The difference between `char *str` and `char str[]` in the declaration of a string

Question

I have the following function in C:

int TransMinMaj(char *c)
{
    if((*c) >= 'a' && (*c) <= 'z')
    {
        *c += 'A' - 'a';
        return 1;
    }
    else if((*c) >= 'A' && (*c) <= 'Z')
        return 1;
    return 0;

}

As you can see, this function:
1. Return 1 if the character tested is a letter
2. Transform the lower case to upper case
3. Otherwise, return 0 (to indicate that it is not a letter)

I have chosen to pass the parameter by address because I want to change the value of the passed parameter c in the memory.

Now my question comes: I tested my function withe the following code :

char str[] = "abcdefg";
printf("Before: %s\n", str);    
TransMinMaj(&str[1]);
printf("After: %s\n", str);

Until now, everything is good, no errors, no warnings. And as you can see, in this test, I transforme the lower case letter b (also the second element of the string (array of characters)) to the upper case letter 'B' in the memory.

But, if change the test code a litte bit:

char *str = "abcdefg";
printf("Before: %s\n", str);    
TransMinMaj(&str[1]);
printf("After: %s\n", str);

There is a segmentation error(core dumped). But I don't know why.
So here comes my question:
1. What is the difference between char *str and char str[] in a declaration ?
2. Where does my segmentation error come from?

Please use search before asking questions. It is very likely that your question has been asked before. This one has been asked since the beginning of time. — 2501, Sep 09 '16 at 07:38
With the pointer you can do: ptr1=ptr2; with the array not. With the pointer you can do ptr1++; with the array not. With the array you can call sizeof(array)/sizeof(array[0]) with the pointer not. And such things — jurhas, Sep 09 '16 at 07:51

score 2 · Accepted Answer · answered Sep 09 '16 at 07:37

char str[] = "abcdefg";

and

char *str = "abcdefg";

are two different things.

The first one is an array, initialized with "abcdefg".
The second one is a pointer pointing to the string literal "abcdefg".

In the second case, when you attempt to modify the string literal, you invoke undefined behavior, as any attempt to modify a string literal is UB.

The difference between `char *str` and `char str[]` in the declaration of a string

1 Answers1