c++ - 初学者C++问题

我在这里遵循了代码示例



并用我自己的代码实现如下

void CharString::MakeUpper()
{
char* str[strlen(m_pString)];
int i=0;
str[strlen(m_pString)]=m_pString;
char* c;
while (str[i])
  {
    c=str[i];
    putchar (toupper(c));
    i++;
  }
}

但这给了我下面的编译器错误

CharString.cpp: In member function 'void CharString::MakeUpper()':
CharString.cpp:276: error: invalid conversion from 'char*' to 'int'
CharString.cpp:276: error:   initializing argument 1of 'int toupper(int)'
CharString.cpp: In member function 'void CharString::MakeLower()':

这是276行

putchar (toupper(c));

我知道toupper在寻找int作为参数，并且还返回int，这是问题吗？如果可以，示例如何工作？

最佳答案

我将假设m_pString是C样式字符串(char *)。您做的事情比需要做的还要麻烦。

void CharString::MakeUpper()
{
   char* str = m_pString; // Since you're not modifying the string, there's no need to make a local copy, just get a pointer to the existing string.
   while (*str) // You can use the string pointer as an iterator over the individual chars
   {
      putchar (toupper(*str)); // Dereference the pointer to get each char.
      str++;   // Move to the next char (you can merge this into the previous line if so desired, but there's no need.
   }
}

在您引用的示例中，它起作用的原因是由于变量的声明方式。

int main ()
{
  int i=0;
  char str[]="Test String.\n";  // This is a compile time string literal, so it's ok to initialize the array with it.  Also, it's an array of `char`s not `char*`s.
  char c;  // Note that this is also a `char`, not a `char *`
  while (str[i])
  {
    c=str[i];
    putchar (toupper(c));
    i++;
  }
  return 0;
}

由于使用C字符串的方式容易出错，因此最好的选择是std::string:

void CharString::MakeUpper()
{
   string str(m_pString);
   transform(str.begin(), str.end(), ostream_iterator<char>(cout), &toupper);
}