@dirkgently的回答非常鼓舞人心，但我想强调的是，出于如下所示的担忧，

与from的所有其他函数一样，如果实参的值既不能表示为unsigned char也不等于EOF，则std::toupper的行为是未定义的。为了安全地将这些函数与普通字符(或有符号字符)一起使用，参数首先应该转换为无符号字符 参考:std:: toupper

由于标准没有指定plain char是带符号的还是无符号的[1]， std::toupper的正确用法应该是:

#include <algorithm>
#include <cctype>
#include <iostream>
#include <iterator>
#include <string>

void ToUpper(std::string& input)
{
    std::for_each(std::begin(input), std::end(input), [](char& c) {
        c = static_cast<char>(std::toupper(static_cast<unsigned char>(c)));
    });
}

int main()
{
    std::string s{ "Hello world!" };
    std::cout << s << std::endl;
    ::ToUpper(s);
    std::cout << s << std::endl;

    return 0;
}

输出:

Hello world!
HELLO WORLD!

这是c++ 11的最新代码

std::string cmd = "Hello World";
for_each(cmd.begin(), cmd.end(), [](char& in){ in = ::toupper(in); });

我的解决方案(清除第6位):

#include <ctype.h>

inline void toupper(char* str)
{
    while (str[i]) {
        if (islower(str[i]))
            str[i] &= ~32; // Clear bit 6 as it is what differs (32) between Upper and Lowercases
        i++;
    }
}

//Since I work on a MAC, and Windows methods mentioned do not work for me, I //just built this quick method.


string str; 
    str = "This String Will Print Out in all CAPS";
    int len = str.size(); 
    char b;

for (int i = 0; i < len; i++){
    b = str[i]; 
    b = toupper(b); 
   // b = to lower(b); //alternately 
     str[i] = b;    
}
    

cout<<str;

尝试toupper()函数(#include <ctype.h>)。它接受字符作为参数，字符串是由字符组成的，所以你必须遍历每个单独的字符，当它们放在一起组成字符串时

使用lambda。

std::string s("change my case");

std::locale locale;
auto to_upper = [&locale] (char ch) { return std::use_facet<std::ctype<char>>(locale).toupper(ch); };

std::transform(s.begin(), s.end(), s.begin(), to_upper);