>> int *numbers = malloc (50 * sizeof *numbers);
you should never dereference a NULL or invalid pointer - sometimes it may work while other times it will trigger an access violation. if you want to make the code easier to change, just typedef the data type, ie:
Code:int size = 1024; typedef int data_t; data_t * data = malloc(size * sizeof(data_t));
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; }
>>sometimes it may work while other times it will trigger an access violation.
The "dereferencing" is done at compile time, as its used as part of a sizeof() call, and is perfectly acceptable, just like shown in the FAQ.
When all else fails, read the instructions.
If you're posting code, use code tags: [code] /* insert code here */ [/code]
If you are talking about the use of sizeof *numbers as an argument to malloc, this will never result in the dereferencing of a null or invalid pointer. sizeof does not evaluate its argument, so the idiom is perfectly safe.Originally Posted by Sebastiani
You are correct, but I have never done that.
Assuming p is a typed pointer (not void), 'sizeof *p' is a constant expression that returns the size in byte of one element of the specified type. You could have used
sizeof p[0]
or
sizeof p[1234]
or
sizeof p["Crusty the clown"[123]]
with the same effect.
http://faq.cprogramming.com/cgi-bin/...&id=1043284351
Emmanuel Delahaye
"C is a sharp tool"
So let me see if I have these two thoughts correct:
1) When malloc returns a void pointer, it is not necessary to cast it to whatever type of data on the left hand side of the assignment operator.
2) The above is true because of type promotion? ie, the void return of malloc is promoted to whatever type is on the left hand side of the assignment operator?
~/
Correct.
The above is true because the C-language guarantees that the conversion from a pointer to an object to a void * is implicit and reverse.2) The above is true because of type promotion? ie, the void return of malloc is promoted to whatever type is on the left hand side of the assignment operator?
Emmanuel Delahaye
"C is a sharp tool"
>> The "dereferencing" is done at compile time, as its used as part of a sizeof() call, and is perfectly acceptable, just like shown in the FAQ.
I never knew that. thanks for setting me straight, guys.
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; }
