C string to uppercase in C and C++

Question

While I was putting together a to-uppercase function in C++ I noticed that I did not receive the expected output in C.

C++ function

#include <iostream>
#include <cctype>
#include <cstdio>

void strupp(char* beg)
{
    while (*beg++ = std::toupper(*beg));
}

int main(int charc, char* argv[])
{
    char a[] = "foobar";
    strupp(a);
    printf("%s\n", a);
    return 0;
}

Output as expected:

FOOBAR

---

C function

#include <ctype.h>
#include <stdio.h>
#include <string.h>

void strupp(char* beg)
{
    while (*beg++ = toupper(*beg));
}

int main(int charc, char* argv[])
{
    char a[] = "foobar";
    strupp(a);
    printf("%s\n", a);
    return 0;
}

The output is the expected result with the first character missing

OOBAR

Does anyone know why the result gets truncated while compiling in C?

Answer 1

with respect to above answer, the 'f' is never passed inside function, you should try using this:

     while ((*beg = (char) toupper(*beg))) beg++;

Answer 2

while (*beg++ = std::toupper(*beg));

leads to undefined behavior.

Whether *beg++ is sequenced before or after std::toupper(*beg) is unspecified.

The simple fix is to use:

while (*beg = std::toupper(*beg))
   ++beg;

Answer 3

The line

while (*beg++ = toupper(*beg));

contains a side effect on an entity that's being used twice. You can not know, whether or not beg++ is executed before or after the *beg (inside the toupper). You're just lucky that both implementations show both behaviors, as I'm pretty sure it's the same for C++. (However, there were some rule changes for c++11, which I'm not certain of - still, it's bad style.)

Just move the beg++ out of the condition:

while (*beg = toupper(*beg)) beg++;

Answer 4

The problem is that there is no sequence point in

while (*beg++ = toupper(*beg));

So we have undefined behavior. What the compiler is doing in this case is evaluating beg++ before toupper(*beg) In C where in C++ it is doing it the other way.