why a[i] is not compiling . array compilation error

Hi,

i tried to compile the below program

1st program

Code:

---

int main()
{
        int i=10;
        int a[2]={i,i+1};
        printf ("hai %d\n",a[0]);
        return 0;
}

---

2nd program

Code:

---

int main()
{
        int i=10;
        int j=2;
        int a[j]={i,i+1};
        printf ("hai %d\n",a[0]);
        return 0;
}

---

Why i am able to compile first program and getting compile error for second program?

Unless you are able to make use of the variable length array feature introduced in the 1999 edition of the C standard, your array a must have a fixed size known at compile time, so j cannot be a variable.

Quote:

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Originally Posted by laserlight

Unless you are able to make use of the variable length array feature introduced in the 1999 edition of the C standard, your array a must have a fixed size known at compile time, so j cannot be a variable.

---

Translation: You are still using Turbo C!


Quzah.

Hi,

is it variable length array features is compiler dependent?

I am using the gcc compiler version (gcc (GCC) 3.4.5 20051201 (Red Hat 3.4.5-2)). my compiler will support variable length array feature?

can you let me know small c program with variable length array feature?

You can compile with compile flag: -std=c99

Quote:

---

Originally Posted by nkrao123@gmail.

is it variable length array features is compiler dependent?

---

In theory, no. In practice, yes, because not all C compilers in present use conform sufficiently to C99.

Quote:

---

Originally Posted by nkrao123@gmail.

I am using the gcc compiler version (gcc (GCC) 3.4.5 20051201 (Red Hat 3.4.5-2)). my compiler will support variable length array feature?

---

It should. You may need to pass -std=c99 as a compiler option.

Quote:

---

Originally Posted by nkrao123@gmail.

can you let me know small c program with variable length array feature?

---

Your second program is such an example program, except that you cannot initialise the array like that. Oh and remember to #include <stdio.h>.

Newer versions of gcc let this slide with a warning. I didn't start C programming until after gcc 4.+ was released, so I'm guessing that 3.4.5 does not do that. Try compiling:

gcc -Wall -std=c99

You can also use "-std=gnu99" which might be slightly more compatible with the default on linux (std=gnu89).

If that doesn't work, post the actual error you are getting.

HI light and Mk27,

i am getting error if i compile with "gcc name.c -std=c99" and "gcc name.c -std=gnu99"

error

gcc 22.c -std=gnu99
22.c: In function `main':
22.c:6: error: variable-sized object may not be initialized
22.c:6: warning: excess elements in array initializer
22.c:6: warning: (near initialization for `a')
22.c:6: warning: excess elements in array initializer
22.c:6: warning: (near initialization for `a')



gcc 22.c -std=c99
22.c: In function `main':
22.c:6: error: variable-sized object may not be initialized
22.c:6: warning: excess elements in array initializer
22.c:6: warning: (near initialization for `a')
22.c:6: warning: excess elements in array initializer
22.c:6: warning: (near initialization for `a')




i fell my compiler may not support this feature. am i right?

I told you precisely what you did wrong: "except that you cannot initialise the array like that", as did your compiler: "variable-sized object may not be initialized".

Quote:

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Originally Posted by nkrao123@gmail.

22.c:6: error: variable-sized object may not be initialized

---

Try actually reading the error message. Also try reading what people actually write to you:

Quote:

---

Originally Posted by laserlight

I told you precisely what you did wrong

---

^ See that?


Quzah.

Quote:

---

It should. You may need to pass -std=c99 as a compiler option.

---

Light used that word as it should. I thought my compiler may have that features. That why i tried. Now i understand that my compilr does not supporting this feature

Quote:

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Originally Posted by nkrao123@gmail.

Now i understand that my compilr does not supporting this feature

---

No, no you don't. You don't understand at all, because you aren't reading what people are telling you over and over.


Quzah.

Quote:

---

Originally Posted by nkrao123@gmail.

Light used that word as it should. I thought my compiler may have that features. That why i tried. Now i understand that my compilr does not supporting this feature

---

You misunderstand because you jumped to a conclusion without reading the rest of what I wrote. Try this program:

Code:

---

#include <stdio.h>

int main(void)
{
    size_t n = 10;
    int numbers[n];

    for (size_t i = 0; i < n; ++i)
    {
        numbers[i] = i * 2;
    }

    for (size_t i = 0; i < n; ++i)
    {
        printf("%d\n", numbers[i]);
    }

    return 0;
}

---

There is no need for numbers to be a variable length array in the above program, but perhaps the value of n could have been obtained from input instead.

Now, this program should compile, once you add braces for the main function:

Code:

---

int main(void)
return 0;

---

Attempt to compile it. Compilation fails. You can thus jump to the conclusion that your C compiler does not support compiling even such a trivial C program ;)

Quote:

---

Originally Posted by nkrao123@gmail.

Light used that word as it should. I thought my compiler may have that features. That why i tried. Now i understand that my compilr does not supporting this feature

---

Sorry, partially my mistake. As per laserlight's post above, what c99 will allow is this:

Code:

---

int array[x];

---

Where x may be determined at runtime. Which means you cannot initialize it to any value, ie.

Code:

---

int array[x] = {1, 2, 3};

---

is no good. You can use a "variable-sized object" but you can't initialize it in the declaration.

And because you still won't understand what they're saying:

Quote:

---

Originally Posted by MK27

Which means you cannot initialize it to any value, ie.

Code:

---

int array[x] = {1, 2, 3};

---

is no good. You can use a "variable-sized object" but you can't initialize it in the declaration.

---

You can do what is in green.
You can't do the part in red.


Quzah.

Quote:

---

Originally Posted by laserlight

Unless you are able to make use of the variable length array feature introduced in the 1999 edition of the C standard, your array a must have a fixed size known at compile time, so j cannot be a variable.

---

Hi light,

I did not get properlly "Unless you are able to make use of the variable length array feature introduced in the 1999 edition of the C standard".

please can you tell once this?

Regards
NKRao

Quote:

---

Originally Posted by nkrao123@gmail.

please can you tell once this?

---

I think we already have at least two or three times. The "variable length" feature in c99 -- the 1999 ANSI C standard -- allows you to declare an array with a variable length.

However, no C standard allows you to initialize such an object with values. You cannot do that, period, end of story.

Here are two declarations:

Code:

---

int x;
int array[n];

---

These variables have not been initialized; they could have any value in them.

Code:

---

x = 12;

---

Now x has been initialized. It could have been initialized in the declaration:

Code:

---

int x = 12;

---

An array of fixed length can also be initialized in its declaration:

Code:

---

int array[3] = {4, 5, 666};

---

However, you cannot:
1) initialize an array after its declaration; you must assign to individual elements.*
2) initialize a variable length array, such as "int array[n]"

Read posts #14 and #15 again if you still do not understand what this means. If you want to put values into a variable length array, you have to do so after the declaration.

* actually you can set more than one at time with memset().

Quote:

---

Originally Posted by nkrao123@gmail.

Light used that word as it should. I thought my compiler may have that features. That why i tried. Now i understand that my compilr does not supporting this feature

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Hint: People are sending messages with more than one line and more than code samples... READ what they are sending to you!