Do I need to cast to unsigned char before calling toupper(), tolower(), et al.?

Yes, the argument to toupper needs to be converted to unsigned char to avoid the risk of undefined behavior.

The types char, signed char, and unsigned char are three distinct types. char has the same range and representation as either signed char or unsigned char. (Plain char is very commonly signed and able to represent values in the range -128..+127.)

The toupper function takes an int argument and returns an int result. Quoting the C standard, section 7.4 paragraph 1:

In all cases the argument is an int, the value of which shall be representable as an unsigned char or shall equal the value of the macro EOF . If the argument has any other value, the behavior is undefined.

(C++ incorporates most of the C standard library, and defers its definition to the C standard.)

The [] indexing operator on std::string returns a reference to char. If plain char is a signed type, and if the value of name[0] happens to be negative, then the expression

toupper(name[0])

has undefined behavior.

The language guarantees that, even if plain char is signed, all members of the basic character set have non-negative values, so given the initialization

string name = "Niels Stroustrup";

the program doesn't risk undefined behavior. But yes, in general a char value passed to toupper (or to any of the functions declared in <cctype> / <ctype.h>) needs to be converted to unsigned char, so that the implicit conversion to int won't yield a negative value and cause undefined behavior.

The <ctype.h> functions are commonly implemented using a lookup table. Something like:

// assume plain char is signed
char c = -2;
c = toupper(c); // undefined behavior

may index outside the bounds of that table.

Note that converting to unsigned:

char c = -2;
c = toupper((unsigned)c); // undefined behavior

doesn't avoid the problem. If int is 32 bits, converting the char value -2 to unsigned yields 4294967294. This is then implicitly converted to int (the parameter type), which probably yields -2.

toupper can be implemented so it behaves sensibly for negative values (accepting all values from CHAR_MIN to UCHAR_MAX), but it's not required to do so. Furthermore, the functions in <ctype.h> are required to accept an argument with the value EOF, which is typically -1.

The C++ standard makes adjustments to some C standard library functions. For example, strchr and several other functions are replaced by overloaded versions that enforce const correctness. There are no such adjustments for the functions declared in <cctype>.

In C, toupper (and many other functions) take ints even though you'd expect them to take chars. Additionally, char is signed on some platforms and unsigned on others.

The advice to cast to unsigned char before calling toupper is correct for C. I don't think it's needed in C++, provided you pass it an int that's in range. I can't find anything specific to whether it's needed in C++.

If you want to sidestep the issue, use the toupper defined in <locale>. It's a template, and takes any acceptable character type. You also have to pass it a std::locale. If you don't have any idea which locale to choose, use std::locale(""), which is supposed to be the user's preferred locale:

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

int main()
{
    std::string name("Bjarne Stroustrup");
    std::string uppercase;

    std::locale loc("");

    std::transform(name.begin(), name.end(), std::back_inserter(uppercase),
                   [&loc](char c) { return std::toupper(c, loc); });

    std::cout << name << '\n' << uppercase << '\n';
    return 0;
}

The reference is referring to the value being representable as an unsigned char, not to it being an unsigned char. That is, the behavior is undefined if the actual value is not between 0 and UCHAR_MAX (typically 255). (Or EOF, which is basically the reason it takes an int instead of a char.)