在c语言中如何统计得出一个字符串中字符的个数？？？ 用c语言语句怎样编写？？？？

Question

如题所述

Answer 1

#include <iostream>
#include <string>
using namespace std;

int main()
{
int Digit=0,Upper=0,Lower=0;
string OriginStr;
cout<<"请输入字符串："<<endl;
cin>>OriginStr;
for (int sz=0;sz<OriginStr.size();++sz)
{
if (isdigit(OriginStr[sz]))
++Digit;
else if (isupper(OriginStr[sz]))
++Upper;
else if (islower(OriginStr[sz]))
++Lower;
else;
}
cout<<"数字个数："<<Digit<<endl
<<"大写个数："<<Upper<<endl
<<"小写个数："<<Lower<<endl;
return 0;
}

Answer 2

在c语言中如何统计得出一个字符串中字符的个数？？？
#include <iostream>
#include <string>
using namespace std;

int main()
{
int Digit=0,Upper=0,Lower=0;
string OriginStr;
cout<<"请输入字符串："<<endl;
cin>>OriginStr;
for (int sz=0;sz<OriginStr.size();++sz)
{
if (isdigit(OriginStr[sz]))
++Digit;
else if (isupper(OriginStr[sz]))
++Upper;
else if (islower(OriginStr[sz]))
++Lower;
else;
}
cout<<"数字个数："<<Digit<<endl
<<"大写个数："<<Upper<<endl
<<"小写个数："<<Lower<<endl;
return 0;
}

Answer 3

int count=0;
for(i=0;i<strlen(str);i++)
if((str[i]>='a'&&str[i]<'z')||(str[i]>='A'&&str[i]<'Z'))
count++;
return count;