[Solved] pointer to a member function of a class
The pointer-to-member operators .* and ->* have lower precedence than function call syntax. You need to do: (b1.*ptr)(); solved pointer to a member function of a class
The pointer-to-member operators .* and ->* have lower precedence than function call syntax. You need to do: (b1.*ptr)(); solved pointer to a member function of a class
Changing a const-value yields undefined behaviour, and the “mysterious” output is just such undefined behaviour. It is actually needless to investigate on why a behaviour in the world of undefined behaviour is as it is then. But in your case it is probably not using a as you declare it as const, so the compiler … Read more
Starting at the left hand side: struct foo *ptr declares ptr to be of type struct foo *, that is a pointer to struct foo. The = initializes that ptr variable to whatever the right hand side evaluates as. And the right hand side (struct foo *) p2 is an expression that casts p2 to … Read more
“Freeing” means “making available for allocation again”. There is no automatic deleting / overwriting of memory contents because it would negatively impact performance. If you want the area to be set to a value, you have to do it yourself before you call free(). That would be bad practice in release code, though (for anything … Read more
Hidden? It’s not hidden. What does that even mean? I do remember that you have to explicitly use it in Java Only in some circumstances. A variable called name could be known inside the class and also be the name of a parameter in a method. Example: class Test { String name; public void test(String … Read more
You can introduce another parameter to count the elements in freeme. bool parse(bool b1, int i, int *j, int *e, char **ptr, int q, char **pString, char *str, char **freeme, int *freeme_len) { for (*e = 0; *(ptr + q + *e); (*e)++) { b1 = true; if (*(ptr + *e + q)) { str … Read more
bool doit(int list[], int size); The function doit takes an array of integers as first parameter, and the size of the array (which is an integer) as second parameter. It returns a boolean (true or false). This sort of function prototype is typically used to access each element of the array within a for loop … Read more
The statement a = new arr***[num]; allocate an array of num pointers to pointers to pointers to arr, and make a point to that memory. The type of a have to be arr****. Which is just silly. 0 solved What is the meaning of “a = new arr ***[num];”? [closed]
‘this’ usually refers to the instance of the object that calls a particular method of a class,union,structure or a function. when you have same names for different variables, then ‘this’ is used to differentiate between them. class stu { int roll_no; string name; public: void input(int roll_no,string name) { name=this->name; roll_no=this->roll_no; } } stu obj=new … Read more
#include <stdio.h> #include <string.h> typedef struct dbl9 { double array[9]; } Dbl9; Dbl9 fun(double *in){ Dbl9 ret; memcpy(ret.array, in, sizeof(ret.array));//There can be no assurance that <in> the correct return ret; } int main(){ double array[9] = {1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0}; Dbl9 dbl9= fun(array); int i; for(i=0;i<9;++i){ printf(“%f\n”, dbl9.array[i]); } return … Read more
&p gets the pointer to the integer p -> That is, the memory address in which p is stored. *p “dereferences” the pointer, i.e. looks at the object at the memory address provided by p and returns that object. Your code above is invalid C, as you cannot dereference an int: error: indirection requires pointer … Read more
Well, this is a problem: fillwordnr(f,p[].ptrletter,p[].numbers,lines,nrofline,a,c,string[]); You have to specify which element of p you want to work with – p[0], p[1], p[i], etc. Assuming nrofline is a valid index for p, you would write fillwordnr( f, p[nrofline].ptrletter, p[nrofline].numbers, lines, nrofline, a, c, string ); Honestly, based on the code you’ve posted, it’s not clear … Read more
Accessing an uninitialized variable results in undefined behaviour. It may be that the program simply printed whatever garbage value was previously in that space in memory. Since C was meant to be a clean and efficient language, it doesn’t automatically fill in a value, it simply allocates an amount of memory. This section of memory … Read more
i intentionally left the previous answer because understanding memory allocation is trivial in programming in c specially. and as i see you have a big issue with that. but still you have issue in nearly every thing. in my actual answer, i will try to simplify you how to use strtok, to split string and … Read more
No it matters. Pointer arithmetic solely depends on what it points to. For example int(*p)[2] and int *p both of them are different. The same way int **p and int ***p is different. You might think even if the sizeof (int*) and sizeof(int**) is same – it is not logical to lose the multiple indirection … Read more