Thread: Convert double to int for array size

Hi,

I want to declare an array size based on a double calculation. Rounding it up or down doesn't really matter. As the array size needs to be an integer, is there any way I can convert the double result to an integer?

Rounding it up or down probably does matter. Depending on round-off error and all that good stuff. If the result was very close to an integer (meant to be an integer but could be +/- some tiny fraction) you have to find the closest integer. Usually something like

Code:

i = (int)(d + 0.5);

I need to use the int value to declare the array size. However when I try the suggested responses it does not work.

It says "integral constant expression expected"


double z = x/y;
int a = (int) z;

int array1 = [a];

Originally Posted by bithub

Code:

int array1 = [a];

That is not how you declare an array. You need to pick up a book, or read through some tutorials online.

At any rate, try this:

Code:

int array1[a];

My apologies, that was a typo. However the code with the correct syntax is not working.

double z = x/y;
int a = (int) z;

int array1[a];


I'm still getting "integral constant expression expected".

check out malloc:

#include <stdlib.h>

int *array1 = NULL;
array1 = malloc(a * sizeof(int));

memory allocated by malloc should be later freed and the pointer set to null:

free(array1);
array1 = NULL;

Originally Posted by laserlight

That is because unless you are compiling with respect to C99, this is wrong:

Code:

int a = (int) z;

int array1[a];

You are trying to create a variable length array, and that is not allowed in standard C prior to the 1999 edition of the C standard.

Thanks. How do I check which compiler I am using?

Originally Posted by millsy5

Thanks. How do I check which compiler I am using?

By ... how you run the compiler? If you type "gcc" you're using gcc, et cetera. If you push a button, then you should know what program you downloaded that has a button you can push, and that program will tell you what the compiler is.