C ++下/上字符char指针

Question

Do you guys know why the following code crash during the runtime?

char* word;
word = new char[20];
word = "HeLlo"; 
for (auto it = word; it != NULL; it++){        
    *it = (char) tolower(*it);

I'm trying to lowercase a char* (string). I'm using visual studio.

Thanks

Answer 1

You cannot compare it to NULL . Instead you should be comparing *it to '\\0' . Or better yet, use std::string and never worry about it :-)

In summary, when looping over a C-style string. You should be looping until the character you see is a '\\0' . The iterator itself will never be NULL , since it is simply pointing a place in the string. The fact that the iterator has a type which can be compared to NULL is an implementation detail that you shouldn't touch directly.

Additionally, you are trying to write to a string literal. Which is a no-no :-).

EDIT : As noted by @Cheers and hth. - Alf, tolower can break if given negative values. So sadly, we need to add a cast to make sure this won't break if you feed it Latin-1 encoded data or similar.

This should work:

char word[] = "HeLlo";
for (auto it = word; *it != '\0'; ++it) {
    *it = tolower(static_cast<unsigned char>(*it));
}

Answer 2

You're setting word to point to the string literal, but literals are read-only, so this results in undefined behavior when you assign to *it . You need to make a copy of it in the dynamically-allocated memory.

char *word = new char[20];
strcpy(word, "HeLlo");

Also in your loop you should compare *it != '\\0' . The end of a string is indicated by the character being the null byte, not the pointer being null.

Answer 3

Given code (as I'm writing this):

char* word;
word = new char[20];
word = "HeLlo"; 
for (auto it = word; it != NULL; it++){        
    *it = (char) tolower(*it);

This code has Undefined Behavior in 2 distinct ways, and would have UB also in a third way if only the text data was slightly different:

• Buffer overrun.
  The continuation condition it != NULL will not be false until the pointer it has wrapped around at the end of the address range, if it does.

• Modifying read only memory.
  The pointer word is set to point to the first char of a string literal, and then the loop iterates over that string and assigns to each char .

• Passing possible negative value to tolower .
  The char classification functions require a non-negative argument, or else the special value EOF . This works fine with the string "HeLlo" under an assumption of ASCII or unsigned char type. But in general, eg with the string "Blåbærsyltetøy" , directly passing each char value to tolower will result in negative values being passed; a correct invocation with ch of type char is (char) tolower( (unsigned char)ch ) .

Additionally the code has a memory leak , by allocating some memory with new and then just forgetting about it.

A correct way to code the apparent intent:

using Byte = unsigned char;

auto to_lower( char const c )
    -> char
{ return Byte( tolower( Byte( c ) ) ); }

// ...
string word = "Hello";
for( char& ch : word ) { ch = to_lower( ch ); }

Answer 4

There are already two nice answers on how to solve your issues using null terminated c-strings and poitners. For the sake of completeness, I propose you an approach using c++ strings:

string word;           // instead of char* 
//word = new char[20]; // no longuer needed: strings take care for themseves
word = "HeLlo";        //  no worry about deallocating previous values: strings take care for themselves
for (auto &it : word)  // use of range for, to iterate through all the string elements      
    it = (char) tolower(it);

Answer 5

它崩溃是因为您正在修改字符串文字。

Answer 6

there is a dedicated functions for this use strupr for making string uppercase and strlwr for making the string lower case.

here is an usage example:

char str[ ] = "make me upper";
printf("%s\n",strupr(str));


char str[ ] = "make me lower";
printf("%s\n",strlwr (str));