What is the use of malloc()?
How to use it? And, what is the return of malloc()?
What is the use of malloc()?
How to use it? And, what is the return of malloc()?
this function reservs a certain amount of memory for the char type and delivers a pointer to the beginneing of the space reserved.
asking for n bytesCode:int n; char *s; .... scanf("%d",&n); s=malloc(n);
What does it mean?
#define Malloc(type,n) (type *)malloc((n)*sizeof(type))
I'm sorry that I haven't read the FAQ before posting question. Thank you for reminding me.
What is n?
Naturally n in this case would represent the number of objects you wish to allocate.Originally posted by siubo
What is n?
Quzah.
Hope is the first step on the road to disappointment.
N is variable.
The reason for using sizeof is because malloc needs to know how many bytes to make room for. If your wondering how malloc handles any data type it is because it's return is of type void, thus no need for a cast like in the macro you posted before. The FAQ explains it better. Goodluck.Code:int *pointer; int N; N = 5; /* Allocates room for 5 integers */ pointer = malloc(N * sizeof int); /* You can use it now like an array */ pointer[0] = 22; pointer[1] = 34;
When I follow the FAQ to change
#define Malloc(type,n) (type *)malloc((n)*sizeof(type))
to
#define Malloc(ptr,n) malloc(n*sizeof *ptr),
however, error messages are come out:
cannot convert 'void *' to 'double *' in initialization
OR
cannot convert 'void *' to 'double *' in assignment.
Why is this so?
Why are you changing it? Just to break it? Congradulations, you've done what you set out to do.
The problem is that you're not using parenthesis correctly. As such you're running into problems with the * operator.
* is for math.
* is for pointers.
Are you dereferencing a pointer, or trying to do math? If it's not broke, don't "fix" it.
Quzah.
Hope is the first step on the road to disappointment.
I change it because I have a segmentation fault when I run my program. I have posted this message before.
Somebody said that I did sth. wrong with the memory. So, I try to change it to fix the segmentaion fault. However, it seems that it's not work.
Not necessarily. If 'new' fails the program exits immediately, whether you had a backup plan or not. Not sure if there's a workaround for that (I'd be interested to here about one), but otherwise you can use malloc() and free() wrappers that ensure proper constructor/destructor invocations.Originally posted by Salem
If you're really writing C++, then you should be using the new operator to allocate memory... [/B]
Code:#include <cmath> #include <complex> bool euler_flip(bool value) { return std::pow ( std::complex<float>(std::exp(1.0)), std::complex<float>(0, 1) * std::complex<float>(std::atan(1.0) *(1 << (value + 2))) ).real() < 0; }
OK Thanks. I totally overlooked catching the exception.
Code:#include <cmath> #include <complex> bool euler_flip(bool value) { return std::pow ( std::complex<float>(std::exp(1.0)), std::complex<float>(0, 1) * std::complex<float>(std::atan(1.0) *(1 << (value + 2))) ).real() < 0; }
