我用这个解。我知道你不应该修改数据区域....但我认为这主要是缓冲区溢出错误和空字符....大写字母就不一样了。

void to_upper(const std::string str) {
    std::string::iterator it;
    int i;
    for ( i=0;i<str.size();++i ) {
        ((char *)(void *)str.data())[i]=toupper(((char *)str.data())[i]);
    }
}

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++;
    }
}

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

Boost字符串算法:

#include <boost/algorithm/string.hpp>
#include <string>

std::string str = "Hello World";

boost::to_upper(str);

std::string newstr = boost::to_upper_copy<std::string>("Hello World");

typedef std::string::value_type char_t;

char_t up_char( char_t ch )
{
    return std::use_facet< std::ctype< char_t > >( std::locale() ).toupper( ch );
}

std::string toupper( const std::string &src )
{
    std::string result;
    std::transform( src.begin(), src.end(), std::back_inserter( result ), up_char );
    return result;
}

const std::string src  = "test test TEST";

std::cout << toupper( src );

std::string value;
for (std::string::iterator p = value.begin(); value.end() != p; ++p)
    *p = toupper(*p);