我认为@ parapura的代码看nice'er like：

I think @parapura's code looks nice'er like this:

while((size_t contain = s.find(c,i)) != string::npos){
    count++;
    i = contain + 1;
}

它可以很好地解决问题; - ）

and it solves the problem nicely ;-)

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