Returning null char array with rewriting Java's String.toUpperCase() [duplicate]

Question

I am originally a Java programmer and I have a deep love of the syntax, especially regarding the String object. With C++, I have tried to recreate the toUpperCase() method that Java has. The only problem is that it always returns a String object that has an empty/NULL char array.

String String::toUpperCase() { char *a = new char[this->length + 1]; memset(a, 0, this->capacity + 1); memcpy(a, this->characters, this->length); for (int i = 0; i < strlen(this->characters); i++) { toupper(a[i]); } return *new String(a); }

You'd be wiser using something from here: stackoverflow.com/questions/735204/… and std::string in general. This function leaks memory every time it's called. Also, read some documentation for toupper. — Qaz
– Qaz, Commented Mar 18, 2014 at 13:30
Also: In Java and C# you use new a lot, in C++ you should avoid it when possible (no GC). — crashmstr
– crashmstr, Commented Mar 18, 2014 at 13:35
As a general rule, do not try to recreate java class APIs in C++ classes (they are optimized for Java, so what you end up with tends to be a variation of the "I can write C code in any language" theme). To convert a C++ string to upper case, see this answer. — utnapistim
– utnapistim, Commented Mar 18, 2014 at 13:37
On another note, stop abusing pointers (i.e. you should write return String(a); instead of 'return *new String(a);' and use a smart pointer on a, or better yet, a std::vector); Also, stop writing this->. It is unnecessary. Also consider using std::fill instead of memset and std::copy instead of memcpy. — utnapistim
– utnapistim, Commented Mar 18, 2014 at 13:39

Zac Howland · Accepted Answer · 2014-03-18 13:35:23Z

2

You have a few memory problems with your attempt, as well as a logical one. All you need to return a copy of a string with the characters being upper case is:

std::string str = "My Original string"; std::string myCopy(str); std::locale loc; std::transform(myCopy.begin(), myCopy.end(), myCopy.begin(), [&](char c) { return std::toupper(c, loc); });

answered Mar 18, 2014 at 13:35

Zac Howland

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5 Comments

japreiss Over a year ago

or just std::transform(myCopy.begin(), myCopy.end(), myCopy.begin(), std::toupper) if you don't need locale sensitivity.

Qaz Over a year ago

@japreiss, That won't always work out so well, although the answer won't either if char is signed and there are negative values. You can check the link I commented for more discussion on this method.

Konrad Rudolph Over a year ago

@japreiss I’d argue that not using a locale is always broken when operating on text – how can you exclude the need for using locales? Then again, locales in C++ are broken anyway (i.e. the above does the wrong thing for the German text “Maß”, regardless of locale).

Zac Howland Over a year ago

@japreiss The primary problem with using the version that does not take a locale is that it is the carry over from the C standard library, which takes, and returns, an int with very specific criteria. If char is signed on a given platform, the call to int std::toupper(int) can lead to undefined behavior.

japreiss Over a year ago

Thanks, I guess my incorrect comment turned out to be a good learning experience :)

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