The system has 10 giga byte RAM. The command: uname -a gives the following output: Linux olympus 2.6.26-2-amd64 #1 SMP Wed Aug 19 22:33:18 UTC 2009 x86_64 GNU/Linux I think this means that the system is 64 bit. So you think that the command: int myvariable = 1000; float *myarray = malloc(pow(myvariable,3)*sizeof(float)); will work if pow(myvariable,3) is smaller than 2 billion? Many Thanks, Anna ----- Original Message ----- From: Ángel González <keisial@xxxxxxxxx> Date: Monday, August 27, 2012 2:14 pm Subject: Re: how to use malloc to reserve space for 1 million floats > On 27/08/12 11:40, Anna Sidera wrote: > > Hello, > > > > I want to use the following command: > > > > int myvariable = 1000; > > float *myarray = malloc(pow(myvariable,3)*sizeof(float)); > > > > but I don't know if it will work because if sizeof(float) is > equal to 4 then pow(myvariable,3)*sizeof(float) is equal to 4 > billion which is larger than the maximum integer which is about 2 > billion.> > > Can you tell me what is the right way to create an array of > pow(myvariable,3) floats? > > > > Many Thanks, > > Anna > > How much memory do you have available? > The parameter to malloc is a size_t, you should have no problems > providing a size of 4000000000 in a 64 bit system. > If you're using a 32 bit system, then you will have problems stating > that size. But the address space is also smaller than that, > so you couldn't reserve so much memory*, even if you were able to > provide that number to malloc(). > > * using conventional methods. But seems silly not to be using a flat > address space nowadays... > >
