Difference between %u and %d in scanf()

Question

In C++, if I read an integer from a string, it seems it does not really matter whether I use u or d as conversion specifier as both accept even negative integers.

#include <cstdio>
using namespace std;

int main() {
    int u, d;
    sscanf("-2", "%u", &u);
    sscanf("-2", "%d", &d);
    puts(u == d ? "u == d" : "u != d");
    printf("u: %u %d\n", u, u);
    printf("d: %u %d\n", d, d);
    return 0;
}

^Ideone.com

I dug deeper to find if there is any difference. I found that

int u, d;
sscanf("-2", "%u", &u);
sscanf("-2", "%d", &d);

is equivalent to

int u, d;
u = strtoul("-2", NULL, 10);
d = strtol("-2", NULL, 10);

according to cppreference.com.

Is there any difference at all between u and d when using these conversion specifiers for parsing, i.e. in format passed to scanf-type functions? What is it?

_{The answer is the same for C and C++, right? If not, I am interested in both.}

Answer 1

%d: Scan an integer as a decimal signed int. A similar conversion specifier, %i, interprets the number as hexadecimal when preceded by 0x and as octal when preceded by 0. Otherwise, it is identical.

%u: Scan an integer as a decimal unsigned int.