size of array

goran00 · 04-02-2008, 08:02 AM

i am trying to have the user chose the size of the array
what am i doing wrong?

Code:

#include <stdio.h>
#include <stdlib.h>

main (){
int size;
int array [size];

printf("Please enter the size of the array", size);
scanf("%i", size);
system("pause");
}

mike_g · 04-02-2008, 08:04 AM

Well for a start 'size' is undefined when you declare the array.

Dino · 04-02-2008, 08:05 AM

Two things are wrong.

First, it's against the "law". When you declare "array", size is uninitialized. If you want a random sized array, you should malloc a chunk of storage from the heap.

Second, your scanf should pass the address of size, not size itself.

Todd

Elysia · 04-02-2008, 08:07 AM

Main returns int.
Size contains garbage at the point you try to allocate the array, so theoretically it could get an undefined size. Secondly, this is only allowed in C99 and not C89.
Thirdly, C is a linear programming language. One statement is executed after the other.
Do the array won't be created after you've asked the size, it will be created before because the line is before the input.
Fourthly, I think you need to take another look at printf.
And avoid system("pause"). Better to use getchar().

matsp · 04-02-2008, 08:07 AM

1. It's not part of C89 standard to use variable size arrays. To be portable to C89 compilers, you should use dynamic memory allocation.

2. You can not declare a variable size array before the size component has been assigned. Right now, when you are saying "int size;" , size contains some "random" value, and then try to create an array of that size. [The "random" value is usually the same for the same piece of code each time, but it's random in the sense that it's not defined by any standard and it may well change if you introduce another variable before it, change your code in some other way, change the compile options, etc, etc]

So, if you want to use the C99 feature of variable sized arrays, you should at least set the size before creating the array.

--
Mats

goran00 · 04-02-2008, 08:09 AM

Code:

#include <stdio.h>
#include <stdlib.h>

main (){
int size;
int array [1];

printf("Please enter the size of the array", size);
scanf("%i", size);
array[size];
system("pause");
}

would this work?
would array size be what the user enters now?

Elysia · 04-02-2008, 08:11 AM

It would be 1.
As I mentioned, it's linear. It won't just jump back and set a new size.
And array[size] just access the element size in the array.

Fix main and system("pause"), then go learn dynamic memory, malloc & free.

goran00 · 04-02-2008, 08:12 AM

i have read my book but its too complicated so i am not understanding much of it

goran00 · 04-02-2008, 08:13 AM

thats y i am trying to get some help here

Dino · 04-02-2008, 08:13 AM

Go learn this too: http://www.cppreference.com/stdio/scanf.html

Elysia · 04-02-2008, 08:15 AM

Well, you need to understand pointers first.
Then you simply call malloc to allocate requested amount and when you don't need it, you call free.

Code:

int main()
{
    int* pArray;
    int size;

    printf("Enter size: ");
    scanf("%i", &size);
    pArray = malloc(size * sizeof(*pArray));
    free(pArray);
    pArray = NULL;

    getchar();
    return 0;
}

matsp · 04-02-2008, 08:18 AM

Can I also point out that allocating or sizing arrays dynamicly is OFTEN not necessary - unless your application is large and/or uses A LOT of memory, just using "sufficiently large" arrays [e.g. an int array[10000]; will use 40KB of memory, which is not very much when your system has 10000x as much memory].

--
Mats

goran00 · 04-02-2008, 08:26 AM

am i on the right path?

Code:

#include <stdio.h>
#include <stdlib.h>

main (){
int size;

printf("Please enter the size of the array", size);
scanf("%i", size);

makeAnArray(size);

system("pause");
}
int* makeAnArray(size){
	int* array;
array=calloc(size,sizeof(int));
return array;
}

what does error C2040: 'makeAnArray' : 'int *()' differs in levels of indirection from 'int ()' mean?
it point to the line: int* makeAnArray(size){
how do i fix that?

Elysia · 04-02-2008, 08:29 AM

Sheesh. Start listening.

Code:

#include <stdio.h>
#include <stdlib.h>

// MAIN RETURNS INT
int main (){
    int size;
    // YOU HAVEN'T SPECIFIED ANYTHING FOR PRINTF TO PRINT, SO REMOVE THE "size" PART
    printf("Please enter the size of the array");
    // SCANF TAKES A POINTER
    scanf("%i", &size);

    makeAnArray(size);
    // DON'T USE SYSTEM("PAUSE")
    getchar();
}

// ALL ARGUMENTS MUST HAVE A TYPE
int* makeAnArray(int size){
    int* array;
    array=calloc(size,sizeof(int));
    return array;
}

You are also missing a prototype for makeAnArray.
And you are not freeing the return from calloc.

matsp · 04-02-2008, 08:34 AM

Quote:

Originally Posted by Elysia

Sheesh. Start listening.

Code:

#include <stdio.h>
#include <stdlib.h>

// MAIN RETURNS INT
int main (){
    int size;
    printf("Please enter the size of the array", size);
    // SCANF TAKES A POINTER
    scanf("%i", &size);

    makeAnArray(size);
    // DON'T USE SYSTEM("PAUSE")
    getchar();
}

// ALL ARGUMENTS MUST HAVE A TYPE
int* makeAnArray(int size){
    int* array;
    array=calloc(size,sizeof(int));
    return array;
}

You are also missing a prototype for makeAnArray.
And you are not freeing the return from calloc.

ANd missing storing the return from makeAnArray().

--
Mats

04-02-2008, 08:02 AM	#1
goran00 Registered User Join Date: Jan 2008 Posts: 124	size of array i am trying to have the user chose the size of the array what am i doing wrong? Code: #include <stdio.h> #include <stdlib.h> main (){ int size; int array [size]; printf("Please enter the size of the array", size); scanf("%i", size); system("pause"); }
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04-02-2008, 08:04 AM	#2
mike_g Wheres the lesbians? Join Date: Oct 2006 Location: UK Posts: 1,219	Well for a start 'size' is undefined when you declare the array.
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04-02-2008, 08:05 AM	#3
Dino Jack of many languages Join Date: Nov 2007 Location: Katy, Texas Posts: 1,929	Two things are wrong. First, it's against the "law". When you declare "array", size is uninitialized. If you want a random sized array, you should malloc a chunk of storage from the heap. Second, your scanf should pass the address of size, not size itself. Todd __________________ Mac and Windows cross platform programmer. Ruby lover. Memorable Quotes From Recent Posts: I can't remember.
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04-02-2008, 08:07 AM	#5
matsp Kernel hacker Join Date: Jul 2007 Location: Farncombe, Surrey, England Posts: 15,686	1. It's not part of C89 standard to use variable size arrays. To be portable to C89 compilers, you should use dynamic memory allocation. 2. You can not declare a variable size array before the size component has been assigned. Right now, when you are saying "int size;" , size contains some "random" value, and then try to create an array of that size. [The "random" value is usually the same for the same piece of code each time, but it's random in the sense that it's not defined by any standard and it may well change if you introduce another variable before it, change your code in some other way, change the compile options, etc, etc] So, if you want to use the C99 feature of variable sized arrays, you should at least set the size before creating the array. -- Mats __________________ Compilers can produce warnings - make the compiler programmers happy: Use them! Please don't PM me for help - and no, I don't do help over instant messengers.
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04-02-2008, 08:09 AM	#6
goran00 Registered User Join Date: Jan 2008 Posts: 124	Code: #include <stdio.h> #include <stdlib.h> main (){ int size; int array [1]; printf("Please enter the size of the array", size); scanf("%i", size); array[size]; system("pause"); } would this work? would array size be what the user enters now?
goran00 is offline

04-02-2008, 08:12 AM	#8
goran00 Registered User Join Date: Jan 2008 Posts: 124	i have read my book but its too complicated so i am not understanding much of it
goran00 is offline

04-02-2008, 08:13 AM	#9
goran00 Registered User Join Date: Jan 2008 Posts: 124	thats y i am trying to get some help here
goran00 is offline

04-02-2008, 08:18 AM	#12
matsp Kernel hacker Join Date: Jul 2007 Location: Farncombe, Surrey, England Posts: 15,686	Can I also point out that allocating or sizing arrays dynamicly is OFTEN not necessary - unless your application is large and/or uses A LOT of memory, just using "sufficiently large" arrays [e.g. an int array[10000]; will use 40KB of memory, which is not very much when your system has 10000x as much memory]. -- Mats __________________ Compilers can produce warnings - make the compiler programmers happy: Use them! Please don't PM me for help - and no, I don't do help over instant messengers.
matsp is offline

04-02-2008, 08:26 AM	#13
goran00 Registered User Join Date: Jan 2008 Posts: 124	am i on the right path? Code: #include <stdio.h> #include <stdlib.h> main (){ int size; printf("Please enter the size of the array", size); scanf("%i", size); makeAnArray(size); system("pause"); } int* makeAnArray(size){ int* array; array=calloc(size,sizeof(int)); return array; } what does error C2040: 'makeAnArray' : 'int ()' differs in levels of indirection from 'int ()' mean? it point to the line: int makeAnArray(size){ how do i fix that?
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