Confused C pointer behavior

Question

#include <stdio.h>

void printa(char *a[])
{
    for (int i = 0; i < 3; ++i) {       
        printf("%s\n", *a);
        a++;
    }   
}

int main(void)
{
    char *a[] = {"The first", "The second", "The third"};

    for (int i = 0; i < 3; ++i) {
        printf("%s\n", *a); 
        a++; // error: cannot increment value of type 'char *[3]'
    }

    printa(a); //OK 

    return 0;
}

Thus, my question is why the code a++ in main function causing compile error (error: cannot increment value of type 'char *[3]'). But if I pass array of pointer a to function printa and call a++ on that pointer, it works perfectly.

Thanks,

A is an array , each element of which is a pointer to a char.... so a++ will fail — Michi, Aug 17 '15 at 16:45

haccks · Accepted Answer · 2015-08-17T16:50:04.330

3

Postfix ++ can't have operand of array type. The type of a in main function is char *[3], i.e. an array of pointers to char while in function printa it is of type char **.

As a function parameter

char *a[]

is equivalent to

char **a

edited Aug 17 '15 at 16:50

answered Aug 17 '15 at 16:44

haccks

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You mean that an array pass to function as a pointer. Thus, `char a[]` is equivalent to `char *a` and `char *a[]` is equivalent to `char **a`. – Linh Aug 17 '15 at 16:47
@Linh Yes it is when passed in function. – ameyCU Aug 17 '15 at 16:47
Yes. Exactly. When an array is passed to a function then a pointer to its first element is passed to that function. – haccks Aug 17 '15 at 16:48
if you speak about function arg then yes, but inside the function *a[] is not the same as a**, no. First is Array and second is pointer – Michi Aug 17 '15 at 16:49
Why we cannot `++` on array type? Is it due to `a` is the base of array, thus we can change it? – Linh Aug 17 '15 at 16:53
@Linh; Read [this](http://www.c-faq.com/aryptr/arrayassign.html). – haccks Aug 17 '15 at 16:55
@haccks: Thanks. I got it :) – Linh Aug 17 '15 at 16:58

score 1 · Answer 2 · answered Aug 17 '15 at 16:43

1

Simply because array names are non-modifiable l-value. Which cannot be used as left operand in any expression. Therefore, you cannot keep it on the left side of a = expression, or increment it using an increment operator

answered Aug 17 '15 at 16:43

Haris

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score 1 · Answer 3 · answered Aug 17 '15 at 16:48

1

When you called printa function, the argument is now another variable of type char **. That variable you can increment.

But inside main, you can not modify a as its the base location of array. Else you get I-Value error.

answered Aug 17 '15 at 16:48

Austin

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Confused C pointer behavior

3 Answers3