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Argument conversion: (normal) pointer to void pointer, cast needed?
(1 answer)
Concept of void pointer in C programming
(16 answers)
Closed 2 years ago.
The pthread_exit function should take a void pointer as input. I'm wondering how come it's possible to pass the address of an integer variable (e.g. pthread_exit(&ret1) here) without performing a cast conversion.
Related
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Caret character between types rather than variables, surrounded by parentheses
(2 answers)
C "block" caret character
(2 answers)
Caret in objective C
(3 answers)
What does this caret ^ syntax, with void on either side mean? [duplicate]
(4 answers)
Closed 2 years ago.
In the macos documentation for qsort, we see the declaration:
void qsort_b(void *base, size_t nel, size_t width, int (^compar)(const void *, const void *));
What is the meaning of the caret?
The ^ in the function declaration indicates that the callback is a block pointer rather than a function pointer.
As stated later in the man page:
The heapsort_b(), mergesort_b(), and qsort_b() routines are like the corresponding routines without the _b suffix, expect that the compar
callback is a block pointer instead of a function pointer.
This question already has answers here:
What is array to pointer decay?
(11 answers)
Why can't we pass arrays to function by value?
(9 answers)
Difference between passing array and array pointer into function in C
(3 answers)
Closed 3 years ago.
In C, if you pass an array as an argument to a function, what actually gets passed? And Why?
See the standard.
On entry to the function, the size expressions of each variably modified parameter are
evaluated and the value of each argument expression is converted to the type of the
corresponding parameter as if by assignment. (Array expressions and function
designators as arguments were converted to pointers before the call.)
This question already has answers here:
How do you read C declarations?
(10 answers)
Closed 4 years ago.
I have this syntax:
double (*(f(double (*)(int))))(int);
Do I understand correctly syntax above that f is function that gets pointer to function that receive int and return a double?
No; it's actually even more complicated than that.
cdecl.org glosses this type definition as:
declare f as function (pointer to function (int) returning double) returning pointer to function (int) returning double
In other words, the function pointed to by f takes a function pointer as an argument, and returns a function pointer. Both of those function pointers must be to functions which take an int as an argument and return a double.
You could simplify this definition a bit using an intermediate typedef as follows:
typedef double (*int_to_double_function)(int);
int_to_double_function (*f)(int_to_double_function);
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How does dereferencing of a function pointer happen?
(5 answers)
Closed 8 years ago.
What is the difference in dereferencing a function pointer with and without parentheses.
both the mechanism are working same on linux Gnu Gcc.
void (*fp)(void); //function pointer
void func(void);
fp = func;
(*fp)(); //dereferencing with parentheses
fp(); // without parentheses
You need to know that
(*fp)() = fp()
The function pointer can be called in both the ways and both are valid and same.
Check the below link:
How does dereferencing of a function pointer happen?
This question already has answers here:
Pointer arithmetic for void pointer in C
(10 answers)
Error: expression must be a pointer to a complete object type (?)
(1 answer)
Closed 8 years ago.
I'm trying to do some basic pointer arithmetic with a void *. My actual code computes an offset by using sizeof and then multiplying. Here is sample code to show an instance of the issue by itself.
void * p;
p = 0;
p = p + 1;
I'm using the MSVC compiler in C (not C++).
The error is:
expression must be a pointer to a complete object type
I'm not understanding what this error is trying to say. There is no object or struct here.
Pointer arithmetic is always in terms of the size of the pointed-to object(s). Incrementing a char* will advance the address by one, whereas for int* it would usually be four (bytes). But void has unknown size, so pointer arithmetic on void* is not allowed by the standard. Cast to the appropriate type first; if you just want to manipulate the address as if it were a number then cast to char* or use intptr_t.