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?

And if you really wanted to do this in C++: std::transform(a, a + strlen(a), a, std::toupper); — PaulMcKenzie
– PaulMcKenzie, Commented Oct 12, 2015 at 16:40
Can you explain why you expected this to convert a string to uppercase? Specifically, why did you expect the right side of the = to be evaluated before the left? — David Schwartz
– David Schwartz, Commented Oct 12, 2015 at 16:41
I'm thankful to all people who gave feedback and for the valuable info provided — Alex Koukoulas
– Alex Koukoulas, Commented Oct 12, 2015 at 16:46
@Schullz I strongly believe that learning from your mistakes is the most efficient way not to make the same mistake again. Hence as this problem stemmed solely from experimentation and not in a team project environment I don't think it was a mistake to write this code snippet to begin with — Alex Koukoulas
– Alex Koukoulas, Commented Oct 12, 2015 at 23:04
@MillieSmith In this case, does it matter the order of which item will be made upper case? Anyway, most, if not all of the classical C++ approaches of mutating a string use std::transform. — PaulMcKenzie
– PaulMcKenzie, Commented Oct 13, 2015 at 10:09

John Kugelman · Accepted Answer · 2015-10-12 19:09:23Z

30

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.

edited Oct 12, 2015 at 19:09

John Kugelman

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answered Oct 12, 2015 at 16:37

NathanOliver

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4 Comments

Mark H Over a year ago

Does this mean that the classic c-string copying one-liner while(*s++ = *t++) ; has undefined behavior?

NathanOliver Over a year ago

@markh No as that is two different variables. This is synonymous with while (*s++ = *s++)

Stack Exchange Broke The Law Over a year ago

@SteveJessop Actually, while(*s++ = *t++); has defined behaviour even if s == t. The reason strcpy is undefined with overlapping regions is that strcpy isn't necessarily implemented with while(*s++ = *t++);.

Steve Jessop Over a year ago

@immibis: fair point, agreed it's defined. However, even if strcpy is implemented with that loop it still has undefined behavior for one "direction" of overlap. If the destination is greater than the source and they overlap, then the terminating nul gets overwritten before it's read, and eventually the buffer is overrun! The reason overlap is forbidden in both "directions" is for possible other implementations.

R Sahu · Accepted Answer · 2015-10-12 16:42:58Z

15

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;

edited Oct 12, 2015 at 16:42

answered Oct 12, 2015 at 16:37

R Sahu

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Comments

J A · Accepted Answer · 2015-10-12 16:40:57Z

10

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++;

answered Oct 12, 2015 at 16:40

J A

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Comments

vishal · Accepted Answer · 2015-10-12 16:44:55Z

3

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

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

edited Oct 12, 2015 at 16:44

answered Oct 12, 2015 at 16:41

vishal

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2 Comments

vishal Over a year ago

casting to char was important as value of type int may not fit into receiver type char

Flexo - Save the data dump Over a year ago

That cast achieves nothing that wouldn't happen without it anyway.

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