doubts about incrementation of array pointer

Question

Here I'm getting the output 'bcde' but since 'a' is a constant array pointer it should not have been incremented inside fun() right? Then why is it allowed here?

    void fun(char a[])
    {
     a++;
     printf("%s",a);
    }

    void main()
    {
     char a[]="abcde";
     fun(a);
    }

You _claim_ `a` is a constant array pointer, but you have two `a` variables: neither is a constant pointer nor pointer to constant. In fact, one isn't even a pointer! — Mooing Duck, Jul 17 '14 at 17:33
possible duplicate of [what is array decaying?](http://stackoverflow.com/questions/1461432/what-is-array-decaying) — Deduplicator, Jul 17 '14 at 17:46

score 3 · Answer 1 · answered Jul 17 '14 at 17:22

3

When you pass an array to a function in C, the array "decays" to a pointer. In other words, an equivalent function can be declared like this:

void fun(char *a)

Now the code inside the function makes perfect sense: the pointer is incremented, so when the result is passed to printf, the original string is printed starting with the second letter.

answered Jul 17 '14 at 17:22

Sergey Kalinichenko

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@user3801801 The pointer is not constant. Moreover, the memory the pointer points to is perfectly modifiable, because you wrote `char a[]="abcde"`, not `char *a="abcde"`. But even a pointer to constant can be modified (as opposed to a constant pointer, which is declared differently). – Sergey Kalinichenko Jul 17 '14 at 17:25
1

No, not a constant pointer. A pointer to constant data is not the same as a constant pointer. Also, what user3801801 said is correct. The data isn't even constant. – Mike Holt Jul 17 '14 at 17:25
okay...and if it was declared char *s="abcde" inside main den it still remains the same ryt? – joel.wilson Jul 17 '14 at 17:29
1

@user3801801 Yes, the code and its behavior would remain the same. – Sergey Kalinichenko Jul 17 '14 at 17:30
In C,if i declare char *s="abcde" then content of s(i.e.*s='o';) is not allowed ryt?den y is it allowed in C++? – joel.wilson Jul 17 '14 at 17:34
1

@user3801801 The assignment is not allowed in both C and in C++. The difference is that in C the code will compile, while in C++ the compiler is going to complain about the assignment of a string literal to a non-const pointer (C++ requires `const char *s = "abcd"`) – Sergey Kalinichenko Jul 17 '14 at 17:37

haccks · Answer 2 · 2014-07-17T17:30:28.920

2

When you pass an array name to your function then you are passing pointer to its first element.

void fun(char a[])  // a is not an array of char

is equivalent to

void fun(char *a)

You can modify a inside function because it is not an array name but a pointer to char.
You can't modify a in main as it is declared as array. Array names are non modifiable l-values.

edited Jul 17 '14 at 17:30

answered Jul 17 '14 at 17:24

haccks

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score 1 · Answer 3 · answered Jul 17 '14 at 17:22

1

When you are in function arguments, char a[] is treated exactly as if you wrote char *a.

answered Jul 17 '14 at 17:22

Bill Lynch

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score 1 · Answer 4 · answered Jul 17 '14 at 17:59

There is no any "constant array pointer"

In this statement

char a[]="abcde";

there is declared a non-const array. You may change it for example as

a[0] = 'A';

When you pass the array to the function as an argument then there is used conversion from the array to a pointer of type char * that points to the first element of the array.

This function declaration

void fun(char a[]);

is equivalent to

void fun(char *a);

The parameter is adjaced to the pointer.

So inside the function you can to change the pointer itself and the object it points to. For example

 a++;
 *a = 'B';

If you want that the pointer would not be changed in the function you could declare it as

void fun(char * const a);

In this case the compiler would issue an error for statement

 a++;

because a is indeed a const pointer.

doubts about incrementation of array pointer

4 Answers4