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

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.

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

关于上面的答案，'f'永远不会传递到函数内部，你应该尝试使用这个：

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