Confusing on pointer array

Question

The following is the C code:

char *ptr[100];
printf("%s\n",*ptr++);

My question is: we know that array name is not a variable (Ritchie's book "The C programming language" Page 99), so if we define

int *pa,a[5];

This is legal:

pa = a; pa++;

while this is illegal:

a++;

Here char *ptr[100] defines a char pointer array, so ptr represents the initial address of the array. For the above code, if *ptr++ means *(ptr++), this is illegal because array name cannot be used as a variable, and also it's meaningless because *(ptr++) still gets an address as opposed to %s. However, if *ptr++ means (*ptr)++, it also looks strange... Can anyone help to understand this?

The above is my thinking process, my question is: how can *ptr++ give us a string as the code printf("%s\n",*ptr++); says?

What do you mean by saying array name cannot be used as a variable? ptr is a variable and it's also an array's name — SHG
– SHG, Commented Mar 21, 2018 at 16:14
I mean if int *pa, a[5]; you cannot use expressions like a=pa; or a++; — Thomas D.
– Thomas D., Commented Mar 21, 2018 at 16:19
Possible duplicate of C/C++ int[] vs int* (pointers vs. array notation). What is the difference? — sg7
– sg7, Commented Mar 21, 2018 at 16:22

Alex Shesterov · Accepted Answer · 2018-03-21 16:35:01Z

1

Postfix increment ++ has a higher precedence than dereference *.

Thus, *ptr++ is interpreted as *(ptr++). And as you've correctly figured out, ptr++ isn't allowed if ptr is an array (of anything, values or pointers).

If ptr is a pointer, then *ptr++ will indeed increment the pointer (so it will point to the next value after the operation), and this expression will return the current pointed-to value (before increment of the pointer). This expression is indeed sometimes used, e.g. for copying memory areas, e.g.:

while (...) {
    *dest++ = *src++; // Copy the current element, then increment both pointers
}

*ptr++ doesn't necessarily give you a string — it gives you a string if and only if ptr is pointing to a string. And in this case, it's not necessary to post-increment just to get the string — *ptr would be enough. ++ is done for another purpose.

For example, it ptr is a pointer to an "array" of strings (i.e. a pointer to a pointer to achar, i.e. char **ptr), then you could print all the strings like this:

int i;
for (i = 0; i < N; ++i) {
    printf("%s\n",*ptr++); // '++' "jumps" to the next string
}

edited Mar 21, 2018 at 16:35

answered Mar 21, 2018 at 16:20

Alex Shesterov

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2 Comments

Thomas D. Over a year ago

Thank you for your reply. My understanding is ptr is a pointer-to-pointer which points to the address of ptr[0], so *ptr == ptr[0], *ptr+1 == ptr[1] etc. So if we want to get the string, it seems that we need to dereference again?

Alex Shesterov Over a year ago

You are right about *ptr == ptr[0] and *(ptr+1) == ptr[1] (note the grouping). But in C, a "string" is a null-terminated "array" of characters. So if you dereference a string, you'll actually get the first character. Having char** ptr - this could be an array of strings; then *ptr or ptr[0] is the first string, *(ptr+1) or ptr[1] is the second string, and so on; **ptr or ptr[0][0] is the first character of the first string, *(*(ptr + 1) + 2) or ptr[1][2] is the third character of the second string, etc.

Vilyanare · Accepted Answer · 2018-03-21 16:20:16Z

0

I believe this link should help you.

C/C++ int[] vs int* (pointers vs. array notation). What is the difference?

The array notation has a bounds limit while the pointer notation does not.

If you wanted to access the next in the array with array notation it would look something like a[n++] n being whatever the current index of the array you are at. Where as assigning ptr to a sets ptr to the first index of the array and you can increment a pointer.

answered Mar 21, 2018 at 16:20

Vilyanare

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Comments

John Bode · Accepted Answer · 2018-03-21 18:15:11Z

0

*ptr++ is parsed as *(ptr++) - apply the ++ operator to ptr, then dereference the result.

As written, this results in a constraint violation - ptr is an array of pointers to char, and an expression of array type is a non-modifiable lvalue, and as such may not be the target of an assignment or the operand of the ++ and -- operators.

The line

printf("%s\n",*ptr++);

should (indeed, must) result in the compiler issuing a diagnostic on the order of "invalid lvalue in increment" or something along those lines.

answered Mar 21, 2018 at 18:15

John Bode

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1 Comment

Thomas D. Over a year ago

Yes, you're right, there is error: lvalue required as increment operand when I use *ptr++. I listed all the possible solutions below. If you have time, could you please look at it for me. Thanks

Thomas D. · Accepted Answer · 2018-03-21 20:56:37Z

0

No. 1:

void main(void)
{
    int i;
    char *ptr[3]={"how","are","you"};
    for (i=0;i<3;i++)
        printf("%s\n",ptr[i]);
}

This prints out correct answer. Good!

No.2:

void main(void)
{
    int i;
    char *ptr[3];
    *ptr[0] = "how";
    *ptr[1] = "are";
    *ptr[2] = "you";
    for (i=0;i<3;i++)
        printf("%s\n",ptr[i]);
}

warning: assignment makes integer from pointer without a cast [enabled by default].....Why can't initialize like this?

No.3:

void main(void)
{
    int i;
    char *ptr[3]={"how","are","you"};
    printf("%s\n",*ptr++);
}

error: lvalue required as increment operand....Why can't we use *ptr++?

answered Mar 21, 2018 at 20:56

Thomas D.

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