CODEWITHSUNDEEP
carrayspointerscharshrink
exliontamer
exliontamer

Reputation: 125

Is it ok to shrink an array in C by changing the pointer to a higher index?

For example, if I have:

char arr[] = "this is a test";

but I decided I wanted to shrink it by 5 so I do:

arr = &arr[5]:

I tried this out and it seemed to work fine, Im just wondering if this would lead to any undefined behavior or "memory issues".

Upvotes: 2

Views: 142

Answers (1)

klutt
klutt

Reputation: 31409

No, this code would not even compile. It gives this error message:

error: assignment to expression with array type
     arr = &arr[5];
         ^

What you could do is this:

char arr[] = "this is a test";
char *ptr = arr;
printf("Before: %s\n", ptr);
ptr = &arr[5];
printf("After: %s\n", ptr);

If this is a good idea or not depends on the situation. Since arrays are allocated on the stack it's pretty safe. It will not lead to memory leaks.

Here is an interesting twister on the subject. How about writing this?

char * arr = "this is a test";

Then you have a pointer and no array, right? Well, this code does indeed allow you to perform the reassignment arr = &arr[5]. However, this is NOT equivalent to this:

char str[] = "this is a test";
char * arr = malloc(strlen(str));
strcpy(arr, str);

It is instead equivalent to this:

static char _unnamed_[] = "this is a test";
char * arr = _unnamed_;

One difference between these two is if you are returning arr from a function. Another is if you are calling free on it.

Arrays vs pointers

In a comment to your post you "thought that an array name was essentially identical to a pointer", which is wrong. It is very easy to make this mistake, and I have done it thousands of times, and I have had my fair share of humble pie here at SO on that matter. But arrays are not pointers. However, an array does in many cases decay to a pointer, which is exactly what's happening on the line char *ptr = arr above.

There are many questions with enlightening answers about this. Here is two:

What is the difference between char array vs char pointer in C?

Why do I get a segmentation fault when writing to a string initialized with "char *s" but not "char s[]"?

Actually, the array decaying to a pointer is also what's happening on the line ptr = &arr[5]. According to the definition of the [] operator, this is the same as writing ptr = &(*(arr + 5))

Why is a[5] the same as 5[a]?

Upvotes: 5

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