CStrings Question: What is the difference between char str[] and char* str

Hi;

I guess I'm a newbie with this question.

My question is about C strings.


I'm trying to figure out why the behavior of these 2 strings are different as it relates to toupper() function ?

char str1[] ="For the perfecting of the saints";
char* str2 ="For the perfecting of the saints";


//when I call ucase(str1) everything works perfectly.

ucase(str1);

//but when i call

ucase(str2)

//it crashes. Can you tell me what's the difference between char str1[] and char* str2?



Here's the ucase() function i'm calling below.

//convert to uppercase
void ucase(char* buffer)
{
int len = strlen(buffer);

for(int i = 0; i < len; i++)
{
buffer[i] = toupper(buffer[i]);
}
}




I can loop through both just fine using something like this
for(int i =0; i<36; i++){
printf("here %c\n", str1[i]);
}

but when it seems to me toupper breaks when you use it with char* x for some reason.

Thanks in Advance.

Stev

char[] is an array of char.

Code:

---

char str1[] ="For the perfecting of the saints";

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str1 is an array of char initialized with a string literal.

char* is a pointer to char

Code:

---

char* str2 ="For the perfecting of the saints";

---

str2 is a pointer to char holding the address of the first char of the string literal (which is immutable, i.e. you cannot change; attempting to do that is yields undefined behaviour). Hence it should rather be:

Code:

---

const char* str2 ="For the perfecting of the saints";

---

http://www.codeguru.com/forum/showth...string+literal
http://www.codeguru.com/forum/showth...string+literal

Everytime you try modifying a string literal, God kills a kitten! And your program as well...

Cheers

Hi all.

In addition what Cilu said, char[] defines an array of char initialized with a string literal and closed by the '\0', the special character that ends the strings;
When you call ucase with char[], the function strlen finds the '\0' and all works fine; instead, when you call ucase with char* strlen don't find the '\0' so your program crashes.

Quote:

---

Originally Posted by davide++

When you call ucase with char[], the function strlen finds the '\0' and all works fine; instead, when you call ucase with char* strlen don't find the '\0' so your program crashes.

---

strlen works fine with char* variables as given in the post.

The problem is that ucase attempts to modify a string literal.
This results in undefined behavior. It might do what you want,
it might not. (For example, ucase would not crash with
VC++ version 5, but will for versions after that.).

Quote:

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Originally Posted by cilu

Hence it should rather be:

Code:

---

const char* str2 ="For the perfecting of the saints";

---

---

no it should be

Code:

---

char const* str2 ="For the perfecting of the saints";

---

just try compiling this:

Code:

---

const char* str1 ="For the perfecting of the saints";
char const* str2 ="For the perfecting of the saints";
str2 = "nooooooooooooot";
str1 = "nooooooooooooot";

---

Quote:

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Originally Posted by Mitsukai

no it should be

Code:

---

char const* str2 ="For the perfecting of the saints";

---

---

What's the difference?

- petter

"const" only matters chich side of the "*" it is on. To the left (either before or after the type), the DATA is const. To the right of the "*", the pointer is const (must always point to the same location), but does not imply that the data is const.

Hi Mitsukai.

Quote:

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Originally Posted by Mitsukai

no it should be

Code:

---

char const* str2 ="For the perfecting of the saints";

---

---

Both char const* and const char* should mean the same thing. What compiler are you using? The important thing is that the asterisc is to the right of the const keyword.