array inside a function

[unoho]

hey guys, im having problem in the following code

code:

int function(int array[], int n){     int b[n]; }

everytime i try that, the IDE gives me an error sometime about c89 ISO etc. etc. anyone knows what that means? but when i change that n to some number, it works fine. any ideas/comments is much appreciated.

[TheGuardian001]

What IDE/Compiler are you using?
I have no problems with similar code using gcc.

Oh, and what specific error is it giving you? C89 is a rule set (and a fairly old one at that,) not an error.

[unoho]

it says
- ISO C89 forbids variable-size array 'a' "
- ISO C89 forbids mixed declarations and code
- unsuccessful build

i tried it on Quicny 2005 and DevC++ and both r same.

so from what i can understand, u can't put a variable as an array size?

EDIT: OMG this is soo stupid, it works fine when i try to do it in command prompt using gcc 3.4.5 but it wont work with the IDEs

[TheGuardian001]

From what I understand, it depends on what version of the C standard your compiler uses.

C89 doesn't allow it, C90 doesn't allow it, however C99 does. Since your compiler uses C89, you won't be able to do this without switching/updating your compiler.

Edit: If your IDE uses GCC (MinGW on windows,) you can force it to use C99 by adding the -std=c99 (case sensitive, c99 not C99,) so long as your version of it supports the c99 standard.

[unoho]

thank you soo much. by default, the IDE had the C99 option unchecked for c programs. but ya..had to spent 4 hours trying to fix this problem :S
but at least i know how to use command promt to get to running gcc:D

[coolgod]

how is the compiler suppose to allocate memory for the program if it doesn't know how much space it need?
this isn't a dynamic array.....

[TheGuardian001]

coolgod @ Sat Oct 09, 2010 12:22 am wrote:

how is the compiler suppose to allocate memory for the program if it doesn't know how much space it need?
this isn't a dynamic array.....

This isn't an attempt to make a dynamic array. The purpose of this is to create an array of size x (useful in a function, if it needs a temporary array of size that will be determined when the function is called.) So long as x has a value when the array is created, the compiler knows exactly how much space to allocate (x*sizeof(type).)
IE:

code:

int x = 4; int y[x]; //same as int y[4];

x resolves to 4, thus this array has 4 elements, and always will.
This is not a problem for most modern compilers. Any compiler that conforms to the C99 specifications (however not one that conforms to C89/ANSI C) can do this.

[DtY]

coolgod @ Sat Oct 09, 2010 12:22 am wrote:

how is the compiler suppose to allocate memory for the program if it doesn't know how much space it need?
this isn't a dynamic array.....

Which is why there was a problem. Nondymanic arrays with a size computed at runtime wasn't a standard feature of C until C99 (and many people still think it's nonstandard).

[ncvitak]

In C to use an array as a function parameter you must pass it as a pointer.
Example:

c:

#include <stdio.h> int a[5] = {0, 1, 2, 3, 4}; int foo(int* array, int n){     return array[n]; } int main() {     printf("%d", foo(a,2)); //prints "2"     return 0; }

[Nathan4102]

Please don't gravedig old posts. This thread has been dead for almost 3 years now.

[ncvitak]

Maybe he still wants help

[Raknarg]

Necroposting is ok as long as the poster has some new content, I believe. Posts such as "wow thats cool" or "thanks" are frowned upon